$\mathbb{Z}[X]/(X^2-1)$ is not isomorphic with $\mathbb{Z}\times \mathbb{Z}$

Question

I have to show that the ring $\mathbb{Z}[X]/(X^2-1)$ is not isomorphic with $\mathbb{Z}\times \mathbb{Z}$.

I know that $(\mathbb{Z}\times\mathbb{Z})^*=\{(\pm1,\pm1)\}$, so I thought I should be looking for elements which have inverses in $\mathbb{Z}[X]/(X^2-1)$ and hopefully find more or less than 4. But I didn't succeed, so I need hints. Thanks.

The only units in $\mathbb{Z}[X]/(X^2-1)$ are $\pm 1, \pm X$. So this won't help here. — Martin Brandenburg, Nov 24 '15 at 08:54

score 18 · Accepted Answer · answered Jul 29 '13 at 01:59

18

Okay, let's start with the polynomial ring and try to define an isomorphism to the product ring. The multiplicative identity needs to go to the multiplicative identity, for starters: \[1 \mapsto (1,1)\] The only remaining question is where $X$ goes. Well, we have $X^2 = 1$, so $X$ has to go to an element that squares to the identity. If it's $(1,1)$ or $(-1,-1)$ the homomorphism won't be injective, so it's either $(1,-1)$ or $(-1,1)$, and it doesn't really matter which: \[X \mapsto (1,-1)\] Now a general element of the polynomial ring can be written $a + bX$ for integers $a$ and $b$, and it'll be mapped to: \[(a + b, a - b)\] But the difference between these two entries is $2b$, so we can never make any element of $\mathbb Z \times \mathbb Z$ whose entries differ by an odd number. In particular, the proposed homomorphism does not hit $(1,0)$ and is not surjective.

answered Jul 29 '13 at 01:59

Ben Millwood

14,211

I think I'd prefer an answer that was along the lines of "this ring has property P but the other ring has not-P, so they are not isomorphic" but this does seem to work. – Ben Millwood Jul 29 '13 at 02:03
7

The property you want it that $\mathbb Z\times\mathbb Z$ has a pair of orthogonal idempotents, while the quotient ring doesn’t. You see this clearly by tensoring with $\mathbb Q$, where the idempotents are $(x+1)/2$ and $(x-1)/2$. – Lubin Jul 29 '13 at 02:17
1

One may also see this in a more elementary way by simply writing $f(x) = ax+b$ with $a,b \in \mathbb{Z}$ and then seeing what must occur for $f(x)^2 = f(x)$. One ends up with $a = 0$ and $b = 1$ as the only solution. – RghtHndSd Jul 29 '13 at 02:27
@rghthndsd the only solution other than $a=b=0$ :) – rschwieb Oct 19 '13 at 23:50
Here's another ring-theoretic property. The minimal prime ideals of $\mathbb{Z}\times\mathbb{Z}$ are coprime, but those of $\mathbb{Z}[X]/(X^2-1)$ are not. – Jul 23 '18 at 10:07
@Lubin As for underlying conceptual properties, the view of the relationship between element and ring factorizations described in my answer has much to recommend. – Bill Dubuque Oct 20 '20 at 22:23

score 9 · Answer 2 · answered Oct 17 '13 at 17:27

9

$\mathbb{Z}[X]/(X^2-1)$ lacks nontrivial idempotents.

answered Oct 17 '13 at 17:27

Harald Hanche-Olsen

31,960

Quick note on the history, for those who come here: I originally posted this as an answer to a question that was then found to be a duplicate of this one. So I posted the same answer here. Very shortly after, a moderator merged the two questions, and deleted my only answer. (You can see it below, if your reputation is high enough.) – Harald Hanche-Olsen Dec 20 '17 at 12:15

score 7 · Answer 3 · answered Jul 29 '13 at 02:07

7

A homomorphism $\mathbb{Z}[X]\to \mathbb{Z}\times \mathbb{Z}$ is equivalent to specifying an element of $\mathbb{Z}\times \mathbb{Z}$.

A homomorphism $\mathbb{Z}[X]/(X^2-1)\to \mathbb{Z}\times \mathbb{Z}$ is equivalent to specifying an element of $\mathbb{Z}\times \mathbb{Z}$ whose square equals $(1,1)$.

What are the elements of $\mathbb{Z}\times \mathbb{Z}$ whose square equals $(1,1)$? Hint: You've only got $4$ possibilities. So, there are $4$ homomorphisms $\mathbb{Z}[X]/(X^2-1)\to \mathbb{Z}\times \mathbb{Z}$.

