Mixing
Quotient ring of Gaussian integers
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A very basic ring theory question, which I am not able to solve. How does one show that
$mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
Extending the result: $mathbb{Z}[i]/(a-ib) cong mathbb{Z}/(a^{2}+b^{2})mathbb{Z}$, if $a,b$ are relatively prime.
My attempt was to define a map, $varphi:mathbb{Z}[i] to mathbb{Z}/10mathbb{Z}$ and show that the kernel is the ideal generated by $langle{3-irangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.
abstract-algebra ring-theory ideals gaussian-integers
edited Sep 1 '16 at 20:13
6005
37k752127
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add a comment |
$begingroup$
A very basic ring theory question, which I am not able to solve. How does one show that
$mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
Extending the result: $mathbb{Z}[i]/(a-ib) cong mathbb{Z}/(a^{2}+b^{2})mathbb{Z}$, if $a,b$ are relatively prime.
My attempt was to define a map, $varphi:mathbb{Z}[i] to mathbb{Z}/10mathbb{Z}$ and show that the kernel is the ideal generated by $langle{3-irangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.
abstract-algebra ring-theory ideals gaussian-integers
edited Sep 1 '16 at 20:13
6005
37k752127
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1
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$(3-i)^2=$ ? this should be a hint.
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– Asaf Karagila♦
Feb 23 '11 at 10:25
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@Asaf: Sorry i am not able to get what exactly you want to say. All i know abt $3-i$ is that its reducible in $mathbb{Z}[i]$.
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– anonymous
Feb 23 '11 at 10:45
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Hint: $3-i=(1-i)cdot(2+i)$
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– lhf
Feb 23 '11 at 12:04
1
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@Chandru1: I'm sorry, I meant $|3-i|^2=$? as a hint.
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– Asaf Karagila♦
Feb 23 '11 at 14:22
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The document math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf might be useful
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– Watson
Jun 19 '18 at 10:06
add a comment |
$begingroup$
A very basic ring theory question, which I am not able to solve. How does one show that
$mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
Extending the result: $mathbb{Z}[i]/(a-ib) cong mathbb{Z}/(a^{2}+b^{2})mathbb{Z}$, if $a,b$ are relatively prime.
My attempt was to define a map, $varphi:mathbb{Z}[i] to mathbb{Z}/10mathbb{Z}$ and show that the kernel is the ideal generated by $langle{3-irangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.
abstract-algebra ring-theory ideals gaussian-integers
edited Sep 1 '16 at 20:13
6005
37k752127
$endgroup$
A very basic ring theory question, which I am not able to solve. How does one show that
$mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
Extending the result: $mathbb{Z}[i]/(a-ib) cong mathbb{Z}/(a^{2}+b^{2})mathbb{Z}$, if $a,b$ are relatively prime.
My attempt was to define a map, $varphi:mathbb{Z}[i] to mathbb{Z}/10mathbb{Z}$ and show that the kernel is the ideal generated by $langle{3-irangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.
abstract-algebra ring-theory ideals gaussian-integers
abstract-algebra ring-theory ideals gaussian-integers
edited Sep 1 '16 at 20:13
6005
37k752127
edited Sep 1 '16 at 20:13
6005
37k752127
edited Sep 1 '16 at 20:13
6005
37k752127
edited Sep 1 '16 at 20:13
6005
37k752127
edited Sep 1 '16 at 20:13
6005
37k752127
37k752127
asked Feb 23 '11 at 10:12
user9413
1
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$(3-i)^2=$ ? this should be a hint.
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– Asaf Karagila♦
Feb 23 '11 at 10:25
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@Asaf: Sorry i am not able to get what exactly you want to say. All i know abt $3-i$ is that its reducible in $mathbb{Z}[i]$.
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– anonymous
Feb 23 '11 at 10:45
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Hint: $3-i=(1-i)cdot(2+i)$
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– lhf
Feb 23 '11 at 12:04
1
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@Chandru1: I'm sorry, I meant $|3-i|^2=$? as a hint.
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– Asaf Karagila♦
Feb 23 '11 at 14:22
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The document math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf might be useful
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– Watson
Jun 19 '18 at 10:06
add a comment |
1
$begingroup$
$(3-i)^2=$ ? this should be a hint.
$endgroup$
– Asaf Karagila♦
Feb 23 '11 at 10:25
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@Asaf: Sorry i am not able to get what exactly you want to say. All i know abt $3-i$ is that its reducible in $mathbb{Z}[i]$.
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– anonymous
Feb 23 '11 at 10:45
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Hint: $3-i=(1-i)cdot(2+i)$
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– lhf
Feb 23 '11 at 12:04
1
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@Chandru1: I'm sorry, I meant $|3-i|^2=$? as a hint.
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– Asaf Karagila♦
Feb 23 '11 at 14:22
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The document math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf might be useful
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– Watson
Jun 19 '18 at 10:06
1
1
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$(3-i)^2=$ ? this should be a hint.
$endgroup$
– Asaf Karagila♦
Feb 23 '11 at 10:25
$begingroup$
$(3-i)^2=$ ? this should be a hint.
$endgroup$
– Asaf Karagila♦
Feb 23 '11 at 10:25
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@Asaf: Sorry i am not able to get what exactly you want to say. All i know abt $3-i$ is that its reducible in $mathbb{Z}[i]$.
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– anonymous
Feb 23 '11 at 10:45
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@Asaf: Sorry i am not able to get what exactly you want to say. All i know abt $3-i$ is that its reducible in $mathbb{Z}[i]$.
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– anonymous
Feb 23 '11 at 10:45
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Hint: $3-i=(1-i)cdot(2+i)$
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– lhf
Feb 23 '11 at 12:04
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Hint: $3-i=(1-i)cdot(2+i)$
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– lhf
Feb 23 '11 at 12:04
1
1
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@Chandru1: I'm sorry, I meant $|3-i|^2=$? as a hint.
$endgroup$
– Asaf Karagila♦
Feb 23 '11 at 14:22
$begingroup$
@Chandru1: I'm sorry, I meant $|3-i|^2=$? as a hint.
$endgroup$
– Asaf Karagila♦
Feb 23 '11 at 14:22
$begingroup$
The document math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf might be useful
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– Watson
Jun 19 '18 at 10:06
$begingroup$
The document math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf might be useful
$endgroup$
– Watson
Jun 19 '18 at 10:06
add a comment |
6 Answers
6
active
oldest
votes
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Define
$$phi: mathbb{Z} rightarrow mathbb{Z}[i]/(3-i) text{ where } phi(z) = z + (3-i)mathbb{Z}[i].$$
It follows simply that
$ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z}$.
So for any such $z in ker phi$, we have $z = (3-i)(a+bi)$ for some $a,b in mathbb{Z}$. But $(3-i)(a+bi) in mathbb{Z}$ happens if and only if $3b-a=0$.
So $$begin{align*}
ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z} &= {(3-i)(3b+bi)mid b in mathbb{Z}}\
&= {(9b + b) + i(3b-3b)mid b in mathbb{Z}}\
&= {10bmid b in mathbb{Z}}\
&= 10mathbb{Z}.
end{align*}$$
To see $phi$ is surjective, let $(a+bi) + (3-i)mathbb{Z}[i] in mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $phi(a+3b) = (a+bi) + (3-i)mathbb{Z}[i]$.
Hence $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 23 '11 at 14:35
Brandon CarterBrandon Carter
7,31822538
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Don't you have to show that $phi$ is also surjective?
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– Greg Graviton
Feb 23 '11 at 17:25
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This is more simple to understand.+1
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– anonymous
Feb 24 '11 at 8:03
add a comment |
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This diagram shows the Gaussian integers modulo $3-i$.
The red points shown are all considered to be $0$ but their locations in $mathbb Z[i]$ are $0$, $3-i$, $i(3-i)$ and $3-i + i(3-i)$. Every congruence class must be inside that box once and you can see there are $10$ of them.
