I know that there is a similar question here but I failed to understand the hints.

My thoughts:

i) Every finite extension $E$ of a finite field $F$ is a simple extension of $F$.

ii) (Primite Element Theorem) Let $E$ be a finite separable extension of a field $F$. Then there exists an $\alpha\in E$ such that $E=F(\alpha)$. This imples that every finite extension is simple in a field of characteristic $0$ because every polynomial in a field of characteristic $0$ is separable.

So we're looking for an infinite field with prime characteristic.

When I think about a prime characteristic field I think of the Galois fields, but those are finite. There's an easy way to provide an example of an infinite prime characteristic field if we take an uncountable set of indeterminates - for instance, $\mathbb{Z}_{2}[S]$, where $S$ is the set of indeterminates and whose cardinality is uncountable. However, I guess this idea isn't of great use here. It seems I have to use a suitable quotient between polynomial rings over Galois fields. My first problem is that I can't make sense of this intuitively. What's the intuition behind taking the quotient of "finite things" to make them infinite?



For a prime $\;p\;$ , take $\;\Bbb F_p(x,y)/\Bbb F_p(x^p,y^p)\;$

=== Show that $\;\left[\Bbb F_p(x,y^p):\Bbb F_p(x^p,y^p)\right]=p\;$ --- Further idea: use the fact that we're in characteristic $\;p\;$ and study a little $\;t^p-x^p\in\Bbb F_p(x^p,y^p)[t]\;$

=== After using symmetry, show that $\;\left[\Bbb F_p(x,y)\,:\,\Bbb F_p(x^p,y^p)\right]=p^2\;$ .

=== Finally, use that $\;\left(\Bbb F_p(x,y)\right)^p\subset\Bbb F_p(x^p,y^p)\;$ to deduce the extension cannot be simple.


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