# $f(x) \in k[x]$ is irreducible with degree $n$ and let $[K:k]=m$, where $(n,m)=1$. Show that $f$ is irreducible in $K[x]$.

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.

• 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. – Arthur Feb 1 '19 at 11:38
• Thank you@Arthur. Isn't the claim true when $K$ is algebraically closed? – Shivering Soldier Feb 1 '19 at 11:47
• In that case, by definition of algebraically closed, any polynomial would factor into linear polynomials. And linear factors correspond to roots. – Arthur Feb 1 '19 at 11:48

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.