The arrows show adding by $1$ each time. Doing that takes you through every equivalence class and then back to the start.
So $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 24 '11 at 16:31
quantaquanta
7,30623974
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Firstly: it is not true in general that $mathbb Z[i]/(a - ib) cong mathbb Z/(a^2 + b^2).$ (Consider the case of $3 - 0cdot i$.)
The claimed isomorphism does hold if $a$ and $b$ are coprime.
Here is a sketch of how to see this:
To begin with, note that it is much easier to consider maps from $mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $mathbb Z$ maps to any ring with unity in a canonical way, by sending $1$ to $1$.
So consider the canonical map $mathbb Z to mathbb Z[i]/(a - i b).$
The target is finite of order $a^2 + b^2$, and so this map factors to give an
injection
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - i b)$
for some $n$ dividing $a^2 + b^2$.
Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $mathbb Z/(n)$. Combining this observation with the equation $a - i b = 0$ (which holds in $mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.
answered Feb 23 '11 at 14:07
Matt EMatt E
105k8220390
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Thanks for this generalization. It is very helpful (to this third party, at least!)
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– The Chaz 2.0
Mar 21 '11 at 16:42
2
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Hello Matt, something I've been trying to wrap my head around recently is why $mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you.
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– yunone
Oct 30 '11 at 1:27
3
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@yunone: Dear yunone, It follows e.g. from the division algorithm in $mathbb Z[i]$. Regards,
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– Matt E
Oct 30 '11 at 2:12
10
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@yunone: Think about the chain of ideals $(a^2+b^2) subset (a+bi) subset {mathbf Z}[i]$. For a positive integer $n$, the index $[{mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{mathbf Z}[i]:(a^2+b^2)] = [{mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${mathbf Z}[i]/(a+bi) cong {mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{mathbf Z}[i]:(a+bi)]^2$. Now take square roots.
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– KCd
Apr 7 '12 at 21:25
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Dear @KCd, thanks for the explanation.
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– yunone
Apr 27 '12 at 21:20
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show 1 more comment
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Go back one step and add the defining equation for $i$ to the ideal. In other words, consider your ring as a quotient of the ring of polynomials $mathbb Z[x]$:
$$ mathbb Z[i] / (3-i) = mathbb Z [x] / (3-x,x^2+1) $$
Manipulating the ideal $(3-x,x^2+1)$ a bit, you will find that the quotient is indeed equal to $mathbb Z/10mathbb Z$.
This makes sense because setting $i^2=-1$ and $i=3$ implies that $9 = -1$, which is true in $mathbb Z/10mathbb Z$.
answered Feb 23 '11 at 11:43
Greg GravitonGreg Graviton
3,69121734
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In general, one knows that if $alpha$ is an integer in the number field $K$, then
$$
{rm N}_{K/{Bbb Q}}(alpha)=left|frac{A}{Aalpha}right|
$$
Here $rm N$ is the norm and $A$ denotes the ring of integers. In the question's situation, $A={Bbb Z}[i]$ is the ring of Gaussian integer and since $K$ is quadratic imaginary, ${rm N}_{K/{Bbb Q}}(alpha)=alphabaralpha$ where the bar denotes complex conjugation.
When $alpha=3-i$, ${rm N}_{K/{Bbb Q}}(alpha)=(3-i)(3+i)=10$, thus
${Bbb Z}[i]/(3-i)$ is a ring with 10 elements, whose representatives are
$$
left{0,1,2,i,i+1,i+2,2i,2i+1,2i+2,-1 right}.
$$
A minute of thought and a brief inspection of these representatives should convince that the ring is indeed isomorphic to the ring of classes modulo 10.
answered Feb 23 '11 at 13:57
Andrea MoriAndrea Mori
20k13466
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Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$...
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– Pete L. Clark
Feb 23 '11 at 14:33
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Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context.
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– Andrea Mori
Feb 23 '11 at 19:25
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my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...)
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– Pete L. Clark
Feb 23 '11 at 19:56
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In this post, I am quoting the solution to a related question from Artin, with the preceding explanation (Section 11.4, p. 337-338). This is really the same as Greg Graviton's answer, but I found the different point of view and the elaborate explanation very useful. (The impatient may skip to Example 11.4.5 directly.)
Adding Relations
We reinterpret the quotient ring construction when the ideal $I$ is principal, say $I = (a)$. In this situation, we think of $overline R = R / I$ as the ring obtained by imposing the relation $a = 0$ on $R$, or by killing the element $a$. For instance, the field $mathbb F_7$ will be thought of as the ring obtained by killing $7$ in the ring $mathbb Z$ of integers.
Let's examine the collapsing that takes place in the map $pi: R to overline R$. Its kernel is the ideal $I$, so $a$ is in the kernel: $pi(a) = 0$. If $b$ is any element of $R$, the elements that have the same image in $overline R$ as $b$ are those in the coset $b + I$ and since $I = (a)$ those elements have the form $b+ra$. We see that imposing the relation $a =0$ in the ring $R$ forces us to set $b = b + ra$ for all $b$ and all $r$ in $R$, and that these are the only consequences of killing $a$.
Any number of relations $a_1 = 0, ldots, a_n = 0$ can be introduced, by working modulo the ideal $I$ generated by $a_1, ldots, a_n$, the set of linear combinations $r_1 a_1 + cdots + r_n a_n$, with coefficients $r_i$ in $R$. The quotient ring $overline R = R/I$ is viewed as the ring obtained by killing the $n$ elements. Two elements $b$ and $b'$ of $R$ have the same image in $overline R$ if and only if $b'$ has the form $b + r_1 a_1 + cdots +r_n a_n$ with $r_i$ in $R$.
The more relations we add, the more collapsing takes place in the map $pi$. If we add relations carelessly, the worst that can happen is that we may end up with $I = R$ and $overline R = 0$. All relations $a = 0$ become true when we collapse $R$ to the zero ring.
Here the Correspondence Theorem asserts something that is intuitively clear: Introducing relations one at a time or all together leads to isomorphic results. To spell this out, let $a$ and $b$ be elements of a ring $R$, and let $overline R = R / (a)$ be the result of killing $a$ in $R$. Let $overline b$ be the residue of $b$ in $overline R$. The Correspondence Theorem tells us that the principal ideal $(overline b)$ of $overline R$ corresponds to the ideal $(a,b)$ of $R$, and that $R/(a,b)$ is isomorphic to $overline R / (overline b)$. Killing $a$ and $b$ in $R$ at the same time gives the same result as killing $overline b$ in the ring $overline R$ that is obtained by killing $a$ first.
Example 11.4.5. We ask to identify the quotient ring $overline R = mathbb Z[i]/(i-2)$, the ring obtained from the Gauss integers by introducing the relation $i-2=0$. Instead of analyzing this directly, we note that the kernel of the map $mathbb Z[x] to mathbb Z[i]$ sending $x mapsto i$ is the principal ideal of $mathbb Z[x]$ generated by $f = x^2 + 1$. The First Isomorphism Theorem tells us that $mathbb Z[x]/(f) approx mathbb Z[i]$. The image of $g = x-2$ is $i-2$, so $overline R$ can also be obtained by introducing the two relations $f = 0$ and $g = 0$ into the integer polynomial ring. Let $I = (f,g)$ be the ideal of $mathbb Z[x]$ generated by the two polynomials $f$ and $g$. Then $overline R =mathbb Z[x]/I$.
To form $overline R$, we may introduce the two relations in the opposite order, first killing $g$ and then $f$. The principal ideal $(g)$ of $mathbb Z[x]$ is the kernel of the homomorphism $mathbb Z[x] to mathbb Z$ that sends $x mapsto 2$. So when we kill $x-2$ in $mathbb Z[x]$, we obtain a ring isomorphic to $mathbb Z$, in which the residue of $x$ is $2$. Then the residue of $f = x^2+1$ becomes $5$. So we can also obtain $overline R$ by killing $5$ in $mathbb Z$, and therefore $overline R approx mathbb F_5$.
