# Determine all maximal and prime ideals of the polynomial ring $\Bbb C[x]$

Determine all maximal and prime ideals of the polynomial ring $\Bbb C[x]$

My attempt:

Note that $\Bbb F[x]$ where $\Bbb F$ is any field is a Euclidean domain, and importantly, that means that it is also a PID. As a PID, prime ideals are the same as maximal ideals. Since $\Bbb C$ is a field, we know that the polynomial ring $\Bbb C[x]$ is a ED and PID.

Prime ideals are ideals such that $ab\in P \implies a\in P$ or $b\in P$. For $a,b\in \Bbb C[x]$.

So then, if a polynomial is in the ideal, then all of its linear factors are. This means that the generator for this ideal must be one of these factors. In fact, any polynomial that is reducible can't be the generator for this ideal, and therefore the prime ideals must be generated by irreducibles.

In $\Bbb C[x]$ it would seem by algebraic closure, the only irreducibles are linear polynomials.

Therefore all prime/maximal ideals of $\Bbb C[x]$ are of the form $(x-\alpha)$, $\alpha\in \Bbb C$.

Note: that although $(0)$ is irreducible, it is not maximal. (I think we just say it is excluded?)

Is my proof valid?

• $0$ is also a prime ideal, as in any integral domain. – Bernard Jun 20 '15 at 10:10
• @Bernard oops indeed. But not forgetting that then – Permute Jun 20 '15 at 10:11
• @permute the last sentence is a little ambiguous. You probably want to make it clear that you know the zero ideal isn't maximal. – rschwieb Jun 20 '15 at 10:56
• @user26857 I just wanted to know if my proof was valid – Permute Jun 21 '15 at 2:24
• @Permute: Try to prove it for $K[x]$, where $K$ is any field. You'll see that the fact that $\mathbb{C}$ is algebraically closed is not relevant for the proof. – math Jan 10 '17 at 23:03

It can be easily solved using this lemma: In a PID $$(a)$$ is maximal ideal iff $$a$$ is an irreducible element.And by Fundamental theorem of algebra the only irreducibles are degree $$1$$ polynomials.