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Let $L/K$ be a field extension and $\alpha,\beta \in L$. Let $f\in K[X]$ be the minimal polynomial of $\alpha$ and $g\in K[X]$ be the minimal polynomial of $\beta$

Show the following:

$$f \text{ is irreducible over } K[\beta] \iff g \text{ is irreducible over } K[\alpha]$$

I actually don't know how to start here, so I will be thankful for any kind of help :)

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    $\begingroup$ What you mean, probably is that «$f$ is irreducible over $K[\beta]$ iff $g$ is irreducible over $K[\alpha]$», which is not quite what you wrote. In particular, «$f\in K[\beta]$ is irreducible» does not mean anything. $\endgroup$ Jan 14, 2015 at 19:55
  • $\begingroup$ Maybe he could say that $f$ is an irreducible element of $K[\beta][X]$. $\endgroup$ Jan 14, 2015 at 19:59
  • $\begingroup$ I will correct it in a second, its exactly what you mean :) $\endgroup$
    – Marc
    Jan 14, 2015 at 19:59

1 Answer 1

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Suppose $f(x)$ is irreducible in $K(\beta)[x]$. We claim that $f(x)$ is the minimal polynomial of $\alpha$ over $K(\beta)$. For clearly $f(\alpha)=0$. Therefore, the minimal polynomial of $\alpha$ over $K(\beta)$ divides $f(x)$. Since $f(x)$ is irreducible over $K(\beta)$, the claim follows.

Thus $[K(\beta,\alpha):K(\beta)]=\deg f$. This gives $[K(\beta,\alpha):K(\beta)][K(\beta):K]=\deg f\deg g$, and thus $[K(\alpha,\beta):K]=\deg f\deg g$.

Frome here we get $[K(\alpha,\beta):K(\alpha)][K(\alpha):K]=\deg f\deg g$.

But $[K(\alpha):K]=\deg f$.

Thus $[K(\alpha,\beta):K(\alpha)]=\deg g$.

Since $\beta$ satisfies $g(x)\in K(\alpha)[x]$, we must have that the minimal polynomial of $\beta$ over $K(\alpha)$ is $g(x)$.

The other direction is symmetrical.

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