Prove that two polynomials $f,g \in k[X]$ have a common root in $\bar k$ iff $f,g$ have a common factor in $k[X])$.

Question

I am here giving the source of the problem how I was stuck, if someone want to skip this history then please concentrate on the heading:

I was reading the book of Milne on elliptic curves. At page $16$ in the proof of Proposition $1.24$ part b he has written that

We know that $r=0$ iff $f$ and $g$ have a common factor in $k(X)[Y]$

Now I tried to elaborate this part a bit, first of all, all the readers might have a look at the definition of resultant at page no $14$ of the same book and the Proposition $1.22$ at page no $15$.

Proposition $1.22$ at page no $15$ states

The resultant $\operatorname{Res}(f,g)=0$ iff

a) both $s_0$ and $t_0$ are zero; or

b)the two polynomials have a common root in $\bar k$(equivalently, a common factor in $k[X])$.

Now if you see the proof of Proposition $1.22$, you will find out that the part

the two polynomials have a common root in $\bar k$

is proved but why it is equivalent to

a common factor in $k[X]$

this part is missing. If someone could give me a proof of this, it would be of great help.

Answer 1

Because both are equivalent to the assertion that the resultant of $f$ and $g$ is $0$.

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Another approach is this one: if they have a common root $\alpha\in\overline k$, then their greatest common divider cannot be $1$, since it must be a multiple of $x-\alpha$. But the greatest common divider is polynomial with coefficients in $k$. So, $\gcd\bigl(f(x),g(x)\bigr)$ is a common factor and it is not a constant polynomial. On the other and, if $f(x)$ and $g(x)$ have a common factor $h(x)\in k[x]$ which is non-constant, then $h(x)$ has a root $\alpha\in\overline k$, which will be a root of both $f(x)$ and $g(x)$.

Answer 2

Hint: If they have common root $a$, consider the Galois group of $G$ $[k(a):k]$, you have for every $g\in G, f^g(a)=f(g(a))=0$, this implies that the $g(a)$ is a root of $a$ for every $g\in G$, we deduce that the minimal polynomial of $a$ divides $f$ and $g$.