Is this using the first isomorphism theorem for rings? Let $F$ be a field and $f(x) = x - 1$ and $g(x) = x^2 - 1$.
1) Show that $F[x]/(f(x)) \cong F$
2) Is ideal $(g(x))$ maximal? Explain your answer. 
** I have a feeling that this uses the first isomorphism for rings but I can't relate the elements of the question two the elements in the isomorphism theorem, maybe there is an other way of doing this?**
 A: Here are some hints:
1) To show $F[x]/(f(x)) \cong F$, it suffices to find a surjective ring homomorphism $\varphi: F[x] \rightarrow F$ whose kernel is $(f(x))$.  (That's the first isomorphism theorem at work.)  Moreover, assuming you map the constant polynomials to $F$ in the obvious way, the homomorphism $\varphi$ is determined by your choice of the value of $\varphi(x)$.  What value of $\varphi(x)$ would result in $f(x) = x - 1$ being in the kernel?
2) The ideal $(g(x))$ is maximal if and only if $F[x] / (g(x))$ is a field.  What does the factorization $x^2 - 1 = (x-1)(x+1)$ tell you about the ring $F[x] / (g(x))$?  (You can also do this problem directly from the definition of a maximal ideal, but the approach suggested is important to understand.)
A: Here are some approaches alternative to MJ's. All are well-worth learning.
$(1)\rm\ \ F\to F[x]/(x\!-\!a)\,$ is onto by $\rm\:f(x)\equiv f(a)\:\!\ (mod\ x\!-\!a).$ It's $\:1\!-\!1$ by $\rm\:x\!-\!a\mid c\in F \Rightarrow\, c = 0$.
$(2)\rm\ \ (f)\supseteq (g)\iff f\ |\ g,\:$ i.e. for principal ideals: contains $\iff$ divides. Therefore, having no proper containing ideal (maximal) is equivalent to having no proper divisor (irreducible).
