I found the following question here.
Let $K $ be an extension field of $k $. Suppose $f(x) \in k[x]$ is irreducible with degree $n$ and let $[K:k]=m$, where $(n,m)=1$. Show that $f$ is irreducible in $K[x]$.
Tentative solution:
For the sake of argument, assume that $f$ is reducible in $K[x]$. Then there exists an $\alpha\in K$ such that $f(\alpha)=0$. Note that $$k\subseteq k (\alpha)\subseteq K .$$ By Tower rule, we have $$[K:k]=[K:k(\alpha)][k (\alpha):k]. $$
In other words, $m=[K:k(\alpha)]\cdot n$. But this contradicts $(n,m)=1$. Hence, $f$ is irreducible over $k $.
But, if $f$ is reducible in $K[x]$, it only means that there exists some irreducible polynomial $p(x)\in K[x]$ such that $p(x)\mid f(x)$. So I don't think my claim
Then there exists an $\alpha\in K$ such that $f(\alpha)=0$
is correct. Also note that the claim will work if $K$ is algebraically closed. As I mentioned above, I don't have full information regarding the problem.
So do I need any further information to complete the solution? If not, is there any possibility of improving this solution?
Hints and alternate solutions are appreciated.
Thank you.
You're right that reducible polynomials need not have roots. For example, take $(x^2 + 1)^2 \in \mathbb R[x]$. However, we may always find a root in some extension of $K$, say the algebraic closure. Indeed, let $\alpha$ be a root of $f$ in an extension of $K$. We would now like to compute $[K(\alpha):k]$.
We have that $[K : k] = m$ by assumption and that $[k(\alpha) : k] = n$ by irreducibility of $f$ over $k$. Then $[K(\alpha) : K] \leq n$ as $\alpha$ is a root of $f$ so its minimal polynomial over $K$ divides $f$ and therefore has lesser degree. Thus, $[K(\alpha) : k] \leq mn$ as $[K : k] = m$ by assumption. Note now that if we prove that this is actually an equality, then we will have shows that $[K(\alpha) : K] = n$. This implies that the minimal polynomial of $\alpha$ over $K$ is degree $n$. As $\alpha$ is a root of $f$, the minimal polynomial of $\alpha$ must divide $f$. As they have the same degree, they must be equal (up to a multiplicative constant), so as the minimal polynomial is irreducible, $f$ must be irreducible.
So it suffices to prove this equality. We can do this by proving the reserve inequality $[K(\alpha) : k] \geq mn$. In fact, we show that $mn \mid [K(\alpha) : k]$. This is where the relative primeness comes into play. As $m$ and $n$ are relatively prime, to prove that $mn \mid [K(\alpha) : k]$ it suffices to prove that $m \mid [K(\alpha) : k]$ and $n \mid [K(\alpha) : k]$. For the first of these, observe that $m = [K : k] \mid [K(\alpha) : k]$. For the second, we have $[k(\alpha) : k] = n$ by irreducibility of $f$ over $k$. Furthermore, $[k(\alpha) : k] \mid [K(\alpha) : k]$ so we are done.