Mixing
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$? [duplicate]
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This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
My method is
$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$
So any $a+bi$ in $Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.
I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.
abstract-algebra ring-theory ideals finite-rings gaussian-integers
asked Jan 3 at 15:21
yLcccyLccc
233
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marked as duplicate by Watson, Lord Shark the Unknown abstract-algebra
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Jan 10 at 3:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 2 more comments
$begingroup$
This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
My method is
$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$
So any $a+bi$ in $Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.
I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.
abstract-algebra ring-theory ideals finite-rings gaussian-integers
asked Jan 3 at 15:21
yLcccyLccc
233
$endgroup$
marked as duplicate by Watson, Lord Shark the Unknown abstract-algebra
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Jan 10 at 3:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Why are you unhappy with your solution? It looks perfectly reasonable to me.
$endgroup$
– Arthur
Jan 3 at 15:31
$begingroup$
I can't prove it directly, which makes me sad:(
$endgroup$
– yLccc
Jan 3 at 16:06
$begingroup$
What does "diectly" mean in this case?
$endgroup$
– Arthur
Jan 3 at 16:17
3
$begingroup$
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
$endgroup$
– Lee Mosher
Jan 3 at 20:31
1
$begingroup$
@Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
$endgroup$
– paul garrett
Jan 4 at 1:11
|
show 2 more comments
$begingroup$
This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
My method is
$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$
So any $a+bi$ in $Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.
I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.
abstract-algebra ring-theory ideals finite-rings gaussian-integers
asked Jan 3 at 15:21
yLcccyLccc
233
$endgroup$
This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
My method is
$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$
So any $a+bi$ in $Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.
I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.
This question already has an answer here:
Quotient ring of Gaussian integers
6 answers
abstract-algebra ring-theory ideals finite-rings gaussian-integers
abstract-algebra ring-theory ideals finite-rings gaussian-integers
asked Jan 3 at 15:21
yLcccyLccc
233
asked Jan 3 at 15:21
yLcccyLccc
233
edited Jan 3 at 16:40
user593746
asked Jan 3 at 15:21
yLcccyLccc
233
asked Jan 3 at 15:21
yLcccyLccc
233
asked Jan 3 at 15:21
yLcccyLccc
233
233
marked as duplicate by Watson, Lord Shark the Unknown abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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Jan 10 at 3:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Watson, Lord Shark the Unknown abstract-algebra
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Jan 10 at 3:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Why are you unhappy with your solution? It looks perfectly reasonable to me.
$endgroup$
– Arthur
Jan 3 at 15:31
$begingroup$
I can't prove it directly, which makes me sad:(
$endgroup$
– yLccc
Jan 3 at 16:06
$begingroup$
What does "diectly" mean in this case?
$endgroup$
– Arthur
Jan 3 at 16:17
3
$begingroup$
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
$endgroup$
– Lee Mosher
Jan 3 at 20:31
1
$begingroup$
@Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
$endgroup$
– paul garrett
Jan 4 at 1:11
|
show 2 more comments
$begingroup$
Why are you unhappy with your solution? It looks perfectly reasonable to me.
$endgroup$
– Arthur
Jan 3 at 15:31
$begingroup$
I can't prove it directly, which makes me sad:(
$endgroup$
– yLccc
Jan 3 at 16:06
$begingroup$
What does "diectly" mean in this case?
$endgroup$
– Arthur
Jan 3 at 16:17
3
$begingroup$
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
$endgroup$
– Lee Mosher
Jan 3 at 20:31
1
$begingroup$
@Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
$endgroup$
– paul garrett
Jan 4 at 1:11
$begingroup$
Why are you unhappy with your solution? It looks perfectly reasonable to me.
$endgroup$
– Arthur
Jan 3 at 15:31
$begingroup$
Why are you unhappy with your solution? It looks perfectly reasonable to me.
$endgroup$
– Arthur
Jan 3 at 15:31
$begingroup$
I can't prove it directly, which makes me sad:(
$endgroup$
– yLccc
Jan 3 at 16:06
$begingroup$
I can't prove it directly, which makes me sad:(
$endgroup$
– yLccc
Jan 3 at 16:06
$begingroup$
What does "diectly" mean in this case?
$endgroup$
– Arthur
Jan 3 at 16:17
$begingroup$
What does "diectly" mean in this case?
