Mixing
Need help to construct a ring isomorphism from $mathbb{Z}_3[sqrt2]$ $rightarrow $ $mathbb{Z}_3[i]$
$begingroup$
I've looked at this problem for some time now and am confused as to where to start. I know that to prove these rings are isomorphic I must to construct a bijective function $phi$ such that:
$phi(a + b)$ = $phi(a) + phi(b)$
and
$phi(ab)$ = $phi(a)phi(b)$.
I'm not exactly sure where to begin. If anyone could give some insight on where to begin or something to think about I would really appreciate it. Thanks!
abstract-algebra ring-theory ring-isomorphism
asked Jan 21 at 20:10
NeilsonsMilkNeilsonsMilk
404
$endgroup$
|
show 2 more comments
$begingroup$
I've looked at this problem for some time now and am confused as to where to start. I know that to prove these rings are isomorphic I must to construct a bijective function $phi$ such that:
$phi(a + b)$ = $phi(a) + phi(b)$
and
$phi(ab)$ = $phi(a)phi(b)$.
I'm not exactly sure where to begin. If anyone could give some insight on where to begin or something to think about I would really appreciate it. Thanks!
abstract-algebra ring-theory ring-isomorphism
asked Jan 21 at 20:10
NeilsonsMilkNeilsonsMilk
404
$endgroup$
3
$begingroup$
Hint: think about what $phi(sqrt{2})$ can be
$endgroup$
– pwerth
Jan 21 at 20:12
2
$begingroup$
How about $sqrt{2} to i$ as a starter.
$endgroup$
– Anurag A
Jan 21 at 20:12
2
$begingroup$
What od s$sqrt 2^2$ and what is $i^2$ and what is the difference between these, if any?
$endgroup$
– Hagen von Eitzen
Jan 21 at 20:22
2
$begingroup$
Hint: $bmod 3!: ,2 equiv-1 $ so $ sqrt 2 equiv sqrt {-1} $
$endgroup$
– Bill Dubuque
Jan 21 at 20:24
$begingroup$
Be careful of ambiguous notations: for you, $Bbb Z_3$ is evidently $Bbb Z/3Bbb Z=Bbb F_3$; for many others, $Bbb Z_3$ is the integers in the three-adic field $Bbb Q_3$.
$endgroup$
– Lubin
Jan 22 at 1:09
|
show 2 more comments
$begingroup$
I've looked at this problem for some time now and am confused as to where to start. I know that to prove these rings are isomorphic I must to construct a bijective function $phi$ such that:
$phi(a + b)$ = $phi(a) + phi(b)$
and
$phi(ab)$ = $phi(a)phi(b)$.
I'm not exactly sure where to begin. If anyone could give some insight on where to begin or something to think about I would really appreciate it. Thanks!
abstract-algebra ring-theory ring-isomorphism
asked Jan 21 at 20:10
NeilsonsMilkNeilsonsMilk
404
$endgroup$
I've looked at this problem for some time now and am confused as to where to start. I know that to prove these rings are isomorphic I must to construct a bijective function $phi$ such that:
$phi(a + b)$ = $phi(a) + phi(b)$
and
$phi(ab)$ = $phi(a)phi(b)$.
I'm not exactly sure where to begin. If anyone could give some insight on where to begin or something to think about I would really appreciate it. Thanks!
abstract-algebra ring-theory ring-isomorphism
abstract-algebra ring-theory ring-isomorphism
asked Jan 21 at 20:10
NeilsonsMilkNeilsonsMilk
404
asked Jan 21 at 20:10
NeilsonsMilkNeilsonsMilk
404
asked Jan 21 at 20:10
NeilsonsMilkNeilsonsMilk
404
asked Jan 21 at 20:10
NeilsonsMilkNeilsonsMilk
404
asked Jan 21 at 20:10
NeilsonsMilkNeilsonsMilk
404
404
3
$begingroup$
Hint: think about what $phi(sqrt{2})$ can be
$endgroup$
– pwerth
Jan 21 at 20:12
2
$begingroup$
How about $sqrt{2} to i$ as a starter.
$endgroup$
– Anurag A
Jan 21 at 20:12
2
$begingroup$
What od s$sqrt 2^2$ and what is $i^2$ and what is the difference between these, if any?