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1
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I like the conceptual image he provided, I wish if I heared about his book earlier
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– SomeOne
Dec 23 '14 at 21:06
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6 Answers
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6 Answers
6
active
oldest
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$begingroup$
Define
$$phi: mathbb{Z} rightarrow mathbb{Z}[i]/(3-i) text{ where } phi(z) = z + (3-i)mathbb{Z}[i].$$
It follows simply that
$ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z}$.
So for any such $z in ker phi$, we have $z = (3-i)(a+bi)$ for some $a,b in mathbb{Z}$. But $(3-i)(a+bi) in mathbb{Z}$ happens if and only if $3b-a=0$.
So $$begin{align*}
ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z} &= {(3-i)(3b+bi)mid b in mathbb{Z}}\
&= {(9b + b) + i(3b-3b)mid b in mathbb{Z}}\
&= {10bmid b in mathbb{Z}}\
&= 10mathbb{Z}.
end{align*}$$
To see $phi$ is surjective, let $(a+bi) + (3-i)mathbb{Z}[i] in mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $phi(a+3b) = (a+bi) + (3-i)mathbb{Z}[i]$.
Hence $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 23 '11 at 14:35
Brandon CarterBrandon Carter
7,31822538
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Don't you have to show that $phi$ is also surjective?
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– Greg Graviton
Feb 23 '11 at 17:25
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This is more simple to understand.+1
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– anonymous
Feb 24 '11 at 8:03
add a comment |
$begingroup$
Define
$$phi: mathbb{Z} rightarrow mathbb{Z}[i]/(3-i) text{ where } phi(z) = z + (3-i)mathbb{Z}[i].$$
It follows simply that
$ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z}$.
So for any such $z in ker phi$, we have $z = (3-i)(a+bi)$ for some $a,b in mathbb{Z}$. But $(3-i)(a+bi) in mathbb{Z}$ happens if and only if $3b-a=0$.
So $$begin{align*}
ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z} &= {(3-i)(3b+bi)mid b in mathbb{Z}}\
&= {(9b + b) + i(3b-3b)mid b in mathbb{Z}}\
&= {10bmid b in mathbb{Z}}\
&= 10mathbb{Z}.
end{align*}$$
To see $phi$ is surjective, let $(a+bi) + (3-i)mathbb{Z}[i] in mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $phi(a+3b) = (a+bi) + (3-i)mathbb{Z}[i]$.
Hence $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 23 '11 at 14:35
Brandon CarterBrandon Carter
7,31822538
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Don't you have to show that $phi$ is also surjective?
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– Greg Graviton
Feb 23 '11 at 17:25
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This is more simple to understand.+1
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– anonymous
Feb 24 '11 at 8:03
add a comment |
$begingroup$
Define
$$phi: mathbb{Z} rightarrow mathbb{Z}[i]/(3-i) text{ where } phi(z) = z + (3-i)mathbb{Z}[i].$$
It follows simply that
$ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z}$.
So for any such $z in ker phi$, we have $z = (3-i)(a+bi)$ for some $a,b in mathbb{Z}$. But $(3-i)(a+bi) in mathbb{Z}$ happens if and only if $3b-a=0$.
So $$begin{align*}
ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z} &= {(3-i)(3b+bi)mid b in mathbb{Z}}\
&= {(9b + b) + i(3b-3b)mid b in mathbb{Z}}\
&= {10bmid b in mathbb{Z}}\
&= 10mathbb{Z}.
end{align*}$$
To see $phi$ is surjective, let $(a+bi) + (3-i)mathbb{Z}[i] in mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $phi(a+3b) = (a+bi) + (3-i)mathbb{Z}[i]$.
Hence $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 23 '11 at 14:35
Brandon CarterBrandon Carter
7,31822538
$endgroup$
Define
$$phi: mathbb{Z} rightarrow mathbb{Z}[i]/(3-i) text{ where } phi(z) = z + (3-i)mathbb{Z}[i].$$
It follows simply that
$ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z}$.
So for any such $z in ker phi$, we have $z = (3-i)(a+bi)$ for some $a,b in mathbb{Z}$. But $(3-i)(a+bi) in mathbb{Z}$ happens if and only if $3b-a=0$.
So $$begin{align*}
ker phi = (3-i)mathbb{Z}[i] cap mathbb{Z} &= {(3-i)(3b+bi)mid b in mathbb{Z}}\
&= {(9b + b) + i(3b-3b)mid b in mathbb{Z}}\
&= {10bmid b in mathbb{Z}}\
&= 10mathbb{Z}.
end{align*}$$
To see $phi$ is surjective, let $(a+bi) + (3-i)mathbb{Z}[i] in mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $phi(a+3b) = (a+bi) + (3-i)mathbb{Z}[i]$.
Hence $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 23 '11 at 14:35
Brandon CarterBrandon Carter
7,31822538
edited Jan 28 at 1:47
answered Feb 23 '11 at 14:35
Brandon CarterBrandon Carter
7,31822538
answered Feb 23 '11 at 14:35
Brandon CarterBrandon Carter
7,31822538
answered Feb 23 '11 at 14:35
Brandon CarterBrandon Carter
7,31822538
7,31822538
$begingroup$
Don't you have to show that $phi$ is also surjective?
$endgroup$
– Greg Graviton
Feb 23 '11 at 17:25
$begingroup$
This is more simple to understand.+1
$endgroup$
– anonymous
Feb 24 '11 at 8:03
add a comment |
$begingroup$
Don't you have to show that $phi$ is also surjective?
$endgroup$
– Greg Graviton
Feb 23 '11 at 17:25
$begingroup$
This is more simple to understand.+1
$endgroup$
– anonymous
Feb 24 '11 at 8:03
$begingroup$
Don't you have to show that $phi$ is also surjective?
$endgroup$
– Greg Graviton
Feb 23 '11 at 17:25
$begingroup$
Don't you have to show that $phi$ is also surjective?
$endgroup$
– Greg Graviton
Feb 23 '11 at 17:25
$begingroup$
This is more simple to understand.+1
$endgroup$
– anonymous
Feb 24 '11 at 8:03
$begingroup$
This is more simple to understand.+1
$endgroup$
– anonymous
Feb 24 '11 at 8:03
add a comment |
$begingroup$
This diagram shows the Gaussian integers modulo $3-i$.
The red points shown are all considered to be $0$ but their locations in $mathbb Z[i]$ are $0$, $3-i$, $i(3-i)$ and $3-i + i(3-i)$. Every congruence class must be inside that box once and you can see there are $10$ of them.
The arrows show adding by $1$ each time. Doing that takes you through every equivalence class and then back to the start.
So $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 24 '11 at 16:31
quantaquanta
7,30623974
$endgroup$
add a comment |
$begingroup$
This diagram shows the Gaussian integers modulo $3-i$.
The red points shown are all considered to be $0$ but their locations in $mathbb Z[i]$ are $0$, $3-i$, $i(3-i)$ and $3-i + i(3-i)$. Every congruence class must be inside that box once and you can see there are $10$ of them.
The arrows show adding by $1$ each time. Doing that takes you through every equivalence class and then back to the start.
So $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 24 '11 at 16:31
quantaquanta
7,30623974
$endgroup$
add a comment |
$begingroup$
This diagram shows the Gaussian integers modulo $3-i$.
The red points shown are all considered to be $0$ but their locations in $mathbb Z[i]$ are $0$, $3-i$, $i(3-i)$ and $3-i + i(3-i)$. Every congruence class must be inside that box once and you can see there are $10$ of them.
The arrows show adding by $1$ each time. Doing that takes you through every equivalence class and then back to the start.
So $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 24 '11 at 16:31
quantaquanta
7,30623974
$endgroup$
This diagram shows the Gaussian integers modulo $3-i$.
