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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.

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    $\begingroup$ Your claim "Then there exists an $\alpha\in K$ such that $f(\alpha)=0$" is not correct at all. $x^4 - 2$ is irreducible in $\Bbb Q$, but in $\Bbb Q(\sqrt2)$ it factors as $(x^2-\sqrt2)(x^2+\sqrt 2)$, still without any roots. $\endgroup$ – Arthur Feb 1 '19 at 11:38
  • $\begingroup$ Thank you@Arthur. Isn't the claim true when $K $ is algebraically closed? $\endgroup$ – Shivering Soldier Feb 1 '19 at 11:47
  • $\begingroup$ In that case, by definition of algebraically closed, any polynomial would factor into linear polynomials. And linear factors correspond to roots. $\endgroup$ – Arthur Feb 1 '19 at 11:48
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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.

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