$endgroup$
– Hagen von Eitzen
Jan 21 at 20:22
2
$begingroup$
Hint: $bmod 3!: ,2 equiv-1 $ so $ sqrt 2 equiv sqrt {-1} $
$endgroup$
– Bill Dubuque
Jan 21 at 20:24
$begingroup$
Be careful of ambiguous notations: for you, $Bbb Z_3$ is evidently $Bbb Z/3Bbb Z=Bbb F_3$; for many others, $Bbb Z_3$ is the integers in the three-adic field $Bbb Q_3$.
$endgroup$
– Lubin
Jan 22 at 1:09
|
show 2 more comments
3
$begingroup$
Hint: think about what $phi(sqrt{2})$ can be
$endgroup$
– pwerth
Jan 21 at 20:12
2
$begingroup$
How about $sqrt{2} to i$ as a starter.
$endgroup$
– Anurag A
Jan 21 at 20:12
2
$begingroup$
What od s$sqrt 2^2$ and what is $i^2$ and what is the difference between these, if any?
$endgroup$
– Hagen von Eitzen
Jan 21 at 20:22
2
$begingroup$
Hint: $bmod 3!: ,2 equiv-1 $ so $ sqrt 2 equiv sqrt {-1} $
$endgroup$
– Bill Dubuque
Jan 21 at 20:24
$begingroup$
Be careful of ambiguous notations: for you, $Bbb Z_3$ is evidently $Bbb Z/3Bbb Z=Bbb F_3$; for many others, $Bbb Z_3$ is the integers in the three-adic field $Bbb Q_3$.
$endgroup$
– Lubin
Jan 22 at 1:09
3
3
$begingroup$
Hint: think about what $phi(sqrt{2})$ can be
$endgroup$
– pwerth
Jan 21 at 20:12
$begingroup$
Hint: think about what $phi(sqrt{2})$ can be
$endgroup$
– pwerth
Jan 21 at 20:12
2
2
$begingroup$
How about $sqrt{2} to i$ as a starter.
$endgroup$
– Anurag A
Jan 21 at 20:12
$begingroup$
How about $sqrt{2} to i$ as a starter.
$endgroup$
– Anurag A
Jan 21 at 20:12
2
2
$begingroup$
What od s$sqrt 2^2$ and what is $i^2$ and what is the difference between these, if any?
$endgroup$
– Hagen von Eitzen
Jan 21 at 20:22
$begingroup$
What od s$sqrt 2^2$ and what is $i^2$ and what is the difference between these, if any?
$endgroup$
– Hagen von Eitzen
Jan 21 at 20:22
2
2
$begingroup$
Hint: $bmod 3!: ,2 equiv-1 $ so $ sqrt 2 equiv sqrt {-1} $
$endgroup$
– Bill Dubuque
Jan 21 at 20:24
$begingroup$
Hint: $bmod 3!: ,2 equiv-1 $ so $ sqrt 2 equiv sqrt {-1} $
$endgroup$
– Bill Dubuque
Jan 21 at 20:24
$begingroup$
Be careful of ambiguous notations: for you, $Bbb Z_3$ is evidently $Bbb Z/3Bbb Z=Bbb F_3$; for many others, $Bbb Z_3$ is the integers in the three-adic field $Bbb Q_3$.
$endgroup$
– Lubin
Jan 22 at 1:09
$begingroup$
Be careful of ambiguous notations: for you, $Bbb Z_3$ is evidently $Bbb Z/3Bbb Z=Bbb F_3$; for many others, $Bbb Z_3$ is the integers in the three-adic field $Bbb Q_3$.
$endgroup$
– Lubin
Jan 22 at 1:09
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Each ring has nine elements. Clearly you need to send $0$ to $0$ and $1$ to $1$, which forces you to send $2$ to $2$ to maintain addition. Now $sqrt 2$ has to go somewhere. As $sqrt 2 cdot sqrt 2=2$ you need it to go somewhere that squares to $2=-1$. Addition and multiplication will then fill out the function.
answered Jan 21 at 20:17
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
So would a potential answer look like this? Define $phi(sqrt2) = i$? I'm confused as to how I should be writing this down. Thank you for your answer by the way.
