Let $L=K(\alpha,\beta)$ be an algebraic field extension, with $\alpha$ separable over K.

Show that $L/K$ is simple.

My attempt:

If we could show that $L/K$ is finite and separable then the claim would follow by the primitive element theorem.

Since $L/K$ is algebraic and finitely generated, it is a finite extension.

Since $\alpha$ is separable the minimal polynomial of $\alpha$ in $\overline{K}$ is of the form:

$m_{\alpha}(X)=(X-\alpha_1)(X-\alpha_2)\cdot....\cdot(X-\alpha_n)$ with pairwise distinct $\alpha_i$ and $\alpha=\alpha_i$ for one $i$.

Then $f_{\alpha,\beta}(X)=(X-\alpha_1)(X-\alpha_2)\cdot....\cdot(X-\alpha_n)\cdot(X-\beta)$ is a separable polynomial with $f_{\alpha,\beta}(\beta)=0$, hence $\beta$ is also separable. The conclusion follows because then $K(\alpha,\beta)$ is separable, since it's generated by separable elements.

Can someone go through it? I also have some questions about the proof:

  1. Our definition of a separable element $\alpha$ is: the minimal polynomial of $\alpha$ is separable. In the proof above I used that $\alpha$ is separable because there exists a polynomial $f$ with $f(\alpha)=0$. Are these two characterizations equivalent?

$(\Leftarrow)$ If $f$ is separable with $f(\alpha)=0$ then $m_\alpha\cdot g=f$, hence $m_\alpha$ can't have a multiple zero, because of the degrees.

But how does the other direction work?

  1. The polynomial $f$ in the proof: is it already the minimal polynomial of $\beta$? If yes, why?

Thanks a lot!

  • 3
    $\begingroup$ $\beta$ doesn't have to be separable. Your polynomial $f_{\alpha,\beta}(X)=m_{\alpha,K}(X)(X-\beta)$ is not in $K[X]$, if $\beta\notin K$. Hence it is not minimal polynomial of $\beta$ over $K$. You can find the proof here: math.cornell.edu/~kbrown/6310/primitive.pdf $\endgroup$
    – SMM
    Jan 17, 2015 at 18:53

1 Answer 1


Your proof is incorrect.

Notice that you have apparently proved that $\beta$ is separable over $K$, when in fact you have not made any initial assumptions on $\beta$. Surely this is a red flag: after all, you could have started with $\beta \not\in K$ and $\beta$ purely inseparable over $K$. Then it is quite impossible for $\beta$ to also be separable over $K$.

Your proof fails because $f_{\alpha,\beta}$ need not be in $K[X]$. In fact, it definitely won't be when $\beta \not\in K$. This also answers (2): it is not the minimal polynomial of $\beta$ over $K$ because it does not lie in $K[X]$ (unless $\beta \in K$, which is a trivial scenario).

Regarding (1): I am not sure what your question is, because you have definitely used the fact that $\alpha$ is separable because its minimal polynomial is separable.

For a full proof of the result, you can see here: math.cornell.edu/~kbrown/6310/primitive.pdf.


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