The red points shown are all considered to be $0$ but their locations in $mathbb Z[i]$ are $0$, $3-i$, $i(3-i)$ and $3-i + i(3-i)$. Every congruence class must be inside that box once and you can see there are $10$ of them.
The arrows show adding by $1$ each time. Doing that takes you through every equivalence class and then back to the start.
So $mathbb{Z}[i]/(3-i) cong mathbb{Z}/10mathbb{Z}$.
answered Feb 24 '11 at 16:31
quantaquanta
7,30623974
edited Apr 27 '13 at 13:59
user26857
answered Feb 24 '11 at 16:31
quantaquanta
7,30623974
answered Feb 24 '11 at 16:31
quantaquanta
7,30623974
answered Feb 24 '11 at 16:31
quantaquanta
7,30623974
7,30623974
add a comment |
add a comment |
$begingroup$
Firstly: it is not true in general that $mathbb Z[i]/(a - ib) cong mathbb Z/(a^2 + b^2).$ (Consider the case of $3 - 0cdot i$.)
The claimed isomorphism does hold if $a$ and $b$ are coprime.
Here is a sketch of how to see this:
To begin with, note that it is much easier to consider maps from $mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $mathbb Z$ maps to any ring with unity in a canonical way, by sending $1$ to $1$.
So consider the canonical map $mathbb Z to mathbb Z[i]/(a - i b).$
The target is finite of order $a^2 + b^2$, and so this map factors to give an
injection
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - i b)$
for some $n$ dividing $a^2 + b^2$.
Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $mathbb Z/(n)$. Combining this observation with the equation $a - i b = 0$ (which holds in $mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.
answered Feb 23 '11 at 14:07
Matt EMatt E
105k8220390
$endgroup$
2
$begingroup$
Thanks for this generalization. It is very helpful (to this third party, at least!)
$endgroup$
– The Chaz 2.0
Mar 21 '11 at 16:42
2
$begingroup$
Hello Matt, something I've been trying to wrap my head around recently is why $mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you.
$endgroup$
– yunone
Oct 30 '11 at 1:27
3
$begingroup$
@yunone: Dear yunone, It follows e.g. from the division algorithm in $mathbb Z[i]$. Regards,
$endgroup$
– Matt E
Oct 30 '11 at 2:12
10
$begingroup$
@yunone: Think about the chain of ideals $(a^2+b^2) subset (a+bi) subset {mathbf Z}[i]$. For a positive integer $n$, the index $[{mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{mathbf Z}[i]:(a^2+b^2)] = [{mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${mathbf Z}[i]/(a+bi) cong {mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{mathbf Z}[i]:(a+bi)]^2$. Now take square roots.
$endgroup$
– KCd
Apr 7 '12 at 21:25
$begingroup$
Dear @KCd, thanks for the explanation.
$endgroup$
– yunone
Apr 27 '12 at 21:20
|
show 1 more comment
$begingroup$
Firstly: it is not true in general that $mathbb Z[i]/(a - ib) cong mathbb Z/(a^2 + b^2).$ (Consider the case of $3 - 0cdot i$.)
The claimed isomorphism does hold if $a$ and $b$ are coprime.
Here is a sketch of how to see this:
To begin with, note that it is much easier to consider maps from $mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $mathbb Z$ maps to any ring with unity in a canonical way, by sending $1$ to $1$.
So consider the canonical map $mathbb Z to mathbb Z[i]/(a - i b).$
The target is finite of order $a^2 + b^2$, and so this map factors to give an
injection
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - i b)$
for some $n$ dividing $a^2 + b^2$.
Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $mathbb Z/(n)$. Combining this observation with the equation $a - i b = 0$ (which holds in $mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.
answered Feb 23 '11 at 14:07
Matt EMatt E
105k8220390
$endgroup$
2
$begingroup$
Thanks for this generalization. It is very helpful (to this third party, at least!)
$endgroup$
– The Chaz 2.0
Mar 21 '11 at 16:42
2
$begingroup$
Hello Matt, something I've been trying to wrap my head around recently is why $mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you.
$endgroup$
– yunone
Oct 30 '11 at 1:27
3
$begingroup$
@yunone: Dear yunone, It follows e.g. from the division algorithm in $mathbb Z[i]$. Regards,
$endgroup$
– Matt E
Oct 30 '11 at 2:12
10
$begingroup$
@yunone: Think about the chain of ideals $(a^2+b^2) subset (a+bi) subset {mathbf Z}[i]$. For a positive integer $n$, the index $[{mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{mathbf Z}[i]:(a^2+b^2)] = [{mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${mathbf Z}[i]/(a+bi) cong {mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{mathbf Z}[i]:(a+bi)]^2$. Now take square roots.
$endgroup$
– KCd
Apr 7 '12 at 21:25
$begingroup$
Dear @KCd, thanks for the explanation.
$endgroup$
– yunone
Apr 27 '12 at 21:20
|
show 1 more comment
$begingroup$
Firstly: it is not true in general that $mathbb Z[i]/(a - ib) cong mathbb Z/(a^2 + b^2).$ (Consider the case of $3 - 0cdot i$.)
The claimed isomorphism does hold if $a$ and $b$ are coprime.
Here is a sketch of how to see this:
To begin with, note that it is much easier to consider maps from $mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $mathbb Z$ maps to any ring with unity in a canonical way, by sending $1$ to $1$.
So consider the canonical map $mathbb Z to mathbb Z[i]/(a - i b).$
The target is finite of order $a^2 + b^2$, and so this map factors to give an
injection
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - i b)$
for some $n$ dividing $a^2 + b^2$.
Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $mathbb Z/(n)$. Combining this observation with the equation $a - i b = 0$ (which holds in $mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.
answered Feb 23 '11 at 14:07
Matt EMatt E
105k8220390
$endgroup$
Firstly: it is not true in general that $mathbb Z[i]/(a - ib) cong mathbb Z/(a^2 + b^2).$ (Consider the case of $3 - 0cdot i$.)
The claimed isomorphism does hold if $a$ and $b$ are coprime.
Here is a sketch of how to see this:
To begin with, note that it is much easier to consider maps from $mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $mathbb Z$ maps to any ring with unity in a canonical way, by sending $1$ to $1$.
So consider the canonical map $mathbb Z to mathbb Z[i]/(a - i b).$
The target is finite of order $a^2 + b^2$, and so this map factors to give an
injection
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - i b)$
for some $n$ dividing $a^2 + b^2$.
Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $mathbb Z/(n)$. Combining this observation with the equation $a - i b = 0$ (which holds in $mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map
$mathbb Z/(n) hookrightarrow mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.
answered Feb 23 '11 at 14:07
Matt EMatt E
105k8220390
answered Feb 23 '11 at 14:07
Matt EMatt E
105k8220390
answered Feb 23 '11 at 14:07
Matt EMatt E
105k8220390
answered Feb 23 '11 at 14:07
Matt EMatt E
105k8220390
105k8220390
2
$begingroup$
Thanks for this generalization. It is very helpful (to this third party, at least!)
$endgroup$
– The Chaz 2.0
Mar 21 '11 at 16:42
2
$begingroup$
Hello Matt, something I've been trying to wrap my head around recently is why $mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you.
$endgroup$
– yunone
Oct 30 '11 at 1:27
3
$begingroup$
@yunone: Dear yunone, It follows e.g. from the division algorithm in $mathbb Z[i]$. Regards,
$endgroup$
– Matt E
Oct 30 '11 at 2:12
10
$begingroup$
@yunone: Think about the chain of ideals $(a^2+b^2) subset (a+bi) subset {mathbf Z}[i]$. For a positive integer $n$, the index $[{mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{mathbf Z}[i]:(a^2+b^2)] = [{mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${mathbf Z}[i]/(a+bi) cong {mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{mathbf Z}[i]:(a+bi)]^2$. Now take square roots.
