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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.

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    $\begingroup$ 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? $\endgroup$ Jun 7 '12 at 20:37
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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 $f$?

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

Suppose on contrary $f $ is reducible over $K$ So $\exists u\in K$ such that $f(u)=0$

$F\subset F(u)\subset K$

By Tower theorem $[F(u):F]|[K:F]$ i.e $n|m$ which is contradiction

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    $\begingroup$ the degree of $f$ could not be $2$ or $3$. $\endgroup$
    – Primavera
    Nov 17 '19 at 14:44

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