# $F[x]/(x^2)\cong F[x]/(x^2 - 1)$ if and only if F has characteristic 2

Artin's Algebra, Chapter 10 problem 5.16 states:

Let $F$ be a field. Prove that the rings $F[x]/(x^2)$ and $F[x]/(x^2-1)$ are isomorphic if and only if $F$ has characteristic 2.

As a pedantic concern: if F has characteristic 0, then surely this isomorphism still holds? So maybe it should be "characteristic at most 2"?

More seriously, it seems like $F[x]/(x^2)=\left\{f_0 + f_1 x\right\}$ since we're just setting $x^2=0$. Similarly, it seems like $f_0 + f_1x / (x^2-1) = f_0 + f_1 x$, which implies that $F[x]/(x^2)=F[x]/(x^2-1)$ independent of the characteristic of $F$. To prove this we need to show that $\text{deg}(fg)=\text{deg}(f)+\text{deg}(g)$, which I believe is true in at least integral domains.

What is my mistake here?

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Why the downvote? What's wrong with the question? – Adrián Barquero Jul 4 '11 at 4:36
@Adrián I didn't downvote it, but perhaps the downvoter wanted the OP to think more deeply about the problem before throwing in the towel. – Bill Dubuque Jul 4 '11 at 4:45
@Bill It could be, but to me that's not a good reason to downvote a question, let alone the good answers which have been provided. – Adrián Barquero Jul 4 '11 at 4:51
@Adrián Why do you jump to the conclusion that there is some correlation between the downvotes? – Bill Dubuque Jul 4 '11 at 4:59

No, the problem is correct as stated: in particular those rings are not isomorphic in characteristic zero. I think your mistake is coming from reading too much into your description of the elements of both quotient rings as (being represented by) $f_0 + f_1 x$. This shows that they are isomorphic as $F$-vector spaces: i.e., they both have dimension $2$. But it doesn't show you that they are isomorphic as rings.

Hint: if the characteristic is not $2$, then $(x-1)$ and $(x+1)$ are comaximal ideals in the polynomial ring $F[x]$, so one can apply the Chinese Remainder Theorem to get a nice description of $F[x]/(x^2-1)$. In particular, you should find that the quotient is a reduced ring, i.e., has no nonzero nilpotent elements, unlike $F[x]/(x^2)$.

I guess you can see why the two rings are isomorphic in characteristic $2$?

Added: After you have solved this problem, it is enlightening to do a more general one: let $F$ be a field, $P(x) \in F[x]$ a polynomial, and try to give as explicit a description as you can of the quotient ring $F[x]/(P(x))$. For instance: when is it an integral domain? A field? A connected ring (i.e., with no idempotents other than $0$ and $1$)? A reduced ring (i.e., with no nilpotents other than $0$)? It turns out that everything depends upon the shape of the factorization of $P(x)$ into powers of distinct irreducible polynomials...

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Neither do I understand the downvote here... – Pete L. Clark Jul 4 '11 at 4:38
I don't understand the downvote here (or with Bill's answer). They are both clear and correct, and have the advantage of using theorems to avoid some of the dirt. – Aaron Jul 4 '11 at 4:41
@Aaron: it's a puzzler. I confess that it crossed my mind that it might have been you (we are given just enough information to let our suspicions run rampant, it seems) but you passed that test in especially dramatic fashion: zero downvotes altogether! Oh, well -- no biggie. – Pete L. Clark Jul 4 '11 at 4:43
@Adrian: I definitely don't agree with the downvoting, but as a citizen of a democracy I recognize and support the right of people to cast a few "wrong" votes. Nobody is really being hurt thus far, so for my part I am not calling for an investigation. – Pete L. Clark Jul 4 '11 at 4:54
@Xod By definition $0\ne 1$ in a domain / field so, being a field, $F \ne 0$. – Bill Dubuque Jul 4 '11 at 13:17

HINT $\$ If $\rm\ char(F) \ne 2\$ then $\rm\:F[x]/(x^2-1)\ \cong\ F[x]/(x-1) + F[x]/(x+1)\ \cong\ F^2\:$ has nontrivial idempotents, e.g. $\rm\:(0,1)\:,\:$ but $\rm\:F[x]/(x^2)\:$ does not (as is easily verified).

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FWIW, I don't understand the downvote. This answer is 100% correct... – Pete L. Clark Jul 4 '11 at 4:34
@Pete It seems somebody downvoted not only the question, but also the answers. Talk about strange behavior. – Adrián Barquero Jul 4 '11 at 4:39
@Pete Perhaps because I denigrate those nasty idempotents far too frequently. – Bill Dubuque Jul 4 '11 at 4:47

The problem is that you are establishing an isomorphism of vector spaces, but not an isomorphism of rings.

In $F[x]/(x^2)$, we have an element that squares to zero. As you have already shown, the elements (equivalence classes) of $F[x]/(x^2-1)$ can be represented by elements of the form $ax+b$. If we had a ring isomorphism $\varphi:F[x]/(x^2)\to F[x]/(x^2-1)$, then we would have $\varphi(x)=ax+b+(x^2-1)F[x]$, and hence $(ax+b)^2=k(x^2-1)$. However, $(ax+b)^2=a^2 x^2 +2abx + b^2$ cannot be a multiple of $x^2-1$ unless $2ab=0$.

In characteristic $2$, we have $F[x]=F[x-1]$ and $x^2-1=(x-1)^2$. This is enough to establish the isomorphism.

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