$endgroup$
– NeilsonsMilk
Jan 21 at 20:23
1
$begingroup$
I would list the entire function $phi$, which you can derive from $phi(sqrt 2)=i$ and include some description of how I got it and ruled out other possibilities. For example, another possibility is that $phi(sqrt 2)=1+i$, but then you would have to have $phi(2)=(1+i)^2=2i$ which contradicts our previous discovery that $phi(2)=2$. You need to do this to show the isomorphism is unique. It depends on what your teacher expects.
$endgroup$
– Ross Millikan
Jan 22 at 3:26
add a comment |
$begingroup$
There is nothing to prove at all. Here $sqrt{2}$ can only be interpreted as an element $u$ in some field extension of $mathbb{Z}_3$ such that $u^2=2$; similarly, $i$ is an element in some field extension such that $i^2=-1=2$. Since both elements are roots of $x^2-2=0$, which is irreducible over $mathbb{Z}_3$, the two fields are obviously isomorphic and an isomorphism is obtained by
$$
a+bsqrt{2}mapsto a+bi
$$
If you use a common extension field, such as the algebraic closure of $mathbb{Z}_3$, then the two fields are actually equal, because $sqrt{2}=i$ or $sqrt{2}=-i$, both being roots of the same polynomial.
The complex numbers $sqrt{2}$ and $i$ have nothing to do with the problem at hand.
answered Jan 21 at 23:22
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
ADDED: It occurred to me that things might have a more pleasant look if I used separate variables, one quotient $F[x] / (x^2 + 1)$ and the other $F[t] / (t^2 -2),$ where $F = mathbb Z / 3 mathbb Z.$ Then the isomorphism has a concrete appearance,
$$ ax+b mapsto at+b $$
ORIGINAL:Well, $x^2 + 1 equiv x^2 - 2 pmod 3,$ which is promising.
Writing as $F[x] / (x^2 + n)$ for some $n$ we are writing elements as $ax+b$ with $a,b in F = mathbb Z / 3 mathbb Z.$
With $x^2 + 1,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd-ac) $$
With $x^2 -2,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd+2ac) $$
These look slightly different, but in $a,b,c,d in F,$ we always have
$$ -ac = 2ac $$
because
$$ -ac equiv 2ac pmod 3 $$
answered Jan 21 at 20:34
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3082345%2fneed-help-to-construct-a-ring-isomorphism-from-mathbbz-3-sqrt2-rightarr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Each ring has nine elements. Clearly you need to send $0$ to $0$ and $1$ to $1$, which forces you to send $2$ to $2$ to maintain addition. Now $sqrt 2$ has to go somewhere. As $sqrt 2 cdot sqrt 2=2$ you need it to go somewhere that squares to $2=-1$. Addition and multiplication will then fill out the function.
answered Jan 21 at 20:17
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
So would a potential answer look like this? Define $phi(sqrt2) = i$? I'm confused as to how I should be writing this down. Thank you for your answer by the way.
$endgroup$
– NeilsonsMilk
Jan 21 at 20:23
1
$begingroup$
I would list the entire function $phi$, which you can derive from $phi(sqrt 2)=i$ and include some description of how I got it and ruled out other possibilities. For example, another possibility is that $phi(sqrt 2)=1+i$, but then you would have to have $phi(2)=(1+i)^2=2i$ which contradicts our previous discovery that $phi(2)=2$. You need to do this to show the isomorphism is unique. It depends on what your teacher expects.
$endgroup$
– Ross Millikan
Jan 22 at 3:26
add a comment |
$begingroup$
Each ring has nine elements. Clearly you need to send $0$ to $0$ and $1$ to $1$, which forces you to send $2$ to $2$ to maintain addition. Now $sqrt 2$ has to go somewhere. As $sqrt 2 cdot sqrt 2=2$ you need it to go somewhere that squares to $2=-1$. Addition and multiplication will then fill out the function.
answered Jan 21 at 20:17
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
So would a potential answer look like this? Define $phi(sqrt2) = i$? I'm confused as to how I should be writing this down. Thank you for your answer by the way.
$endgroup$
– NeilsonsMilk
Jan 21 at 20:23
1
$begingroup$
I would list the entire function $phi$, which you can derive from $phi(sqrt 2)=i$ and include some description of how I got it and ruled out other possibilities. For example, another possibility is that $phi(sqrt 2)=1+i$, but then you would have to have $phi(2)=(1+i)^2=2i$ which contradicts our previous discovery that $phi(2)=2$. You need to do this to show the isomorphism is unique. It depends on what your teacher expects.
