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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.

Jack M
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3 Answers3

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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.

André Nicolas
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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).

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Bill Dubuque
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Let $R$ be the minimal polynomial of the common root. Then $R$ divides every polynomial having this root, esp. $R$ divides both $P$ and $Q$. So $P=RS$, $Q=RT$ with some polynomials $S,T$. As $P$ is irreducible, one of $R,S$ must be a unit and $R$ cannot be a unit. Thus $Q=RTS^{-1}$ shows that $P$ divides $Q$.

Hagen von Eitzen
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