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.

Key Idea $\ $ The gcd $\,(a,b) = c\,$ persists in extension domains when it is linearly representable, i.e. when $\,ar+bs = c,\,$ for some $\,r,s\,$ i.e. when a Bezout identity exists. In particular,  if $\,a,b\,$ are coprime in $D\,$ then they remain coprime in every extension domain $\,E\supset D.\,$ Indeed, if $\,\gcd(a,b) = 1\,$ in $D\,$ then by hypothesis  $\,ar+bs = 1\,$ for some $\,r,s\in D.\,$ Thus if $\,d\mid a,b\,$ in $E\,$ then $\,d\mid ar+bs = 1,\,$ so $\,a,b\,$  remain coprime in $\,E,\,$ i.e. coprimality persists in extensions.
In your case $\,P,Q\,$ are not coprime in an extension domain since they have a common root $\,\alpha\,$ hence  a common factor $\,X-\alpha.\,$ Therefore, by above, $\,P\,$ and $\,Q\,$ are not coprime in $\,D = \Bbb K[X],\,$ i.e. $\,\gcd(P,Q)\neq 1\,$ thus, since $P$ is irreducible, $\,\gcd(P,Q) = P,\,$  hence $\,P\mid Q.$
See this answer for further discussion of the key idea of gcd persistence in extension rings of Euclidean domains (or, more generally, Bezout domains or PIDs).
