12
$\begingroup$

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?

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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). $\endgroup$ – Jack Schmidt Sep 3 '12 at 1:15
  • 1
    $\begingroup$ @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. $\endgroup$ – Bill Dubuque Sep 3 '12 at 1:24
  • 1
    $\begingroup$ 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. $\endgroup$ – Jack Schmidt Sep 3 '12 at 1:29
  • $\begingroup$ I don't get it: what are $\,\alpha\,,\,\alpha'\,$ in the argument about the second ring? $\endgroup$ – DonAntonio Sep 3 '12 at 2:53
  • 1
    $\begingroup$ @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. $\endgroup$ – Bill Dubuque Sep 3 '12 at 3:09
10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is my answer correct too? $\endgroup$ – user198044 Dec 19 '18 at 12:44
5
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ @AsafKaragila I don't claim $B$ is not an integral domain. Regards, $\endgroup$ – Makoto Kato Sep 2 '12 at 23:40
  • 2
    $\begingroup$ @AsafKaragila So? I didn't claim $B$ is not an integral domain. Regards, $\endgroup$ – Makoto Kato Sep 2 '12 at 23:44
  • 2
    $\begingroup$ @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. $\endgroup$ – user38268 Sep 2 '12 at 23:45
  • 1
    $\begingroup$ @MakotoKato +1 For your nice answer. $\endgroup$ – user38268 Sep 2 '12 at 23:46
  • $\begingroup$ @GeorgesElencwajg Please help me understand this solution: since $2\beta^2 + 7 = 0$, I think $\beta$ is integral over integers (without the assumption that $A$ and $B$ are isomorphic). Second, after “$\beta^2 = -\frac 72$” how did he deduced that $-\frac 72$ in integral over integers? $\endgroup$ – Sara.T May 24 at 6:39
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.