Let $E/F$ a field extension. If $x\in E$ is separable over $F$ is $E/F$ separable?

Question

Let $E/F$ a field extension. If $x\in E$ is separable over $F$ is $E/F$ separable ?

I would say yes since the fact that $x$ separable over $F$ implies $E(x)/F$ separable, an since $E=E(x)$ then $E/F$ would be separable. But I'm not sure if this argument is valid.

Answer 1

The answer is more or less contained in the comments by Lubin and Menag. I simply give a specific example of a situation they described.

Let $F=\Bbb{F}_p(x)$ be the field of rational functions with coefficients in $\Bbb{F}_p$. Of course, $p$ is a prime number. For convenience I assume that $p>3$. Let then $y$ be a solution of $y^2=x^3+1$. Then the minimal polynomial of $y$ is $T^2-(x^3+1)\in F[T]$. Its zeros are $\pm y$. These are distinct, so $y$ is separable. It follows that the extension $F[y]/F$ is separable.

But if you do not assume that $E=F[y]$, then we have the following possibility. We may have $E=F[y,z]$, where $z$ is a zero of $z^p=x$. Then $E/F$ will not be separable even though there are separable elements in $E$.

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OTOH if you assume that $E=F[x]$ is an algebraic extension such that $x$ is separable over $F$, then the extension $E/F$ is separable itself. IOW all the elements of $E$ have separable minimal polynomials over $F$. This follows from the fact that separable elements of an algebraic extension form an intermediate field. See, for example, Section 8.7 in Jacobson's Basic Algebra II for the details.

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An algebraic extension $E/F$ is called purely inseparable if no element from $E\setminus F$ has a separable minimal polynomial. When this happens all the elements $z\in E$ satisfy $z^{p^n}\in F$ for some positive integer $n$.

Answer 2

Consider $E=\mathbb{Q}(2^{1/6})$, which is clearly not separable, since it's a real field and the minimal polynomial $x^6-2$ has non-real roots. However, $\sqrt{2}\in E$ and is separable, since all the roots of $x^2-2$ are in $E$. Thus the claim is false.

What's going on here is that the selected element isn't necessarily enough to generate the whole field; it might generate a subfield. If you add the hypothesis that $F(x)=E$, then it works.