Polynomial irreducible in $F[x]$ implies irreducible in $F[x,y]$?

Question

Let say we have a polynomial $p(x)$ that is irreducible in $F[x]$. Is it also irreducible in $F[x,y]$? ($x$, $y$ are both indeterminates.)

It seems intuitive that this is true, but how do we prove it?

Thanks.

Is this attempted proof ok?

Attempted proof:

Suppose $p(x)$ is not irreducible in $F[x,y]$. Then $p(x)=f(x,y)g(x,y)$ for some non-constant $f,g\in F[x,y]$.

Since $p$ is solely a function of $x$, so are $f, g$. This contradicts that $p$ is irreducible in $F[x]$.

Answer 1

In $p(x)=f(x,y)g(x,y)$, think of everything as a polynomial in $y$ with coefficients from $F[x]$. The only way the product of two polynomials in $y$ can be constant is if they are both constant (as long as there are no zero divisors).

Concretely: if the highest power of $y$ in $f$ is $y^n$ and in $g$ is $y^m$ then $fg$ must have $y^{n+m}$.

Answer 2

If $F$ is a factorial ring (a field is a factorial ring), we can invoke the Gauß lemma to see that $F[x,y]$ is factorial, i.e. irreducible and prime coincide and we can test irreducibility by considering the quotient ring. Then we can argue as follows:

$F[x,y]/ \langle p(x) \rangle \cong \Big(F[x]/ \langle p(x) \rangle \Big)[y]$ is a domain, since $F[x]/ \langle p(x) \rangle$ is a domain by the assumption and the polynomial ring over a domain is a domain.