$endgroup$
– Ross Millikan
Jan 22 at 3:26
add a comment |
$begingroup$
Each ring has nine elements. Clearly you need to send $0$ to $0$ and $1$ to $1$, which forces you to send $2$ to $2$ to maintain addition. Now $sqrt 2$ has to go somewhere. As $sqrt 2 cdot sqrt 2=2$ you need it to go somewhere that squares to $2=-1$. Addition and multiplication will then fill out the function.
answered Jan 21 at 20:17
Ross MillikanRoss Millikan
299k24200374
$endgroup$
Each ring has nine elements. Clearly you need to send $0$ to $0$ and $1$ to $1$, which forces you to send $2$ to $2$ to maintain addition. Now $sqrt 2$ has to go somewhere. As $sqrt 2 cdot sqrt 2=2$ you need it to go somewhere that squares to $2=-1$. Addition and multiplication will then fill out the function.
answered Jan 21 at 20:17
Ross MillikanRoss Millikan
299k24200374
answered Jan 21 at 20:17
Ross MillikanRoss Millikan
299k24200374
answered Jan 21 at 20:17
Ross MillikanRoss Millikan
299k24200374
answered Jan 21 at 20:17
Ross MillikanRoss Millikan
299k24200374
299k24200374
$begingroup$
So would a potential answer look like this? Define $phi(sqrt2) = i$? I'm confused as to how I should be writing this down. Thank you for your answer by the way.
$endgroup$
– NeilsonsMilk
Jan 21 at 20:23
1
$begingroup$
I would list the entire function $phi$, which you can derive from $phi(sqrt 2)=i$ and include some description of how I got it and ruled out other possibilities. For example, another possibility is that $phi(sqrt 2)=1+i$, but then you would have to have $phi(2)=(1+i)^2=2i$ which contradicts our previous discovery that $phi(2)=2$. You need to do this to show the isomorphism is unique. It depends on what your teacher expects.
$endgroup$
– Ross Millikan
Jan 22 at 3:26
add a comment |
$begingroup$
So would a potential answer look like this? Define $phi(sqrt2) = i$? I'm confused as to how I should be writing this down. Thank you for your answer by the way.
$endgroup$
– NeilsonsMilk
Jan 21 at 20:23
1
$begingroup$
I would list the entire function $phi$, which you can derive from $phi(sqrt 2)=i$ and include some description of how I got it and ruled out other possibilities. For example, another possibility is that $phi(sqrt 2)=1+i$, but then you would have to have $phi(2)=(1+i)^2=2i$ which contradicts our previous discovery that $phi(2)=2$. You need to do this to show the isomorphism is unique. It depends on what your teacher expects.
$endgroup$
– Ross Millikan
Jan 22 at 3:26
$begingroup$
So would a potential answer look like this? Define $phi(sqrt2) = i$? I'm confused as to how I should be writing this down. Thank you for your answer by the way.
$endgroup$
– NeilsonsMilk
Jan 21 at 20:23
$begingroup$
So would a potential answer look like this? Define $phi(sqrt2) = i$? I'm confused as to how I should be writing this down. Thank you for your answer by the way.
$endgroup$
– NeilsonsMilk
Jan 21 at 20:23
1
1
$begingroup$
I would list the entire function $phi$, which you can derive from $phi(sqrt 2)=i$ and include some description of how I got it and ruled out other possibilities. For example, another possibility is that $phi(sqrt 2)=1+i$, but then you would have to have $phi(2)=(1+i)^2=2i$ which contradicts our previous discovery that $phi(2)=2$. You need to do this to show the isomorphism is unique. It depends on what your teacher expects.
$endgroup$
– Ross Millikan
Jan 22 at 3:26
$begingroup$
I would list the entire function $phi$, which you can derive from $phi(sqrt 2)=i$ and include some description of how I got it and ruled out other possibilities. For example, another possibility is that $phi(sqrt 2)=1+i$, but then you would have to have $phi(2)=(1+i)^2=2i$ which contradicts our previous discovery that $phi(2)=2$. You need to do this to show the isomorphism is unique. It depends on what your teacher expects.
