If $K$ is a field, prove the polynomial ring $K[x]$ has infinitely many maximal ideals.

Question

I know $(x)$ is a maximal ideal, but I don't know how to construct more maximal ideals.

If $K= \mathbf{F}q$ is a finite field then for each $n$ there is the splitting field of $x^{q^n}-x$ which is $\mathbf{F}{q^n} \cong K[x]/(f_n(x))$ where $f_n$ is irreducible of degree $n$. If $K$ is not a finite field then $K[x]$ has infinitely many maximal ideals of the form $(x-a),a \in K$. — reuns, Dec 15 '17 at 20:26

score 3 · Accepted Answer · edited Dec 15 '17 at 19:58

3

HINT: $K[x]$ is principal ideal domain. Now consider the ideals generated by irreducible polynomials in $K[x]$

edited Dec 15 '17 at 19:58

user26857

answered Dec 15 '17 at 19:44

Stefan4024

Out of interest, do you have an easy argument to show that $K[x]$ must contain infinitely many irreducible polynomials (modulo associates)? I can only think of the argument implicit in reuns's comment above. – Rob Arthan Dec 15 '17 at 21:06
3

@RobArthan Euclid's proof of infinitude of primes can be altered to give the wanted proof. – Stefan4024 Dec 15 '17 at 21:13

1 Answers1