# Show that the algebraic extension $K(\alpha,\beta)/K$ is simple if $\alpha$ is separable over $K$ [duplicate]

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!

## marked as duplicate by Brahadeesh, YuiTo Cheng, José Carlos Santos proof-verification StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 1 at 18:09

• $\beta$ 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 – SMM Jan 17 '15 at 18:53

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.