Let $K/k$ be a field extension.Show that if $[K:k] = p$ with p prime and $f(x)\in k[x]$ has degree $p+1$ then $f(x)$ has a root in $K$ $\iff$ $f(x)$ has a root in $k$.
My attempt : $[K:k] = p = [K:k(\alpha)] [k(\alpha):k]$. Since p is prime we have that $[K:k(\alpha)] =1$ or $[k(\alpha):k]=1$ ,which implies that $K=k(\alpha)$ or $k(\alpha)=k$ .
I also know that $[k(\alpha):k]$ is equal to the degree of the minimal polynomial of $\alpha$ over $k$ that is $\leq p+1$ .
I am not seeing how to proceed with this. Any hints ?
Best Answer
If $f(x)$ has a root in $k$ then it is in $K$ in particular.
To prove the other implication use what you tried. In that case, $K=k(\alpha)$ for $\alpha$ a root in $K$ but not in $k$. You know that the minimal polynomial $m(\alpha, k)$ is such that $m(\alpha, k)| f$. You also know that the minimal polynomial is irreducible. That is the factorization of $f$ in $k$ is the following, $$f = m(\alpha, k) (x-\beta) $$ as the degree of $f$ is exactly $p+1$ and the degree of the minimal $m(\alpha, k)$ is exactly $p$ because that's the order of the extension. This means that $\beta \in k$ so that $f$ has a root in $k$. We have proved that if $f$ has a root in $K$ then it has one in $k$.
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