Is $K[X]$ not a semilocal ring?

Question

Let $K$ be a field. We will write $K[X]$ to denote the set of all polynomials in one variable over the field $K$ and $\mathrm{Maxspec}(K[X])$, the set of all maximal ideals of $K[X].$ Also, we call a ring semilocal, if $|\mathrm{Maxspec}(K[X])|<\infty$.

I found in a book, that the ring $K[X]$ is not semilocal. But, why does this happen?

Thoughts:

1. One can prove that $$\mathrm{Maxspec}(K[X])=\{\langle p(X) \rangle \triangleleft K[X]: p(X) \text{ is irreducible over } K[X] \}.$$ And this, drives as directly to count the irreducible polynomials over $K[X]$. But, how can we find this cardinality in general?

2. If $K=\mathbb{Z}_p$, does this topic help us?

Thank you in advance.

Answer 1

A simpler proof.

We can prove that there are infinitely many irreducibles by adapting Euclid's proof from the integers.

If $k$ is a field, then $k[x]$ has infinitely many irreducibles.

Proof:

Suppose there are not infinitely many irreducible polynomials, suppose there are only $p_1,\ldots,p_n$. Consider $P=p_1p_2\cdots p_n+1$. Some irreducible $p_i$ must divide it since it has positive degree and is therefore not a unit. (See the note)

But then we have $P=p_iq+1=p_ik$ where $q=p_1p_2\cdots \hat{p_i}\cdots p_n$ and $k$ is the polynomial such that $P=p_ik$, so $p_i(k-q)=1$, and $p_i$ is a unit. Contradiction. $k[x]$ has infinitely many irreducibles. (This proof of infinitely many irreducibles generalizes to $k$ any Noetherian domain) $\blacksquare$

Note

For the fewest assumptions, we can prove that any positive degree polynomial over a field is divisible by an irreducible by induction.

If $P$ has positive degree and is not irreducible itself, then $P=ab$ for $ab$ both nonunits, and hence both of positive degree (necessarily less than that of $P$). Thus one of $a$ or $b$ is divisible by an irreducible by induction. (Note that the base case here is if $P$ is irreducible).

General case

For the case of $k$ a Noetherian domain, we use the fact that $k$ Noetherian implies $k[x]$ is Noetherian, and the fact that in any Noetherian domain any nonunit is divisible by some irreducible, which is proved in the following manner.

Let $p_0$ be a nonunit. If $p_0$ is irreducible, then we are done. Otherwise $p_0=p_1a_1$ with $p_1,a_1$ nonunits. Now if $p_1$ is irreducible, we are done. Otherwise, we repeat to get $p_1=p_2a_2$, and $p_n=p_{n+1}a_{n+1}$. Then either some $p_{n+1}$ is irreducible, or this continues forever, but if it continued forever, we get this strictly increasing chain of ideals: $$(p_0)\subsetneq (p_1)\subsetneq (p_2)\subsetneq \cdots \subsetneq(p_n)\subsetneq \cdots,$$ which is impossible. Thus some $p_n$ is irreducible, and $p_n\mid p_0$.

That the chain is strictly increasing follows from the fact that if $p_n\in (p_{n+1})$, then $p_n\mid p_{n+1}$ and $p_{n+1}\mid p_n$, which in a domain would imply that $a_{n+1}$ is a unit, which would be a contradiction.

Answer 2

This is a very strange claim, and it is very likely to be false.

It states basically that the ring $K[X]$ is semilocal iff it has finitely many irreducible monic polynomials.

Since all $X+a$, $a \in K$, are monic irreducible polynomials, if $K[X]$ is semilocal, then $K$ is finite.

Now assume that $K$ is finite.

I claim that $X^{|K|^n}-X$ is the product $\Pi$ of all monic irreducible polynomials over $K$ such that their degree divides $n$.

First, let $P$ denote some irreducible factor of $X^{|K|^n}-X$, let $d$ be its degree. Then $L=K[X]/(P)$ is a field extension of degree $d$ of $K$. Let $F : u \in L \longmapsto u^{|K|}$, it is a field automorphism of $L$, and $F^{\circ n}$ is identity on $K$ and the image of $X$ in $L$ (Because $P|X^{|K|^n}-X$). These elements span $L$ as a ring, thus $F^{\circ n}$ is the identity on $L$.

Now, let $r$ be the gcd of $n$ and $d$, then $|K|^r-1$ is the gcd of $|K|^n-1$ and $|K|^d-1=|L^*|$. Since every element of $L^*$ is a root of $X^{|K|^n-1}-1$ and of $X^{|K|^d-1}-1$, the polynomial $X^{|K|^r-1}-1$ vanishes at every element of $L^*$ so has at least $|K|^d-1$ roots. Thus $|K|^d-1 \leq |K|^r -1$ hence $d | n$ thus $P | \Pi$.

Conversely, let $P$ be an irreducible factor of $\Pi$ with degree $d$, and $P \neq X$. Then $L=K[X]/(P)$ is a field extension of $K$ with degree $d$, so $L^*$ is a group of cardinality $|K|^d-1$, thus the identity $X^{|K|^d-1}=1$ holds in $L$, thus $X^{|K|^n-1}=1$ in $L$ as well, which translates to $P | X^{|K|^n-1}-1$ in $K[X]$.

Since $X^{|K|^n}-X$ and $\Pi$ have the same irreducible factors, are monic, and do not have double factors (check the derivative for the first one), $\Pi=X^{|K|^n-1}$.

As a consequence: there exists a monic irreducible polynomial in $K$ of degree $2^n$ for every $n$. Thus there are infinitely many irreducible polynomials over $K$ and $K[X]$ is not semilocal.