I'm having a hard time showing this:

If $K$ is an extension of $\mathbb{Q}$ with degree $m$ and $f(x)$ an irreducible polynomial over the rationals with degree $n$, such that $\gcd(m, n)=1$, then $f(x)$ is irreducible over $K$.

I have tried it by writing $f(x)=a(x)b(x)$ and then looking at the coefficients of those polynomials (some of them must belong to $K-\mathbb{Q}$ which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime.

Any help would be appreciated.

  • 2
    $\begingroup$ You're missing a hypothesis. $f$ needs to be irreducible over $\mathbb{Q}$ first, or else it could just be a product of linear rational factors. With this extra hypothesis, consider the extension of $\mathbb{Q}$ obtained by adjoining a root of $f$. What is its degree? $\endgroup$ – Qiaochu Yuan Jun 7 '12 at 20:37

Let $u$ be a root of $f(x)$. Then $[\mathbb{Q}(u):\mathbb{Q}] = n$. Now consider $K(u)$. We know that $[K(u):K]\leq n$, since the monic irreducible of $u$ over $K$ divides $f$. Since $$[K(u):\mathbb{Q}] = [K(u):K][K:\mathbb{Q}] = m[K(u):K]\leq mn$$ and $$[K(u):\mathbb{Q}] = [K(u):\mathbb{Q}(u)][\mathbb{Q}(u):\mathbb{Q}] = n[K(u):\mathbb{Q}(u)]$$ then $[K(u):\mathbb{Q}]$ is at most $nm$, is a multiple of $n$, and is a multiple of $m$. Since $\gcd(n,m)=1$, it follows that $[K(u):\mathbb{Q}]=nm$, so the degree $[K(u):K]$ is equal to $n$.

What does that imply about the monic irreducible of $u$ over $K$ and about $f$?

  • $\begingroup$ It implies that the monic irreducible poly of u over K is equal to $f$ right? $\endgroup$ – PaulDavis Apr 29 '18 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.