$endgroup$
– Ross Millikan
Jan 22 at 3:26
add a comment |
$begingroup$
There is nothing to prove at all. Here $sqrt{2}$ can only be interpreted as an element $u$ in some field extension of $mathbb{Z}_3$ such that $u^2=2$; similarly, $i$ is an element in some field extension such that $i^2=-1=2$. Since both elements are roots of $x^2-2=0$, which is irreducible over $mathbb{Z}_3$, the two fields are obviously isomorphic and an isomorphism is obtained by
$$
a+bsqrt{2}mapsto a+bi
$$
If you use a common extension field, such as the algebraic closure of $mathbb{Z}_3$, then the two fields are actually equal, because $sqrt{2}=i$ or $sqrt{2}=-i$, both being roots of the same polynomial.
The complex numbers $sqrt{2}$ and $i$ have nothing to do with the problem at hand.
answered Jan 21 at 23:22
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
There is nothing to prove at all. Here $sqrt{2}$ can only be interpreted as an element $u$ in some field extension of $mathbb{Z}_3$ such that $u^2=2$; similarly, $i$ is an element in some field extension such that $i^2=-1=2$. Since both elements are roots of $x^2-2=0$, which is irreducible over $mathbb{Z}_3$, the two fields are obviously isomorphic and an isomorphism is obtained by
$$
a+bsqrt{2}mapsto a+bi
$$
If you use a common extension field, such as the algebraic closure of $mathbb{Z}_3$, then the two fields are actually equal, because $sqrt{2}=i$ or $sqrt{2}=-i$, both being roots of the same polynomial.
The complex numbers $sqrt{2}$ and $i$ have nothing to do with the problem at hand.
answered Jan 21 at 23:22
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
There is nothing to prove at all. Here $sqrt{2}$ can only be interpreted as an element $u$ in some field extension of $mathbb{Z}_3$ such that $u^2=2$; similarly, $i$ is an element in some field extension such that $i^2=-1=2$. Since both elements are roots of $x^2-2=0$, which is irreducible over $mathbb{Z}_3$, the two fields are obviously isomorphic and an isomorphism is obtained by
$$
a+bsqrt{2}mapsto a+bi
$$
If you use a common extension field, such as the algebraic closure of $mathbb{Z}_3$, then the two fields are actually equal, because $sqrt{2}=i$ or $sqrt{2}=-i$, both being roots of the same polynomial.
The complex numbers $sqrt{2}$ and $i$ have nothing to do with the problem at hand.
answered Jan 21 at 23:22
egregegreg
183k1486205
$endgroup$
There is nothing to prove at all. Here $sqrt{2}$ can only be interpreted as an element $u$ in some field extension of $mathbb{Z}_3$ such that $u^2=2$; similarly, $i$ is an element in some field extension such that $i^2=-1=2$. Since both elements are roots of $x^2-2=0$, which is irreducible over $mathbb{Z}_3$, the two fields are obviously isomorphic and an isomorphism is obtained by
$$
a+bsqrt{2}mapsto a+bi
$$
If you use a common extension field, such as the algebraic closure of $mathbb{Z}_3$, then the two fields are actually equal, because $sqrt{2}=i$ or $sqrt{2}=-i$, both being roots of the same polynomial.
The complex numbers $sqrt{2}$ and $i$ have nothing to do with the problem at hand.
answered Jan 21 at 23:22
egregegreg
183k1486205
answered Jan 21 at 23:22
egregegreg
183k1486205
answered Jan 21 at 23:22
egregegreg
183k1486205
answered Jan 21 at 23:22
egregegreg
183k1486205
183k1486205
add a comment |
add a comment |
$begingroup$
ADDED: It occurred to me that things might have a more pleasant look if I used separate variables, one quotient $F[x] / (x^2 + 1)$ and the other $F[t] / (t^2 -2),$ where $F = mathbb Z / 3 mathbb Z.$ Then the isomorphism has a concrete appearance,
$$ ax+b mapsto at+b $$
ORIGINAL:Well, $x^2 + 1 equiv x^2 - 2 pmod 3,$ which is promising.
