An irreducible polynomial cannot share a root with a polynomial without dividing it There is a lemma of Galois stating, "An irreducible equation can have no common root with a rational equation without dividing it".
His definitions are a little bit imprecise, but I think he means:

Let $P, Q\in \Bbb K[X]$, $P$ irreducible. Then if $P$ shares a root with $Q$, $P$ divides $Q$.

His only gesture towards a proof is "Since the greatest common divisor of $P$ and $Q$ would be rational [would be in $\Bbb K[X]$] , therefore, etc.".
Can I have a modern proof, or at least a stronger hint at a proof? All I can see is that if $a$ is a common root to $P$ and $Q$, then $X-a$ would divide them both.

Let $D$ be "the" gcd of $P$ and $Q$. Since $P$ is irreducible, $D$ is either $P$ or $1$. 
We show that $D$ cannot be $1$. If $D=1$, there are polynomials $X$ and $Y$ such that $PX+QY=1$. This is impossible, since any common root of $P$ and $Q$ is a root of $PX+QY$, and the polynomial $1$ has no root. 