$endgroup$
– KCd
Apr 7 '12 at 21:25
$begingroup$
Dear @KCd, thanks for the explanation.
$endgroup$
– yunone
Apr 27 '12 at 21:20
|
show 1 more comment
2
$begingroup$
Thanks for this generalization. It is very helpful (to this third party, at least!)
$endgroup$
– The Chaz 2.0
Mar 21 '11 at 16:42
2
$begingroup$
Hello Matt, something I've been trying to wrap my head around recently is why $mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you.
$endgroup$
– yunone
Oct 30 '11 at 1:27
3
$begingroup$
@yunone: Dear yunone, It follows e.g. from the division algorithm in $mathbb Z[i]$. Regards,
$endgroup$
– Matt E
Oct 30 '11 at 2:12
10
$begingroup$
@yunone: Think about the chain of ideals $(a^2+b^2) subset (a+bi) subset {mathbf Z}[i]$. For a positive integer $n$, the index $[{mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{mathbf Z}[i]:(a^2+b^2)] = [{mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${mathbf Z}[i]/(a+bi) cong {mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{mathbf Z}[i]:(a+bi)]^2$. Now take square roots.
$endgroup$
– KCd
Apr 7 '12 at 21:25
$begingroup$
Dear @KCd, thanks for the explanation.
$endgroup$
– yunone
Apr 27 '12 at 21:20
2
2
$begingroup$
Thanks for this generalization. It is very helpful (to this third party, at least!)
$endgroup$
– The Chaz 2.0
Mar 21 '11 at 16:42
$begingroup$
Thanks for this generalization. It is very helpful (to this third party, at least!)
$endgroup$
– The Chaz 2.0
Mar 21 '11 at 16:42
2
2
$begingroup$
Hello Matt, something I've been trying to wrap my head around recently is why $mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you.
$endgroup$
– yunone
Oct 30 '11 at 1:27
$begingroup$
Hello Matt, something I've been trying to wrap my head around recently is why $mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you.
$endgroup$
– yunone
Oct 30 '11 at 1:27
3
3
$begingroup$
@yunone: Dear yunone, It follows e.g. from the division algorithm in $mathbb Z[i]$. Regards,
$endgroup$
– Matt E
Oct 30 '11 at 2:12
$begingroup$
@yunone: Dear yunone, It follows e.g. from the division algorithm in $mathbb Z[i]$. Regards,
$endgroup$
– Matt E
Oct 30 '11 at 2:12
10
10
$begingroup$
@yunone: Think about the chain of ideals $(a^2+b^2) subset (a+bi) subset {mathbf Z}[i]$. For a positive integer $n$, the index $[{mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{mathbf Z}[i]:(a^2+b^2)] = [{mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${mathbf Z}[i]/(a+bi) cong {mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{mathbf Z}[i]:(a+bi)]^2$. Now take square roots.
$endgroup$
– KCd
Apr 7 '12 at 21:25
$begingroup$
@yunone: Think about the chain of ideals $(a^2+b^2) subset (a+bi) subset {mathbf Z}[i]$. For a positive integer $n$, the index $[{mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{mathbf Z}[i]:(a^2+b^2)] = [{mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${mathbf Z}[i]/(a+bi) cong {mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{mathbf Z}[i]:(a+bi)]^2$. Now take square roots.
$endgroup$
– KCd
Apr 7 '12 at 21:25
$begingroup$
Dear @KCd, thanks for the explanation.
$endgroup$
– yunone
Apr 27 '12 at 21:20
$begingroup$
Dear @KCd, thanks for the explanation.
$endgroup$
– yunone
Apr 27 '12 at 21:20
|
show 1 more comment
$begingroup$
Go back one step and add the defining equation for $i$ to the ideal. In other words, consider your ring as a quotient of the ring of polynomials $mathbb Z[x]$:
$$ mathbb Z[i] / (3-i) = mathbb Z [x] / (3-x,x^2+1) $$
Manipulating the ideal $(3-x,x^2+1)$ a bit, you will find that the quotient is indeed equal to $mathbb Z/10mathbb Z$.
This makes sense because setting $i^2=-1$ and $i=3$ implies that $9 = -1$, which is true in $mathbb Z/10mathbb Z$.
answered Feb 23 '11 at 11:43
Greg GravitonGreg Graviton
3,69121734
$endgroup$
add a comment |
$begingroup$
Go back one step and add the defining equation for $i$ to the ideal. In other words, consider your ring as a quotient of the ring of polynomials $mathbb Z[x]$:
$$ mathbb Z[i] / (3-i) = mathbb Z [x] / (3-x,x^2+1) $$
Manipulating the ideal $(3-x,x^2+1)$ a bit, you will find that the quotient is indeed equal to $mathbb Z/10mathbb Z$.
This makes sense because setting $i^2=-1$ and $i=3$ implies that $9 = -1$, which is true in $mathbb Z/10mathbb Z$.
answered Feb 23 '11 at 11:43
Greg GravitonGreg Graviton
3,69121734
$endgroup$
add a comment |
$begingroup$
Go back one step and add the defining equation for $i$ to the ideal. In other words, consider your ring as a quotient of the ring of polynomials $mathbb Z[x]$:
$$ mathbb Z[i] / (3-i) = mathbb Z [x] / (3-x,x^2+1) $$
Manipulating the ideal $(3-x,x^2+1)$ a bit, you will find that the quotient is indeed equal to $mathbb Z/10mathbb Z$.
This makes sense because setting $i^2=-1$ and $i=3$ implies that $9 = -1$, which is true in $mathbb Z/10mathbb Z$.
answered Feb 23 '11 at 11:43
Greg GravitonGreg Graviton
3,69121734
$endgroup$
Go back one step and add the defining equation for $i$ to the ideal. In other words, consider your ring as a quotient of the ring of polynomials $mathbb Z[x]$:
$$ mathbb Z[i] / (3-i) = mathbb Z [x] / (3-x,x^2+1) $$
Manipulating the ideal $(3-x,x^2+1)$ a bit, you will find that the quotient is indeed equal to $mathbb Z/10mathbb Z$.
This makes sense because setting $i^2=-1$ and $i=3$ implies that $9 = -1$, which is true in $mathbb Z/10mathbb Z$.
answered Feb 23 '11 at 11:43
Greg GravitonGreg Graviton
3,69121734
answered Feb 23 '11 at 11:43
Greg GravitonGreg Graviton
3,69121734
answered Feb 23 '11 at 11:43
Greg GravitonGreg Graviton
3,69121734
answered Feb 23 '11 at 11:43
Greg GravitonGreg Graviton
3,69121734
3,69121734
add a comment |
add a comment |
$begingroup$
In general, one knows that if $alpha$ is an integer in the number field $K$, then
$$
{rm N}_{K/{Bbb Q}}(alpha)=left|frac{A}{Aalpha}right|
$$
Here $rm N$ is the norm and $A$ denotes the ring of integers. In the question's situation, $A={Bbb Z}[i]$ is the ring of Gaussian integer and since $K$ is quadratic imaginary, ${rm N}_{K/{Bbb Q}}(alpha)=alphabaralpha$ where the bar denotes complex conjugation.
When $alpha=3-i$, ${rm N}_{K/{Bbb Q}}(alpha)=(3-i)(3+i)=10$, thus
${Bbb Z}[i]/(3-i)$ is a ring with 10 elements, whose representatives are
$$
left{0,1,2,i,i+1,i+2,2i,2i+1,2i+2,-1 right}.
$$
A minute of thought and a brief inspection of these representatives should convince that the ring is indeed isomorphic to the ring of classes modulo 10.
answered Feb 23 '11 at 13:57
Andrea MoriAndrea Mori
20k13466
$endgroup$
2
$begingroup$
Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$...
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– Pete L. Clark
Feb 23 '11 at 14:33
$begingroup$
Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context.
$endgroup$
– Andrea Mori
Feb 23 '11 at 19:25
5
$begingroup$
my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...)
