Ring of polynomials over a field has infinitely many primes

Let $F$ be a field. Why does $F[x]$ have infinitely many irreducible elements?

For the case F has characteristic 0

Then x-a is irreducible for all a $\in F$ since x satisfies no non-trivial relations in F.

Obviously this argument fails for a finite field since there are only finitely many a to choose from.

So how may I construct irreducible polynomials in a finite field?

I figure it must involve higher powers of x, maybe $x^n-a$ ?

Solution 1:

You can copy Euclid's proof. Let $p_1, \dots, p_n$ be a finite collection of prime polynomials in $F[X]$. Consider $f=p_1 p_2 \cdots p_n +1$. Let $p$ be a prime factor of $f$. Then $p$ cannot be any of $p_1, \dots, p_n$ because otherwise $p$ would divide $1$. Hence, no finite collection of prime polynomials exhausts the set of prime polynomials and so the set of prime polynomials is infinite.

Solution 2:

Let's talk about fields of characteristic $p$. Every finite field $k$ (of this characteristic) has $q=p^r$ elements, and they all are roots of the polynomial $f_q(X)=X^q-X$, which you see has no repeated roots, so that this polynomial identifies the elements of $k$. Now let's apply this to the question of irreducible polynomials over ${\mathbb{F}}_p$. Each of the polynomials $f_q$ factors into irreducibles over ${\mathbb{F}}_p$, always including $X$ and $X-1$, of course, but if you look closely, you'll see that all the irreducibles of degree $r$ over $\mathbb{F}_p$ divide $f_q$. Conversely, the irreducible ${\mathbb{F}}_p$-polynomials dividing $f_q$ are precisely the irreducibles of degree dividing $r$. So, for example, the irreducible polynomials of degree $4$ over the field with $2$ elements are the quartic irreducibles dividing $X^{16}-X$, whose total factorization is $X(X+1)(X^2 +X+1)(X^4+X+1)(X^4+X^3+1)(X^4+X^3+X^2+X+1)$.

Generalization to the irreducibles over $\mathbb{F}_q$ left to you.

Solution 3:

If $F$ is finite, we show that for any prime $q$, there exists an irreducible polynomial with degree $q$ over $F$.

Suppose $F=GF(p^n)$. For any prime $q$, consider $E=GF((p^{n})^q)$. Since $E$ is a finite extension of $F$, so is an algebraic extension. For any $\alpha\in E-F$, let $m(x)$ be the minimal polynomial of $\alpha$ over $F$. Then we have the tower of fields $$\underbrace{F\overbrace{\leq}^{=\deg{m(x)}}F(\alpha)\leq E}_{q}.$$ Since $\alpha\notin F$, $\deg{m(x)}$ must be $q$. Thus, $m(x)$ is an irreducible polynomial of degree $q$ over $F$.