Hope this isn't a duplicate.
I was trying to answer the following questions:
(i) Let $k$ be any field. Then prove that $k[X]$ has infinitely many maximal ideals.
(ii) Using (i) prove that, given any commutative ring $R$ with unity, $R[X]$ has infinitely many maximal ideals.
My attempt:
(i) If $k$ is infinite, then the collection of ideals of the form $(x-a) \forall a \in k$ will suffice.
For finite fields I argue by contradiction. Begin by assuming that there are only finitely many prime polynomials, a listing of them in $k[X]$ say, $p_1,\ldots,p_n$ . Next we define $p_{n+1} := p_{1} \cdots p_{n} +1$ then $p_{n+1} > p_{j} \forall 1\leq j\leq n $ and none of them divide it i. e. proceed like Euclid's argument for proving infinitely many primes.
(ii) By Krull's theorem, any non-zero commutative ring with unity has a maximal ideal. Let $\mathscr{M}$ be a maximal ideal of $R$. Then $\frac{R}{\mathscr{M}}$ is a field, we consider $\frac{R}{\mathscr{M}} [X]$ , then by part (i), $\frac{R}{\mathscr{M}} [X]$ has infinitely many maximal ideals. Since,$\frac{R}{\mathscr{M}} [X] \cong \frac{R[X]}{\mathscr{M}[X]}$ , thus $\frac{R[X]}{\mathscr{M}[X]}$ must have maximal ideals.
Now any ideal of $\frac{R[X]}{\mathscr{M}[X]}$ is of the form $\frac{I}{\mathscr{M}[X]}$ where $I$ is an ideal of $R[X]$ containing $\mathscr{M}[X]$. So any maximal ideal of $\frac{R[X]}{\mathscr{M}[X]}$ is an ideal of $R[X]$ containing ${\mathscr{M}[X]}$, say $J$. But I don't know whether $J$ is maximal in $R[X]$ or not.
How to cross the hurdle? Also if there are mistakes in my arguments please point them out.
Best Answer
By the ideal correspondence theorem, a maximal ideal of $R[X]/\mathscr{M}[X]$ gives a maximal ideal of $R[X]$.