$endgroup$
– Pete L. Clark
Feb 23 '11 at 19:56
add a comment |
$begingroup$
In general, one knows that if $alpha$ is an integer in the number field $K$, then
$$
{rm N}_{K/{Bbb Q}}(alpha)=left|frac{A}{Aalpha}right|
$$
Here $rm N$ is the norm and $A$ denotes the ring of integers. In the question's situation, $A={Bbb Z}[i]$ is the ring of Gaussian integer and since $K$ is quadratic imaginary, ${rm N}_{K/{Bbb Q}}(alpha)=alphabaralpha$ where the bar denotes complex conjugation.
When $alpha=3-i$, ${rm N}_{K/{Bbb Q}}(alpha)=(3-i)(3+i)=10$, thus
${Bbb Z}[i]/(3-i)$ is a ring with 10 elements, whose representatives are
$$
left{0,1,2,i,i+1,i+2,2i,2i+1,2i+2,-1 right}.
$$
A minute of thought and a brief inspection of these representatives should convince that the ring is indeed isomorphic to the ring of classes modulo 10.
answered Feb 23 '11 at 13:57
Andrea MoriAndrea Mori
20k13466
$endgroup$
2
$begingroup$
Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$...
$endgroup$
– Pete L. Clark
Feb 23 '11 at 14:33
$begingroup$
Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context.
$endgroup$
– Andrea Mori
Feb 23 '11 at 19:25
5
$begingroup$
my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...)
$endgroup$
– Pete L. Clark
Feb 23 '11 at 19:56
add a comment |
$begingroup$
In general, one knows that if $alpha$ is an integer in the number field $K$, then
$$
{rm N}_{K/{Bbb Q}}(alpha)=left|frac{A}{Aalpha}right|
$$
Here $rm N$ is the norm and $A$ denotes the ring of integers. In the question's situation, $A={Bbb Z}[i]$ is the ring of Gaussian integer and since $K$ is quadratic imaginary, ${rm N}_{K/{Bbb Q}}(alpha)=alphabaralpha$ where the bar denotes complex conjugation.
When $alpha=3-i$, ${rm N}_{K/{Bbb Q}}(alpha)=(3-i)(3+i)=10$, thus
${Bbb Z}[i]/(3-i)$ is a ring with 10 elements, whose representatives are
$$
left{0,1,2,i,i+1,i+2,2i,2i+1,2i+2,-1 right}.
$$
A minute of thought and a brief inspection of these representatives should convince that the ring is indeed isomorphic to the ring of classes modulo 10.
answered Feb 23 '11 at 13:57
Andrea MoriAndrea Mori
20k13466
$endgroup$
In general, one knows that if $alpha$ is an integer in the number field $K$, then
$$
{rm N}_{K/{Bbb Q}}(alpha)=left|frac{A}{Aalpha}right|
$$
Here $rm N$ is the norm and $A$ denotes the ring of integers. In the question's situation, $A={Bbb Z}[i]$ is the ring of Gaussian integer and since $K$ is quadratic imaginary, ${rm N}_{K/{Bbb Q}}(alpha)=alphabaralpha$ where the bar denotes complex conjugation.
When $alpha=3-i$, ${rm N}_{K/{Bbb Q}}(alpha)=(3-i)(3+i)=10$, thus
${Bbb Z}[i]/(3-i)$ is a ring with 10 elements, whose representatives are
$$
left{0,1,2,i,i+1,i+2,2i,2i+1,2i+2,-1 right}.
$$
A minute of thought and a brief inspection of these representatives should convince that the ring is indeed isomorphic to the ring of classes modulo 10.
answered Feb 23 '11 at 13:57
Andrea MoriAndrea Mori
20k13466
answered Feb 23 '11 at 13:57
Andrea MoriAndrea Mori
20k13466
answered Feb 23 '11 at 13:57
Andrea MoriAndrea Mori
20k13466
answered Feb 23 '11 at 13:57
Andrea MoriAndrea Mori
20k13466
20k13466
2
$begingroup$
Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$...
$endgroup$
– Pete L. Clark
Feb 23 '11 at 14:33
$begingroup$
Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context.
$endgroup$
– Andrea Mori
Feb 23 '11 at 19:25
5
$begingroup$
my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...)
$endgroup$
– Pete L. Clark
Feb 23 '11 at 19:56
add a comment |
2
$begingroup$
Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$...
$endgroup$
– Pete L. Clark
Feb 23 '11 at 14:33
$begingroup$
Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context.
$endgroup$
– Andrea Mori
Feb 23 '11 at 19:25
5
$begingroup$
my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...)
$endgroup$
– Pete L. Clark
Feb 23 '11 at 19:56
2
2
$begingroup$
Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$...
$endgroup$
– Pete L. Clark
Feb 23 '11 at 14:33
$begingroup$
Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$...
$endgroup$
– Pete L. Clark
Feb 23 '11 at 14:33
$begingroup$
Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context.
$endgroup$
– Andrea Mori
Feb 23 '11 at 19:25
$begingroup$
Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context.
$endgroup$
– Andrea Mori
Feb 23 '11 at 19:25
5
5
$begingroup$
my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...)
$endgroup$
– Pete L. Clark
Feb 23 '11 at 19:56
$begingroup$
my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...)
$endgroup$
– Pete L. Clark
Feb 23 '11 at 19:56
add a comment |
$begingroup$
In this post, I am quoting the solution to a related question from Artin, with the preceding explanation (Section 11.4, p. 337-338). This is really the same as Greg Graviton's answer, but I found the different point of view and the elaborate explanation very useful. (The impatient may skip to Example 11.4.5 directly.)
Adding Relations
We reinterpret the quotient ring construction when the ideal $I$ is principal, say $I = (a)$. In this situation, we think of $overline R = R / I$ as the ring obtained by imposing the relation $a = 0$ on $R$, or by killing the element $a$. For instance, the field $mathbb F_7$ will be thought of as the ring obtained by killing $7$ in the ring $mathbb Z$ of integers.
Let's examine the collapsing that takes place in the map $pi: R to overline R$. Its kernel is the ideal $I$, so $a$ is in the kernel: $pi(a) = 0$. If $b$ is any element of $R$, the elements that have the same image in $overline R$ as $b$ are those in the coset $b + I$ and since $I = (a)$ those elements have the form $b+ra$. We see that imposing the relation $a =0$ in the ring $R$ forces us to set $b = b + ra$ for all $b$ and all $r$ in $R$, and that these are the only consequences of killing $a$.
Any number of relations $a_1 = 0, ldots, a_n = 0$ can be introduced, by working modulo the ideal $I$ generated by $a_1, ldots, a_n$, the set of linear combinations $r_1 a_1 + cdots + r_n a_n$, with coefficients $r_i$ in $R$. The quotient ring $overline R = R/I$ is viewed as the ring obtained by killing the $n$ elements. Two elements $b$ and $b'$ of $R$ have the same image in $overline R$ if and only if $b'$ has the form $b + r_1 a_1 + cdots +r_n a_n$ with $r_i$ in $R$.
The more relations we add, the more collapsing takes place in the map $pi$. If we add relations carelessly, the worst that can happen is that we may end up with $I = R$ and $overline R = 0$. All relations $a = 0$ become true when we collapse $R$ to the zero ring.
Here the Correspondence Theorem asserts something that is intuitively clear: Introducing relations one at a time or all together leads to isomorphic results. To spell this out, let $a$ and $b$ be elements of a ring $R$, and let $overline R = R / (a)$ be the result of killing $a$ in $R$. Let $overline b$ be the residue of $b$ in $overline R$. The Correspondence Theorem tells us that the principal ideal $(overline b)$ of $overline R$ corresponds to the ideal $(a,b)$ of $R$, and that $R/(a,b)$ is isomorphic to $overline R / (overline b)$. Killing $a$ and $b$ in $R$ at the same time gives the same result as killing $overline b$ in the ring $overline R$ that is obtained by killing $a$ first.
