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Let $K$ be a field, $f(X)\in K[X]$ be a polynomial of prime degree. Assume that for all extension $L$ of $K$, if $f$ has roots in $L$ then $f$ splits over $L$. Prove that either $f$ is irreducible over $K$ or $f$ splits over $K$.

How do I prove it?

Thanks a lot.

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If $f$ has a root in $K$ then $f$ splits over $K$. If not, assume that $f$ is reducible over $K$, take an irreducible factor of $f$ of minimal degree and $a$ a root of this factor. If $L=K(a)$ then any other irreducible factor of $f$ has a root (in fact all roots) in $L$. Now use the degree of these extensions and the prime degree of $f$ to finish the proof.

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The hypothesis implies that all irreducible factors$~f_i$ of$~f$ have the same splitting field, which coindices with their rupture field $K[X]/(f_i)$. This implies in particular that all $f_i$ have the same degree, and it therefore divides $\deg(f)=p$, so one can conclude.

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