# If $E/k$ is not separable, but satisfies the property that there exists a positive integer $n$ for which every $\alpha \in E$ of degree less than $n$

In chapter V of Lang's graduate algebra, there is a Lemma 1.7 saying that if $E/k$ is separable and satisfies the property that there exists a positive integer $n$ for which $\alpha \in E$ has degree $\leq n$ over $k$, then we have $i)$ $[E:k]$ is finite and $ii)$ $[E:k]\leq n$

This lemma gives me a further question, is the conclusions of the lemma are still true if one drops the hypothesis that $E$ be separable over $k$?

I think it cannot hold any longer but I cannot get the counterexamples. I think perhaps I can get an example that $[E:k]$ is finite but strictly greater than $n$, and another one could be where $[E:k]$ is infinite, but I cannot think of anyone.

Any hints and detailed explanations are highly appreciated!!!!!

First, remember the classic example of a non-separable extension, $\mathbb{F}_p(\sqrt[p]{X})/\mathbb{F}_p(X)$. It's non-separable because the minimal polynomial for $\sqrt[p]{X}$ has repeated roots: $T^p-X=(T-\sqrt[p]{X})^p$. Observe that the $p$th power of any element of $\mathbb{F}_p(\sqrt[p]{X})$ is in $\mathbb{F}_p(X)$, because the Frobenius map is a homomorphism.

However, this isn't so surprising, since that extension is just degree $p$. So to make something more like what you're looking for, an example would be $\mathbb{F}_p(\sqrt[p]{X_1},\sqrt[p]{X_2},\ldots)/\mathbb{F}_p(X_1,X_2,\ldots)$, an extension which still has the property that the $p$th power of every element in the top field lies in the bottom field (again because the Frobenius map is a homomorphism), but which is of infinite degree.

Maybe this is the example you're thinking of?

• Yes!!! Could you also give me an example where the degree is finite but greater than $n$? – JacobsonRadical Mar 25 '18 at 17:01
• Certainly! We can just modify the same idea, consider $\mathbb{F}_p(\sqrt[p]{X},\sqrt[p]{Y})/\mathbb{F}_p(X,Y)$, which is of degree $p^2$. – Zev Chonoles Mar 25 '18 at 17:06
• OMG! Thank you so much! – JacobsonRadical Mar 25 '18 at 17:09
• Could I ask you the question please: I see that irreducible polynomial should divide $X^p-a$. But why it should be equal to that? – ZFR Mar 25 at 19:16