Example 11.4.5. We ask to identify the quotient ring $overline R = mathbb Z[i]/(i-2)$, the ring obtained from the Gauss integers by introducing the relation $i-2=0$. Instead of analyzing this directly, we note that the kernel of the map $mathbb Z[x] to mathbb Z[i]$ sending $x mapsto i$ is the principal ideal of $mathbb Z[x]$ generated by $f = x^2 + 1$. The First Isomorphism Theorem tells us that $mathbb Z[x]/(f) approx mathbb Z[i]$. The image of $g = x-2$ is $i-2$, so $overline R$ can also be obtained by introducing the two relations $f = 0$ and $g = 0$ into the integer polynomial ring. Let $I = (f,g)$ be the ideal of $mathbb Z[x]$ generated by the two polynomials $f$ and $g$. Then $overline R =mathbb Z[x]/I$.
To form $overline R$, we may introduce the two relations in the opposite order, first killing $g$ and then $f$. The principal ideal $(g)$ of $mathbb Z[x]$ is the kernel of the homomorphism $mathbb Z[x] to mathbb Z$ that sends $x mapsto 2$. So when we kill $x-2$ in $mathbb Z[x]$, we obtain a ring isomorphic to $mathbb Z$, in which the residue of $x$ is $2$. Then the residue of $f = x^2+1$ becomes $5$. So we can also obtain $overline R$ by killing $5$ in $mathbb Z$, and therefore $overline R approx mathbb F_5$.
$endgroup$
1
$begingroup$
I like the conceptual image he provided, I wish if I heared about his book earlier
$endgroup$
– SomeOne
Dec 23 '14 at 21:06
add a comment |
$begingroup$
In this post, I am quoting the solution to a related question from Artin, with the preceding explanation (Section 11.4, p. 337-338). This is really the same as Greg Graviton's answer, but I found the different point of view and the elaborate explanation very useful. (The impatient may skip to Example 11.4.5 directly.)
Adding Relations
We reinterpret the quotient ring construction when the ideal $I$ is principal, say $I = (a)$. In this situation, we think of $overline R = R / I$ as the ring obtained by imposing the relation $a = 0$ on $R$, or by killing the element $a$. For instance, the field $mathbb F_7$ will be thought of as the ring obtained by killing $7$ in the ring $mathbb Z$ of integers.
Let's examine the collapsing that takes place in the map $pi: R to overline R$. Its kernel is the ideal $I$, so $a$ is in the kernel: $pi(a) = 0$. If $b$ is any element of $R$, the elements that have the same image in $overline R$ as $b$ are those in the coset $b + I$ and since $I = (a)$ those elements have the form $b+ra$. We see that imposing the relation $a =0$ in the ring $R$ forces us to set $b = b + ra$ for all $b$ and all $r$ in $R$, and that these are the only consequences of killing $a$.
Any number of relations $a_1 = 0, ldots, a_n = 0$ can be introduced, by working modulo the ideal $I$ generated by $a_1, ldots, a_n$, the set of linear combinations $r_1 a_1 + cdots + r_n a_n$, with coefficients $r_i$ in $R$. The quotient ring $overline R = R/I$ is viewed as the ring obtained by killing the $n$ elements. Two elements $b$ and $b'$ of $R$ have the same image in $overline R$ if and only if $b'$ has the form $b + r_1 a_1 + cdots +r_n a_n$ with $r_i$ in $R$.
The more relations we add, the more collapsing takes place in the map $pi$. If we add relations carelessly, the worst that can happen is that we may end up with $I = R$ and $overline R = 0$. All relations $a = 0$ become true when we collapse $R$ to the zero ring.
Here the Correspondence Theorem asserts something that is intuitively clear: Introducing relations one at a time or all together leads to isomorphic results. To spell this out, let $a$ and $b$ be elements of a ring $R$, and let $overline R = R / (a)$ be the result of killing $a$ in $R$. Let $overline b$ be the residue of $b$ in $overline R$. The Correspondence Theorem tells us that the principal ideal $(overline b)$ of $overline R$ corresponds to the ideal $(a,b)$ of $R$, and that $R/(a,b)$ is isomorphic to $overline R / (overline b)$. Killing $a$ and $b$ in $R$ at the same time gives the same result as killing $overline b$ in the ring $overline R$ that is obtained by killing $a$ first.
Example 11.4.5. We ask to identify the quotient ring $overline R = mathbb Z[i]/(i-2)$, the ring obtained from the Gauss integers by introducing the relation $i-2=0$. Instead of analyzing this directly, we note that the kernel of the map $mathbb Z[x] to mathbb Z[i]$ sending $x mapsto i$ is the principal ideal of $mathbb Z[x]$ generated by $f = x^2 + 1$. The First Isomorphism Theorem tells us that $mathbb Z[x]/(f) approx mathbb Z[i]$. The image of $g = x-2$ is $i-2$, so $overline R$ can also be obtained by introducing the two relations $f = 0$ and $g = 0$ into the integer polynomial ring. Let $I = (f,g)$ be the ideal of $mathbb Z[x]$ generated by the two polynomials $f$ and $g$. Then $overline R =mathbb Z[x]/I$.
To form $overline R$, we may introduce the two relations in the opposite order, first killing $g$ and then $f$. The principal ideal $(g)$ of $mathbb Z[x]$ is the kernel of the homomorphism $mathbb Z[x] to mathbb Z$ that sends $x mapsto 2$. So when we kill $x-2$ in $mathbb Z[x]$, we obtain a ring isomorphic to $mathbb Z$, in which the residue of $x$ is $2$. Then the residue of $f = x^2+1$ becomes $5$. So we can also obtain $overline R$ by killing $5$ in $mathbb Z$, and therefore $overline R approx mathbb F_5$.
$endgroup$
1
$begingroup$
I like the conceptual image he provided, I wish if I heared about his book earlier
$endgroup$
– SomeOne
Dec 23 '14 at 21:06
add a comment |
$begingroup$
In this post, I am quoting the solution to a related question from Artin, with the preceding explanation (Section 11.4, p. 337-338). This is really the same as Greg Graviton's answer, but I found the different point of view and the elaborate explanation very useful. (The impatient may skip to Example 11.4.5 directly.)
Adding Relations
We reinterpret the quotient ring construction when the ideal $I$ is principal, say $I = (a)$. In this situation, we think of $overline R = R / I$ as the ring obtained by imposing the relation $a = 0$ on $R$, or by killing the element $a$. For instance, the field $mathbb F_7$ will be thought of as the ring obtained by killing $7$ in the ring $mathbb Z$ of integers.
Let's examine the collapsing that takes place in the map $pi: R to overline R$. Its kernel is the ideal $I$, so $a$ is in the kernel: $pi(a) = 0$. If $b$ is any element of $R$, the elements that have the same image in $overline R$ as $b$ are those in the coset $b + I$ and since $I = (a)$ those elements have the form $b+ra$. We see that imposing the relation $a =0$ in the ring $R$ forces us to set $b = b + ra$ for all $b$ and all $r$ in $R$, and that these are the only consequences of killing $a$.
Any number of relations $a_1 = 0, ldots, a_n = 0$ can be introduced, by working modulo the ideal $I$ generated by $a_1, ldots, a_n$, the set of linear combinations $r_1 a_1 + cdots + r_n a_n$, with coefficients $r_i$ in $R$. The quotient ring $overline R = R/I$ is viewed as the ring obtained by killing the $n$ elements. Two elements $b$ and $b'$ of $R$ have the same image in $overline R$ if and only if $b'$ has the form $b + r_1 a_1 + cdots +r_n a_n$ with $r_i$ in $R$.
The more relations we add, the more collapsing takes place in the map $pi$. If we add relations carelessly, the worst that can happen is that we may end up with $I = R$ and $overline R = 0$. All relations $a = 0$ become true when we collapse $R$ to the zero ring.