Writing as $F[x] / (x^2 + n)$ for some $n$ we are writing elements as $ax+b$ with $a,b in F = mathbb Z / 3 mathbb Z.$
With $x^2 + 1,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd-ac) $$
With $x^2 -2,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd+2ac) $$
These look slightly different, but in $a,b,c,d in F,$ we always have
$$ -ac = 2ac $$
because
$$ -ac equiv 2ac pmod 3 $$
answered Jan 21 at 20:34
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
ADDED: It occurred to me that things might have a more pleasant look if I used separate variables, one quotient $F[x] / (x^2 + 1)$ and the other $F[t] / (t^2 -2),$ where $F = mathbb Z / 3 mathbb Z.$ Then the isomorphism has a concrete appearance,
$$ ax+b mapsto at+b $$
ORIGINAL:Well, $x^2 + 1 equiv x^2 - 2 pmod 3,$ which is promising.
Writing as $F[x] / (x^2 + n)$ for some $n$ we are writing elements as $ax+b$ with $a,b in F = mathbb Z / 3 mathbb Z.$
With $x^2 + 1,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd-ac) $$
With $x^2 -2,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd+2ac) $$
These look slightly different, but in $a,b,c,d in F,$ we always have
$$ -ac = 2ac $$
because
$$ -ac equiv 2ac pmod 3 $$
answered Jan 21 at 20:34
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
ADDED: It occurred to me that things might have a more pleasant look if I used separate variables, one quotient $F[x] / (x^2 + 1)$ and the other $F[t] / (t^2 -2),$ where $F = mathbb Z / 3 mathbb Z.$ Then the isomorphism has a concrete appearance,
$$ ax+b mapsto at+b $$
ORIGINAL:Well, $x^2 + 1 equiv x^2 - 2 pmod 3,$ which is promising.
Writing as $F[x] / (x^2 + n)$ for some $n$ we are writing elements as $ax+b$ with $a,b in F = mathbb Z / 3 mathbb Z.$
With $x^2 + 1,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd-ac) $$
With $x^2 -2,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd+2ac) $$
These look slightly different, but in $a,b,c,d in F,$ we always have
$$ -ac = 2ac $$
because
$$ -ac equiv 2ac pmod 3 $$
answered Jan 21 at 20:34
Will JagyWill Jagy
104k5102201
$endgroup$
ADDED: It occurred to me that things might have a more pleasant look if I used separate variables, one quotient $F[x] / (x^2 + 1)$ and the other $F[t] / (t^2 -2),$ where $F = mathbb Z / 3 mathbb Z.$ Then the isomorphism has a concrete appearance,
$$ ax+b mapsto at+b $$
ORIGINAL:Well, $x^2 + 1 equiv x^2 - 2 pmod 3,$ which is promising.
Writing as $F[x] / (x^2 + n)$ for some $n$ we are writing elements as $ax+b$ with $a,b in F = mathbb Z / 3 mathbb Z.$
With $x^2 + 1,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd-ac) $$
With $x^2 -2,$ we get multiplication
$$ (ax+b)(cx+d) = (ad+bc)x + (bd+2ac) $$
These look slightly different, but in $a,b,c,d in F,$ we always have
$$ -ac = 2ac $$
because
$$ -ac equiv 2ac pmod 3 $$
answered Jan 21 at 20:34
Will JagyWill Jagy
104k5102201
edited Jan 21 at 23:09
answered Jan 21 at 20:34
Will JagyWill Jagy
104k5102201
answered Jan 21 at 20:34
Will JagyWill Jagy
104k5102201
answered Jan 21 at 20:34
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3082345%2fneed-help-to-construct-a-ring-isomorphism-from-mathbbz-3-sqrt2-rightarr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
Hint: think about what $phi(sqrt{2})$ can be
$endgroup$
– pwerth
Jan 21 at 20:12
2
$begingroup$
How about $sqrt{2} to i$ as a starter.
$endgroup$
– Anurag A
Jan 21 at 20:12
2
$begingroup$
What od s$sqrt 2^2$ and what is $i^2$ and what is the difference between these, if any?
$endgroup$
– Hagen von Eitzen
Jan 21 at 20:22
2
$begingroup$
Hint: $bmod 3!: ,2 equiv-1 $ so $ sqrt 2 equiv sqrt {-1} $
$endgroup$
– Bill Dubuque
Jan 21 at 20:24
$begingroup$
Be careful of ambiguous notations: for you, $Bbb Z_3$ is evidently $Bbb Z/3Bbb Z=Bbb F_3$; for many others, $Bbb Z_3$ is the integers in the three-adic field $Bbb Q_3$.
$endgroup$
– Lubin
Jan 22 at 1:09