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Let $f(x)$ be an irreducible polynomial of degree $6$ over field $K$.

If $L$ is a field extension of $K$ and $[L: K]=2$ then show that either $f$ is irreducible or $f$ factors into irreducible polynomials of degree $3$ over $K$.

Attempt:

If $f$ is irreducible then we are done.

Otherwise $f$ factors into either $1+5$ or $2+4$ or $3+3$ degree polynomials.

I am unable to derive a contradiction for the first two cases.

Please give some hints.

Community
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Learnmore
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2 Answers2

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Proof that $f$ can't have a linear factor in $L$: Assume for contradiction that it has one, and let that linear factor be $x - \alpha$. Then $\alpha \in L$, and $f$ is the minimal polynomial of $\alpha$ over $K$. Can you get a contradiction from this?

The 2+4 works similarily, except we take the 2 factor, and find its roots (which might not be in $L$, but at the very least are contained in a degree 2 extension of $L$, which is then a degree 4 extension of $K$). Let $\beta$ be one of those roots, and derive a contradiction like above.

Arthur
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You might want to know the following general fact:

Let $F_1/F$ be field extension where $F_1$ is the splitting of some $f(x)\in F[x]$. For an irreducible $g(x)\in F[x]$, if $$g(x) = g_1 (x) g_2 (x) ... g_r (x) \quad \quad \text{ in } F_1[x]$$ where each $g_i$ is irreducible in $F_1[x]$, then $\deg g_1 = \deg g_2 = \cdots = \deg g_r$.

In your example, $[L:K]=2$, so $L$ is a splitting field over $K$, the above result immediately addresses your concern.

pisco
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  • From where did you get this ? i.e books/reference? – Learnmore Sep 01 '17 at 16:29
  • It is an exercise from Advanced Modern Algebra by Rotman. The book suggest to prove it by constructing some automorphisms, although it can also be proven via symmetric polynomials. – pisco Sep 01 '17 at 16:31
  • You can also search the site, and you may find this. There they do assume Galois rather, so normal and separable rather than just normal. – Jyrki Lahtonen Sep 01 '17 at 22:34