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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?**

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up vote 5 down vote accepted

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.)

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So perhaps the isomorphism defined by $$ \phi (a_0 + a_1x + ... + a_nx^n ) = \sum_{i=0}^n a_i$$ as then $\phi (x-1) = 1-1 = 0$ and so is in the kernal but it is not ALL of the elements in the kernal, surely for the isomorphism theorem to work you need it to be the whole kernal, not just one element of it? – user26069 Mar 2 '12 at 15:07
True, but the remainder theorem will tell you something about all the elements of the kernel, in relation to the element $x-1$... – Matthew Pressland Mar 2 '12 at 15:11
Ahh I see so all the polynomials with x-1 as a factor will be in the kernal – user26069 Mar 2 '12 at 15:17
If you have a ring homomorphism $\phi : R \rightarrow S$ and an element $r \in R$ that is in the kernel, then $(r) \subseteq \text{ker}(\phi)$. (Exercise: prove this!) Now use Matt's hint to prove the reverse inclusion, that $\text{ker}(\phi) \subseteq (r)$. – Michael Joyce Mar 2 '12 at 15:19
You could also look at the composition $F \to F[x] \to F[x]/(f(x))$ and see if that's an inverse. – Dylan Moreland Mar 2 '12 at 15:28

Here are some approaches alternative to MJ's. All are well-worth learning.

$(1)\rm\ \ F\to F[x]/(x-1)$ is onto by $\rm\:f(x)\equiv f(1)\:\!\ (mod\ x\!-\!1).$ It's $\:1\!-\!1$ by $\rm\:x\!-\!1\: |\ c\in F\: \Rightarrow\: c = 0$.

$(2)\rm\ \ (f)\supseteq (g)\iff f\ |\ g,\:$ i.e. for principal ideals: contains $\iff$ divides. Therefore, having no proper containing ideal (maximal) is equivalent to having no proper divisor (irreducible).

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