# Are these two quotient rings of $\Bbb Z[x]$ isomorphic?

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?

-

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.

-
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
@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
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
@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

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.

-

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.

-
@AsafKaragila I don't claim $B$ is not an integral domain. Regards, – Makoto Kato Sep 2 '12 at 23:40
@AsafKaragila So? I didn't claim $B$ is not an integral domain. Regards, – Makoto Kato Sep 2 '12 at 23:44
@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
@MakotoKato +1 For your nice answer. – user38268 Sep 2 '12 at 23:46

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.

-