12

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?

MJD
  • 65,394
  • 39
  • 298
  • 580
Alexander
  • 157

5 Answers5

14

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.

Bill Dubuque
  • 272,048
  • 2
    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 03 '12 at 01:15
  • 1
    @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 03 '12 at 01:24
  • 1
    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 03 '12 at 01:29
  • I don't get it: what are $,\alpha,,,\alpha',$ in the argument about the second ring? – DonAntonio Sep 03 '12 at 02:53
  • 1
    @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 03 '12 at 03:09
11

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.

Georges Elencwajg
  • 150,790
5

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.

Makoto Kato
  • 42,602
0

Suppose they are isomorphic. Then $\left(\mathbb{Z}[x]/(x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2x^2+7)\right)/(2)$. Ravi helpfully pointed out that considering ideals in either ring in terms of $x$ will often give us different ideals, but we do not suffer from this problem when using ideals generated by integers such as $(2)$. Letting overlines represent the canonical homomorphism from $\mathbb{Z}[x]$ to $\mathbb{F}_2[x]$, applying the correspondence theorem for rings on the left side tells us that $$\left(\mathbb{Z}[x]/(x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2)\right)/(\overline{x^2 + 7}) \cong \mathbb{F}_2[x]/(x^2 + 1)\,.$$ Similarly, on the right, we get $$\left(\mathbb{Z}[x]/(2x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2)\right)/(\overline{2x^2 + 7}) \cong \mathbb{F}_2[x]/(1) = \{0\}\,.$$

One can verify the former ring contains four distinct elements. The latter is trivial, which is absurd.

EDIT: To further bolster intuition for an approach like this, see Bill's mention of it in his first comment on his own post.

sporterfield
  • 1
  • 4
    This answer seems to assume that the isomorphism between the two rings sends $x$ to $x$, which isn't necessarily the case. – Ravi Fernando Feb 26 '22 at 21:21
  • 1
    I didn't comment it before, but for the record I fixed the post accordingly to remove this bad implicit assumption. – sporterfield Feb 27 '22 at 14:42
  • 1
    A slight further correction: since $(x^2 + 1) = (x+1)^2 \in \mathbb F_2[x]$ isn't irreducible, the first ring is isomorphic not to $\mathbb F_4$ but to $\mathbb F_2[\varepsilon]/(\varepsilon^2)$. – Ravi Fernando Feb 27 '22 at 16:33
  • Funny that I missed that, given that I spoke to someone about exactly the same thing before posting. It was a very silly mistake, much appreciated! – sporterfield Feb 27 '22 at 17:30
0

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.

user26857
  • 52,094
dragoboy
  • 1,891