Let $P(X),Q(X)\in K[X]$, with $P(X)$ irreducible. Suppose there is $\alpha$ in an extension of $K$ such that $P(\alpha)=Q(\alpha)=0$.

Question

"Let $P(X),Q(X)$ $\in K[X]$, with $P(X)$ irreducible. Suppose there is $\alpha$ in an extension of $K$ such that $P(\alpha)=Q(\alpha)=0$. Show that Q(X) is multiple of P(X). If $Q(x)$ is also irreducible, then $P(X)$ and $Q(X)$ are associated."

right, if $\alpha$ is the root of $P(X)$ and $Q(X)$, then there are $t_{1}(X)$ and $t_{2}(X)$ in $K[X]$ so that $$P(X)=(x-\alpha)t_{1}(X)$$ and $$Q(X)=(x-\alpha)t_{2}(X)$$ And from here, I can't show that $P(X)|Q(X)$. Someone can help me?

Answer 1

In the extension $F$, $X-\alpha$ is a common divisor of $P$ and $Q$ and therefore a divisor of the GCD $P\wedge Q$. Since the GCD is the same computed in $K[X]$ and $F[X]$, we see that $P$ and $Q$ have a common factor in $K[X]$.

But of course, since $P$ is irreducible, this means that $P|Q$.

Answer 2

Hint: The point is that $Q$ is a multiple of $P$ in $K[X]$. Write $Q=AP+R$ with $R=0$ or $\deg R < \deg P$. Then $R(\alpha)=0$. Now $P$ is irreducible and $P(\alpha)=0$. What does that imply about the degree of $\alpha$ over $K$?