# Does an irreducible polynomial in K(t)[x] give an irreducible polynomial in K[t][x]

Let $K[t]$ be the ring of polynomials over a field $K$. Let $K(t)$ be its fraction field. Let $f$ be an irreducible polynomial in $K(t)[x]$. There exists an element $a\in K[t]$ such that $af$ is in $K[t,x]$.

Does there exist $a \in K[t]$ such that $a f$ is irreducible in $K[t,x]$?

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Have a look at Gauss's lemma for polynomials over UFD : en.wikipedia.org/wiki/Gauss's_lemma_(polynomial) – Lierre Nov 5 '12 at 11:34
Ow that looks helpful. I think it implies that if I take $a$ to be the least common multiple of the denominators of the coefficients of f (which lie in $K(t)$) we get that the polynomial $af$ is irreducible. In fact, it is irreducible in $K(t)[x]$ (because we're just multiplying by an element in $K(t)$) and it is primitive in $K[t,x]$. – Harry Nov 5 '12 at 11:41
You are almost right : to get a primitive polynomial, you have to multiply by the l.c.m. of the denominators of the coefficients AND to divide by the g.c.d. of their numerators. – Lierre Nov 5 '12 at 11:45