Exercise: Check that none of these homomorphisms is surjective.

I hope this helps!

answered Jul 29 '13 at 02:07

Amitesh Datta

20,779

Whee, universal properties :) – Ben Millwood Jul 29 '13 at 02:11
Those properties are no more universal than Ben's. Actually, it is the same idea. – Tomas Jul 29 '13 at 02:13
@Tomas Yes, you're right. I typed my answer independently of Ben's answer though. – Amitesh Datta Jul 29 '13 at 02:15
No offense! This was merely a note to the above comment, which (I now noticed) is also from Ben :) – Tomas Jul 29 '13 at 02:18
Yes, I wondered if my comment was accurate. I stand by the "Whee", though, I like this answer better than mine :P – Ben Millwood Jul 29 '13 at 02:19
@Tomas OK, thanks for your comment! (It will be useful to future readers to know that the two answers are, in fact, equivalent.) – Amitesh Datta Jul 29 '13 at 02:21
@Ben Thanks! Though it's clear from reading your answer that you had the same idea as me even though you didn't refer to it as "universal property". I think our two answers complement each other very well! – Amitesh Datta Jul 29 '13 at 02:24

Martin Brandenburg · Answer 4 · 2015-11-24T06:35:05.993

6

$\mathbb{Z}[X]/(X^2-1) \cong \mathbb{Z} \times \mathbb{Z}$ would imply (after modding out the ideal $(2)$) that $$\mathbb{F}_2[X]/(X-1)^2 \cong \mathbb{F}_2 \times \mathbb{F}_2.$$The first ring has a nontrivial nilpotent element, whereas the second one does not.

edited Nov 24 '15 at 06:35

answered Oct 17 '13 at 21:21

Martin Brandenburg

163,620

score 1 · Answer 5 · answered Nov 24 '15 at 09:04

The ring $\mathbb{Z} \times \mathbb{Z} \cong \mathbb{Z}[X]/(X^2-X)$ has the universal property $$\mathrm{Hom}_{\mathsf{Ring}}(\mathbb{Z} \times \mathbb{Z},R) \cong \{a \in R : a^2=a\},$$ and the ring $\mathbb{Z}[X]/(X^2-1)$ has the universal property $$\mathrm{Hom}_{\mathsf{Ring}}(\mathbb{Z}[X]/(X^2-1),R) \cong \{a \in R : a^2=1\}.$$ Thus, if $\mathbb{Z} \times \mathbb{Z}$ and $\mathbb{Z}[X]/(X^2-1)$ were isomorphic, this would imply that every ring $R$ has the same numbers of idempotents and of involutions. This is of course wrong, consider $R=\mathbb{Z}/2 \times \mathbb{Z}/2$ for instance; this ring has $4$ idempotents, but only $1$ involution.

Bill Dubuque · Answer 6 · 2023-09-05T17:58:38.443

By below Lemma $\,\Bbb Z[x]/(x^2\!-\!1)\cong\:\! \Bbb Z\!\times\! \Bbb Z\,$ $\Rightarrow\, \overbrace{(x\!-\!1)\,a(x)\!+\!(x\!+\!1)\,b(x) = 1}^{\large x-1,\ x+1\ \ {\rm are\ \color{#0a0}{comaximal}\ in}^{\phantom{}}\ \Bbb Z[x]}\ \ \smash{\overset{x\,\to\, 1}\Longrightarrow}\ \ \bbox[4px,border:1px solid #0a0]{2\mid 1}\,$ in $\Bbb Z$

Idea $ $ proper factorizations of $\,R/f\,$ $\rm\small\color{#f60}{induce}$ coprime $\small\rm\color{#0a0}{(i.e. comaximal)}$ proper factorizations of $f\,$ by

Lemma $ $ If $\ \begin{align}f\in R\,\ \rm a\ UFD,\\ {\rm rings}\ G,H\!\neq 0\end{align}\,\ $ then $\, \begin{align} R/f\,\ &\cong\,\ G\!\times\! H\\ \color{#f60}{\bf\large \Rightarrow} f\ \ &\!\!=\ g\ h\end{align}\, $ with $\ \begin{align} &\color{#0a0}{(g)\!+\!(h)=(1)^{\phantom{|^|}}\!\!}\\ &\,(g),\, (h)\neq\:\! (1)\end{align}\,$ for some $\,g,h\in R^{\phantom{|^|}}\!\!\!$

Proof $\ $ See this answer for a simple $\,4\,$ line proof.

$\mathbb{Z}[X]/(X^2-1)$ is not isomorphic with $\mathbb{Z}\times \mathbb{Z}$

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