$endgroup$
– Arthur
Jan 3 at 16:17
3
3
$begingroup$
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
$endgroup$
– Lee Mosher
Jan 3 at 20:31
$begingroup$
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
$endgroup$
– Lee Mosher
Jan 3 at 20:31
1
1
$begingroup$
@Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
$endgroup$
– paul garrett
Jan 4 at 1:11
$begingroup$
@Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
$endgroup$
– paul garrett
Jan 4 at 1:11
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
$endgroup$
$begingroup$
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
$endgroup$
– yLccc
Jan 3 at 16:55
1
$begingroup$
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
$endgroup$
– user593746
Jan 3 at 20:00
1
$begingroup$
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
$endgroup$
– user593746
Jan 3 at 20:00
add a comment |
$begingroup$
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
answered Jan 3 at 22:02
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
answered Jan 3 at 18:38
Chris CusterChris Custer
11.3k3824
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add a comment |
$begingroup$
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
answered Jan 3 at 15:41
Mark BennetMark Bennet
80.8k981179
$endgroup$
$begingroup$
So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
$endgroup$
– yLccc
Jan 3 at 16:05
$begingroup$
@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
$endgroup$
– Mark Bennet
Jan 3 at 16:09
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
$endgroup$
$begingroup$
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
$endgroup$
– yLccc
Jan 3 at 16:55
1
$begingroup$
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
$endgroup$
– user593746
Jan 3 at 20:00
1
$begingroup$
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
$endgroup$
– user593746
Jan 3 at 20:00
add a comment |
$begingroup$
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
$endgroup$
$begingroup$
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
$endgroup$
– yLccc
Jan 3 at 16:55
1
$begingroup$
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
$endgroup$
– user593746
Jan 3 at 20:00
1
$begingroup$
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
$endgroup$
– user593746
Jan 3 at 20:00
add a comment |
$begingroup$
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
$endgroup$
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
edited Jan 3 at 20:02
answered Jan 3 at 16:25
user593746
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Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
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– yLccc
Jan 3 at 16:55
1
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Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
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– user593746
Jan 3 at 20:00
1
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In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
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– user593746
Jan 3 at 20:00
add a comment |
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Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
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– yLccc
Jan 3 at 16:55
1
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Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
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– user593746
Jan 3 at 20:00
1
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In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
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– user593746
Jan 3 at 20:00
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Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
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– yLccc
Jan 3 at 16:55
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Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
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– yLccc
Jan 3 at 16:55
1
1
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Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
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– user593746
Jan 3 at 20:00
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Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
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– user593746
Jan 3 at 20:00
1
1
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In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
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– user593746
Jan 3 at 20:00
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In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
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– user593746
Jan 3 at 20:00
add a comment |
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A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
answered Jan 3 at 22:02
egregegreg
180k1485202
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add a comment |
$begingroup$
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
answered Jan 3 at 22:02
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
answered Jan 3 at 22:02
egregegreg
180k1485202
$endgroup$
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
answered Jan 3 at 22:02
egregegreg
180k1485202
answered Jan 3 at 22:02
egregegreg
180k1485202
answered Jan 3 at 22:02
egregegreg
180k1485202
answered Jan 3 at 22:02
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
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Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
answered Jan 3 at 18:38
Chris CusterChris Custer
11.3k3824
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add a comment |
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Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
answered Jan 3 at 18:38
Chris CusterChris Custer
11.3k3824
$endgroup$
add a comment |
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Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
answered Jan 3 at 18:38
Chris CusterChris Custer
11.3k3824
$endgroup$
Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
answered Jan 3 at 18:38
Chris CusterChris Custer
11.3k3824
edited Jan 4 at 1:07
answered Jan 3 at 18:38
Chris CusterChris Custer
11.3k3824
answered Jan 3 at 18:38
Chris CusterChris Custer
11.3k3824
answered Jan 3 at 18:38
Chris CusterChris Custer
11.3k3824
11.3k3824
add a comment |
add a comment |
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If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
answered Jan 3 at 15:41
Mark BennetMark Bennet
80.8k981179
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So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
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– yLccc
Jan 3 at 16:05
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@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
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– Mark Bennet
Jan 3 at 16:09
add a comment |
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If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
answered Jan 3 at 15:41
Mark BennetMark Bennet
80.8k981179
$endgroup$
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So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
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– yLccc
Jan 3 at 16:05
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@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
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– Mark Bennet
Jan 3 at 16:09
add a comment |
$begingroup$
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
answered Jan 3 at 15:41
Mark BennetMark Bennet
80.8k981179
$endgroup$
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
answered Jan 3 at 15:41
Mark BennetMark Bennet
80.8k981179
answered Jan 3 at 15:41
Mark BennetMark Bennet
80.8k981179
answered Jan 3 at 15:41
Mark BennetMark Bennet
80.8k981179
answered Jan 3 at 15:41
Mark BennetMark Bennet
80.8k981179
80.8k981179
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So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
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– yLccc
Jan 3 at 16:05
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@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
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– Mark Bennet
Jan 3 at 16:09
add a comment |
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So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
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– yLccc
Jan 3 at 16:05
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@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
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– Mark Bennet
Jan 3 at 16:09
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So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
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– yLccc
Jan 3 at 16:05
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So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
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– yLccc
Jan 3 at 16:05
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@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
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– Mark Bennet
Jan 3 at 16:09
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@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
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– Mark Bennet
Jan 3 at 16:09
add a comment |
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Why are you unhappy with your solution? It looks perfectly reasonable to me.
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– Arthur
Jan 3 at 15:31
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I can't prove it directly, which makes me sad:(
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– yLccc
Jan 3 at 16:06
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What does "diectly" mean in this case?
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– Arthur
Jan 3 at 16:17
3
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$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
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– Lee Mosher
Jan 3 at 20:31
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@Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
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– paul garrett
Jan 4 at 1:11