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 didnt succeed, so I need hints. Thanks.

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    $\begingroup$ Why did you not succeed in computing the units? $\endgroup$ – Mariano Suárez-Álvarez Jul 29 '13 at 2:10
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    $\begingroup$ The only units in $\mathbb{Z}[X]/(X^2-1)$ are $\pm 1, \pm X$. So this won't help here. $\endgroup$ – Martin Brandenburg Nov 24 '15 at 8:54

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.

  • $\begingroup$ 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. $\endgroup$ – Ben Millwood Jul 29 '13 at 2:03
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    $\begingroup$ 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 $(x-1)/2$. $\endgroup$ – Lubin Jul 29 '13 at 2:17
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    $\begingroup$ 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. $\endgroup$ – RghtHndSd Jul 29 '13 at 2:27
  • $\begingroup$ @rghthndsd the only solution other than $a=b=0$ :) $\endgroup$ – rschwieb Oct 19 '13 at 23:50
  • $\begingroup$ 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. $\endgroup$ – user501746 Jul 23 '18 at 10:07

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

  • $\begingroup$ 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.) $\endgroup$ – Harald Hanche-Olsen Dec 20 '17 at 12:15

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!

  • $\begingroup$ Whee, universal properties :) $\endgroup$ – Ben Millwood Jul 29 '13 at 2:11
  • $\begingroup$ Those properties are no more universal than Ben's. Actually, it is the same idea. $\endgroup$ – Tomas Jul 29 '13 at 2:13
  • $\begingroup$ @Tomas Yes, you're right. I typed my answer independently of Ben's answer though. $\endgroup$ – Amitesh Datta Jul 29 '13 at 2:15
  • $\begingroup$ No offense! This was merely a note to the above comment, which (I now noticed) is also from Ben :) $\endgroup$ – Tomas Jul 29 '13 at 2:18
  • $\begingroup$ Yes, I wondered if my comment was accurate. I stand by the "Whee", though, I like this answer better than mine :P $\endgroup$ – Ben Millwood Jul 29 '13 at 2:19

$\mathbb{Z}[X]/(X^2-1) \cong \mathbb{Z} \times \mathbb{Z}$ 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.


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.


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