# Question about Irreducible Polynomials Being Relatively Prime.

Question: Let $k$ be any field, and $f; g \in k[x; y]$ irreducible polynomials, not multiples of one another. Write $K = k(x)$; prove that $f; g$ have no common factors in the PID $K[y]$.

My interpretation: I assume by no common factors, it is meant that $f,g$ are co-prime in $K[y]$ and thus we can find $a,b\in K[y]$ such that $af+bg=1$.

My thoughts: I would try to deduce it from definitions but I'm not sure how to manipulate the information that $f,g$ are irreducible polynomials of two variables because I cannot just say consider some field extension where they both factor linearly.

My only seemingly legitimate attempt would be: Let $h$ be the gcd of $f,g$ in $K[y]$ then there exist $a,b \in K[y]$ such that $af+bg=h$. I'm not sure where to go from here since I am not sure what I know about $f,g$ in $K[y]$. Are they irreducible still, etc...?

Extra Questions: Is $K[y]$ which is $k(x)[y]$ an extension of $k[x,y]$. How should I think about $k(x)[y]?$

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Is $k(x)$ different from $k[x]$ ? (What does $k(x)$ mean ?) –  Amr Feb 10 '13 at 14:22
$k(x)$ is the field of rational expressions/functions. –  Steven-Owen Feb 10 '13 at 14:28

By Gauss' Lemma $f,g$ remain irreducible in $K[y]$, hence are coprime.
By the way, this easily implies that $V(f,g) \subseteq \mathbb{A}^2$ is finite.
@ricky: It's the same Lemma that tells us that a univariate polynomial that is irreducible in $\mathbb{Z}[x]$ remains so in $\mathbb{Q}[x]$. –  Jyrki Lahtonen Feb 10 '13 at 14:52
So the idea here is that since $k[x,y]$ is the same as $k[x][y]$ we think of $f,g$ as irreducible in $k[x][y]$ which implies irreducible in $k(x)[y]$ where $k[x]$ is the ring playing the role of the integers and $k(x)$ is its field of fractions playing the role of the rationals? –  Steven-Owen Feb 10 '13 at 14:55