$F[x]/(x^2)\cong F[x]/(x^2 - 1)$ 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?
 A: 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).
A: 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.
A: 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...
