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I am self studying Artin's algebra and stuck with this exercise which asks 'Are $\mathbb{Z}[x] /\langle x^2 +7 \rangle$ and $\mathbb{Z}[x] /\langle 2x^2 + 7\rangle$ isomorphic?'

My sense tells answer is no but to show non-isomorphism , we should execute one property that one ring has and other hasn't ... I couldn't figure out the property... Thanks in advance for any help/ hints ..

Pritam Roy
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2 Answers2

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I think you're right. The first ring $\Bbb Z[X]/(x^2+7)\cong \Bbb Z[\alpha] $ where $\alpha^2+7=0$. As abelian groups, we have $\Bbb Z[X]/(x^2+7)\cong\{a+b\alpha \mid a,b\in\Bbb Z\}\cong \Bbb Z^2$, since we can divide any polynomial by $x^2+7$ and get a remainder of degree less than $2$.

This is not true in the case of $\Bbb Z[X]/(2x^2+7)$. We can't perform division of any polynomial in $\Bbb Z[X]$ by $2x^2+7$, because of its not being monic. In particular, we can only divide by $2x^2+7$, inside $\Bbb Z[X]$, when the dividend is a polynomial whose leading coefficient is a multiple of $2$. (Apparently you learn more about this sort of thing when you study localization.) But the upshot is that you get some higher degree polynomials.

In fact, we can conclude that $\Bbb Z[x]/(f)$ is not a finitely generated abelian group if $f$ is not monic. See this. So it can't be isomorphic as an abelian group to $\Bbb Z^2$.

calc ll
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  • Do you want to say isomorphic rings preserves division rule ? I couldn't clearly understand what is the relationship between doing division and isomorphism here. Also I want to know about the element of the ring Z[x] /<2x^2 + 7> whereas you told element of Z[x] /<x^2 + 7> will be of the form a+b€ where €^2 + 7 = 0. – Pritam Roy Aug 07 '22 at 09:20
  • Yeah I don't know a canonical form for $\Bbb Z[X]/(2x^2+7)$ because of not being able to do that division. Sorry I couldn't be a little more rigorous. I'll look into it. In fact, it isn't finitely generated. See this https://math.stackexchange.com/q/2230921/1070376 – calc ll Aug 07 '22 at 10:03
  • $\mathbb{Z}[x]/(x^2+7)$ is not isomorphic to $\mathbb{Z}^2$ as rings. You want to specify that this is an isomorphism of abelian groups and that $\mathbb{Z}[x]/(2x^2+7)$ is not finitely generated as abelian group. – egreg Aug 07 '22 at 10:19
  • @egreg thanks. I made a mistake. So, the problem is about rings; and, we'll do it by proving they're not isomorphic as groups. Correct? – calc ll Aug 07 '22 at 10:49
  • @Cpc If you prove that the underlying abelian groups are not isomorphic, then the rings cannot be isomorphic (as rings) and the proof is good. Note that both rings are finitely generated (as rings), so what notion of “finitely generated” is used should be mentioned. – egreg Aug 07 '22 at 11:53
  • Yes, nontrivial localizations invert a nonunit so they cannot be integral (see the Remark in my answer). More generally every proper ideal survives in an integral extension, and this property can be used to characterize integral extensions - see here. – Bill Dubuque Aug 07 '22 at 17:40
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Hint $ $ Easy way: only one ring has a sense of parity (i.e. an image $\cong \Bbb Z/2).\,$

Or $\ \underbrace{\frac{1}2\in \Bbb Z[x]/(2x^2\!+\!7)}_{\small\textstyle 2(x^2\!+\!4)=1}\,$ but $\frac{1}2\not\in \Bbb Z[x]/(x^2\!+\!7)\,$ else $\,2w = 1\underset{\rm Norm}\Longrightarrow 4w\bar w = 1\in \Bbb Z.\,$

Remark $ $ Above a discriminating property is integrality. A crucial property of integral ring extensions is that they don't alter unit properties in the base ring, i.e. a nonunit $\,r\in R\,$ remains a nonunit in any integral extension ring $S\,$ (so they cannot introduce fractions like $1/2\,$ above). For example, this may allow us to deduce that Diophantine equations in $\Bbb Z$ are unsolvable by deducing a parity contradiction $\,2\mid 1\,$ in some convenient extension ring of algebraic integers. See here for further discussion.

Bill Dubuque
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  • 1/2 belongs to the first quotient ring but doesn't belong to next. Ok , but to be explicit, how does it imply that the aforesaid rings are non-isomorphic ? Also how can I take norm in this ring ? I haven't studied yet. Any help ? – Pritam Roy Aug 07 '22 at 13:22
  • Ring homs $,h,$ preserve roots of polynomials, so a root of $2x=1$ persists as a root in any isomorphic ring, i.e. $,1= h(1) = h(2x) = 2h(x).,$ The norm of $w = a+b\sqrt{-7}$ is $,N(w) := w\bar w = a^2+7b^2,$ where $,\bar w = a-b\sqrt 7,$ is the conjugate. It's immediate that the norm is multiplicative $,N(\alpha\beta) = N(\alpha) N(\beta).\ \ $ – Bill Dubuque Aug 07 '22 at 13:43