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

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.

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 $n$.
