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Are the rings $\mathbb{Z}[x]/(x^2+7)$ and $\mathbb{Z}[x]/(2x^2+7)$ isomorphic?

Attempted Solution:

My guess is that they are not isomorphic. I am having trouble demonstrating this. Any hints, as to how i should approach this?

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3 Answers 3

The ring $\mathbb{Z}[x]/(x^2+7)$ is free of rank $2$ over $\mathbb{Z}$ with basis $(1,\bar x)$ (by euclidean division), and thus integral over $\mathbb{Z}$ (i.e. all its elements are integral over $\mathbb{Z}$).

On the other hand the ring $\mathbb{Z}[x]/(2x^2+7)=\mathbb{Z}[\xi]$ contains the element $\xi^2=-\frac {7}{2}$ which is not integral over $\mathbb Z$.

Hence these two rings are not isomorphic.

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Hint $\ \ 2\:$ is invertible in $\rm\ \Bbb{Z}[x]/(2x^2\!+7),\:$ but not in $\rm\, \Bbb{Z}[x]/(x^2\!+7)\,\cong\, \Bbb Z[\sqrt{-7}].\:$ Indeed, in the first ring $\rm\:2(x^2\!+4) = 1.\:$ In the second, $\rm\:2\alpha = 1\:\Rightarrow\:2\alpha'=1\:$ $\Rightarrow$ $\rm\:4\alpha\alpha' = 1,\:$ $\rm\: \alpha \alpha'\in \Bbb Z,\:$ contradiction.

Remark $\ $ The proof is accessible at high-school level by eliminating use of the conjugation automorphism in $\rm\,R \cong \Bbb Z[\sqrt{-7}].\:$ If $\,2\,$ is invertible in $\rm\,R\,$ then $\rm\:2\,(a\!+\!b\sqrt{-7})= 1,\,$ for $\rm\, a,b\in \Bbb Z.\:$ Therefore $\rm\:b\ne 0\ $ (else $\rm\:2a=1,\ a\in\Bbb Z)\ $ hence $\rm\,\sqrt{-7}\, =\, (1\!-\!2a)/(2b)\in \Bbb Q,\,$ contradiction.

This elementary approach may be helpful for readers not familiar with the more advanced techniques applied in the other answers.

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Another low tech solution looks at the rings mod 3 (any ring isomorphism will take 3 to 3, so the quotient mod 3 still makes sense up to isomorphism; one quotient ring is a field, the other is a direct product of two fields). –  Jack Schmidt Sep 3 '12 at 1:15
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@Jack I was tempted to post the similar argument mod $2$, i.e $\,\Bbb Z[\sqrt{-7}]\,$ is a ring admitting parity, but the other ring is not. But I thought the above would be more accessible. –  Bill Dubuque Sep 3 '12 at 1:24
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I think your posted answer is a much better way to describe what happens mod 2 (rather than quotienting out by the whole ring)! Your parity answer is a good read (elementary and natural; shouldn't everyone want to know how "even" generalizes?). Mod 3 isn't worth an answer: it is dull and straightforward, but I think it is a common situation. –  Jack Schmidt Sep 3 '12 at 1:29
    
I don't get it: what are $\,\alpha\,,\,\alpha'\,$ in the argument about the second ring? –  DonAntonio Sep 3 '12 at 2:53
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@Don $\rm\ \alpha = a+b\,\sqrt{-7}\:$ is an element of $\rm\,R\,\cong\,\Bbb Z[\sqrt{-7}]\,$ and $\rm\,\alpha'= a-b\,\sqrt{-7}\:$ is its conjugate. –  Bill Dubuque Sep 3 '12 at 3:09

Let $A = \mathbb{Z}[x]/(x^2+7)$

Let $B = \mathbb{Z}[x]/(2x^2+7)$

$A$ and $B$ are clearly integral domains. Let $\alpha$ be the image of $x$ by the canonical homomorphism $\mathbb{Z}[x] \rightarrow A$. Let $\beta$ be the image of $x$ by the canonical homomorphism $\mathbb{Z}[x] \rightarrow B$. Since $\alpha$ is integral over $\mathbb{Z}$, $A$ is integral over $\mathbb{Z}$.

Suppose $A$ and $B$ are isomorphic. Then $\beta$ is integral over $\mathbb{Z}$. Let $K$ be the field of fractions of $B$. Since $2\beta^2+7 = 0$, $\beta^2 = -\frac{7}{2}$ in $K$. Hence $-\frac{7}{2}$ must be integral over $\mathbb{Z}$. This is a contradiction.

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@AsafKaragila I don't claim $B$ is not an integral domain. Regards, –  Makoto Kato Sep 2 '12 at 23:40
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@AsafKaragila So? I didn't claim $B$ is not an integral domain. Regards, –  Makoto Kato Sep 2 '12 at 23:44
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@AsafKaragila I think he claimed that $B$ is an integral domain, but where does he then proceed to claim it is not? I think his contradiction at the end is that we have an element $-7/2$ being in $\Bbb{Q}$ that is integral over $\Bbb{Z}$ and since $\Bbb{Z}$ is integrally closed in its field of fractions, we must necessarily have that $-7/2$ be an integer, a contradiction. –  user38268 Sep 2 '12 at 23:45
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@MakotoKato +1 For your nice answer. –  user38268 Sep 2 '12 at 23:46

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