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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?

4 Answers 4

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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, where $\,\rm (a+b\sqrt{-7})' = a-b\sqrt{-7}\,$ is the conjugation automorphism.

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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    @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.2012-09-03
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a) The ring $\mathbb{Z}[x]/(x^2+7)=\mathbb Z[\xi]$ is finitely generated as a $\mathbb Z$-module, since $\mathbb Z[\xi]=\mathbb Z\cdot1\oplus \mathbb Z\cdot \xi$ Thus the ring $\mathbb Z[\xi]$ integral over $\mathbb{Z}$.

b) On the other hand the ring $\mathbb{Z}[x]/(2x^2+7)=\mathbb{Z}[\eta]$ contains the element $\eta^2=-\frac {7}{2}$ which is not integral over $\mathbb Z$, since the the only numbers in $\mathbb Q$ integral over $\mathbb Z$ are the elements of $\mathbb Z$ .
So the ring $\mathbb Z[\eta]$ is not integral over $\mathbb Z$.

Hence our two rings are not isomorphic.

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    Is my answer correct too?2018-12-19
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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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Another approach, let $f:\mathbb{Z}[x]/(2x^2+7)\to \mathbb{Z}[x]/(x^2+7)$ be an isomorphism.

Now as per condition $x^2+7|f(2x^2+7)$. Let $f(x)=ax+b+(x^2+7)$.

So, $f(2x^2+7)=2(ax+b)^2+7+(x^2+7)$.

Now that implies $x^2+7|−14a^2+2b^2+7+4abx \implies 7+2b^2=14a^2$. Absurd.