Prime ideals in $\mathbb{Q}[X]$

Question

Could you tell me why prime ideals in $\mathbb{Q}[X]$ are of the form $(q(x))$ where $q \in \mathbb{Q}[X]$ is irreducible?

Answer 1

This is a case of a general theorem that states: $R[x]$ is a PID when $R$ is a field

Here we have $R=\mathbb{Q}$ is a field.

A proof for this can be found for example at this link

Answer 2

Hints:

1) Because $\;\Bbb Q[x]\;$ is a PID (in fact, even an Euclidean domain) and thus all its ideals are principal.

2) Becasue in $\;\Bbb Q[x]\;$ , a non zero ideal is prime iff it is maximal (this is far from being true in general rings, though).

3) Finally, because a principal ideal (in this cases) is prime iff it is generated by a prime element, and in the ring of polynomials an element is prime iff it is irreducible.

Answer 3

Hint: If $F$ is a field and $I$ is a non-zero ideal in $F[x]$, and $g(x)$ is an element of $F[x]$. Then, $I=\langle g(x)\rangle$ iff $g(x)$ is a non-zero polynimial of minimum degree in $I$.

And $g(x)$ is irreducible iff $\langle g(x)\rangle$ is maximal

In $\mathbb{Q}[x]$, maximal ideals $\iff$ prime ideals