Here the Correspondence Theorem asserts something that is intuitively clear: Introducing relations one at a time or all together leads to isomorphic results. To spell this out, let $a$ and $b$ be elements of a ring $R$, and let $overline R = R / (a)$ be the result of killing $a$ in $R$. Let $overline b$ be the residue of $b$ in $overline R$. The Correspondence Theorem tells us that the principal ideal $(overline b)$ of $overline R$ corresponds to the ideal $(a,b)$ of $R$, and that $R/(a,b)$ is isomorphic to $overline R / (overline b)$. Killing $a$ and $b$ in $R$ at the same time gives the same result as killing $overline b$ in the ring $overline R$ that is obtained by killing $a$ first.
Example 11.4.5. We ask to identify the quotient ring $overline R = mathbb Z[i]/(i-2)$, the ring obtained from the Gauss integers by introducing the relation $i-2=0$. Instead of analyzing this directly, we note that the kernel of the map $mathbb Z[x] to mathbb Z[i]$ sending $x mapsto i$ is the principal ideal of $mathbb Z[x]$ generated by $f = x^2 + 1$. The First Isomorphism Theorem tells us that $mathbb Z[x]/(f) approx mathbb Z[i]$. The image of $g = x-2$ is $i-2$, so $overline R$ can also be obtained by introducing the two relations $f = 0$ and $g = 0$ into the integer polynomial ring. Let $I = (f,g)$ be the ideal of $mathbb Z[x]$ generated by the two polynomials $f$ and $g$. Then $overline R =mathbb Z[x]/I$.
To form $overline R$, we may introduce the two relations in the opposite order, first killing $g$ and then $f$. The principal ideal $(g)$ of $mathbb Z[x]$ is the kernel of the homomorphism $mathbb Z[x] to mathbb Z$ that sends $x mapsto 2$. So when we kill $x-2$ in $mathbb Z[x]$, we obtain a ring isomorphic to $mathbb Z$, in which the residue of $x$ is $2$. Then the residue of $f = x^2+1$ becomes $5$. So we can also obtain $overline R$ by killing $5$ in $mathbb Z$, and therefore $overline R approx mathbb F_5$.
$endgroup$
In this post, I am quoting the solution to a related question from Artin, with the preceding explanation (Section 11.4, p. 337-338). This is really the same as Greg Graviton's answer, but I found the different point of view and the elaborate explanation very useful. (The impatient may skip to Example 11.4.5 directly.)
Adding Relations
We reinterpret the quotient ring construction when the ideal $I$ is principal, say $I = (a)$. In this situation, we think of $overline R = R / I$ as the ring obtained by imposing the relation $a = 0$ on $R$, or by killing the element $a$. For instance, the field $mathbb F_7$ will be thought of as the ring obtained by killing $7$ in the ring $mathbb Z$ of integers.
Let's examine the collapsing that takes place in the map $pi: R to overline R$. Its kernel is the ideal $I$, so $a$ is in the kernel: $pi(a) = 0$. If $b$ is any element of $R$, the elements that have the same image in $overline R$ as $b$ are those in the coset $b + I$ and since $I = (a)$ those elements have the form $b+ra$. We see that imposing the relation $a =0$ in the ring $R$ forces us to set $b = b + ra$ for all $b$ and all $r$ in $R$, and that these are the only consequences of killing $a$.
Any number of relations $a_1 = 0, ldots, a_n = 0$ can be introduced, by working modulo the ideal $I$ generated by $a_1, ldots, a_n$, the set of linear combinations $r_1 a_1 + cdots + r_n a_n$, with coefficients $r_i$ in $R$. The quotient ring $overline R = R/I$ is viewed as the ring obtained by killing the $n$ elements. Two elements $b$ and $b'$ of $R$ have the same image in $overline R$ if and only if $b'$ has the form $b + r_1 a_1 + cdots +r_n a_n$ with $r_i$ in $R$.
The more relations we add, the more collapsing takes place in the map $pi$. If we add relations carelessly, the worst that can happen is that we may end up with $I = R$ and $overline R = 0$. All relations $a = 0$ become true when we collapse $R$ to the zero ring.
Here the Correspondence Theorem asserts something that is intuitively clear: Introducing relations one at a time or all together leads to isomorphic results. To spell this out, let $a$ and $b$ be elements of a ring $R$, and let $overline R = R / (a)$ be the result of killing $a$ in $R$. Let $overline b$ be the residue of $b$ in $overline R$. The Correspondence Theorem tells us that the principal ideal $(overline b)$ of $overline R$ corresponds to the ideal $(a,b)$ of $R$, and that $R/(a,b)$ is isomorphic to $overline R / (overline b)$. Killing $a$ and $b$ in $R$ at the same time gives the same result as killing $overline b$ in the ring $overline R$ that is obtained by killing $a$ first.
Example 11.4.5. We ask to identify the quotient ring $overline R = mathbb Z[i]/(i-2)$, the ring obtained from the Gauss integers by introducing the relation $i-2=0$. Instead of analyzing this directly, we note that the kernel of the map $mathbb Z[x] to mathbb Z[i]$ sending $x mapsto i$ is the principal ideal of $mathbb Z[x]$ generated by $f = x^2 + 1$. The First Isomorphism Theorem tells us that $mathbb Z[x]/(f) approx mathbb Z[i]$. The image of $g = x-2$ is $i-2$, so $overline R$ can also be obtained by introducing the two relations $f = 0$ and $g = 0$ into the integer polynomial ring. Let $I = (f,g)$ be the ideal of $mathbb Z[x]$ generated by the two polynomials $f$ and $g$. Then $overline R =mathbb Z[x]/I$.
To form $overline R$, we may introduce the two relations in the opposite order, first killing $g$ and then $f$. The principal ideal $(g)$ of $mathbb Z[x]$ is the kernel of the homomorphism $mathbb Z[x] to mathbb Z$ that sends $x mapsto 2$. So when we kill $x-2$ in $mathbb Z[x]$, we obtain a ring isomorphic to $mathbb Z$, in which the residue of $x$ is $2$. Then the residue of $f = x^2+1$ becomes $5$. So we can also obtain $overline R$ by killing $5$ in $mathbb Z$, and therefore $overline R approx mathbb F_5$.
edited Jun 21 '15 at 20:33
community wiki
2 revs, 2 users 97%
Srivatsan
1
$begingroup$
I like the conceptual image he provided, I wish if I heared about his book earlier
$endgroup$
– SomeOne
Dec 23 '14 at 21:06
add a comment |
1
$begingroup$
I like the conceptual image he provided, I wish if I heared about his book earlier
$endgroup$
– SomeOne
Dec 23 '14 at 21:06
1
1
$begingroup$
I like the conceptual image he provided, I wish if I heared about his book earlier
$endgroup$
– SomeOne
Dec 23 '14 at 21:06
$begingroup$
I like the conceptual image he provided, I wish if I heared about his book earlier
$endgroup$
– SomeOne
Dec 23 '14 at 21:06
add a comment |
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1
$begingroup$
$(3-i)^2=$ ? this should be a hint.
$endgroup$
– Asaf Karagila♦
Feb 23 '11 at 10:25
$begingroup$
@Asaf: Sorry i am not able to get what exactly you want to say. All i know abt $3-i$ is that its reducible in $mathbb{Z}[i]$.
$endgroup$
– anonymous
Feb 23 '11 at 10:45
$begingroup$
Hint: $3-i=(1-i)cdot(2+i)$
$endgroup$
– lhf
Feb 23 '11 at 12:04
1
$begingroup$
@Chandru1: I'm sorry, I meant $|3-i|^2=$? as a hint.
$endgroup$
– Asaf Karagila♦
Feb 23 '11 at 14:22
$begingroup$
The document math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf might be useful
$endgroup$
– Watson
Jun 19 '18 at 10:06