# A characterization of finite purely inseparable extensions of fields

Let $K/k$ be a finite extension of fields. Let $A=K \otimes_k K$.

An exercise: Show that $K/k$ is purely inseparable $\Leftrightarrow A/J(A) \cong k$, where $J(A)$ is the Jacobson radical of $A$.

It doesn't seem right to me. I can show that $A/J(A)$ is a product of copies of fields, each containing $K$. So that must mean that $A/J(A) \cong k$ implies $k=K$, making the $\Leftarrow$ direction trivial and $\Rightarrow$ direction wrong.

Am I correct?

Maybe $A/J(A) \cong K$ instead of $A/A(J) \cong k$ would make it correct?

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Where did you get that exercise whose statement is false and which is quite difficult when corrected? Students are sometimes handled roughly here by users who criticize the way they ask questions. Fair enough, but I think that teachers' questions should not be above criticism either... –  Georges Elencwajg Jan 21 '12 at 13:03

The equivalence in the exercise is indeed false: since $A$ is a $K$-algebra (multiply elements of $K\otimes_k K$ by elements of $K$ through the left factor, say) , so is $A/J(A)$ which can thus never equal $k$ if the extension $k\subset K$ is not trivial.
You are also right that for any finite-dimensional extension $k\to K$, purely inseparable or not, $A/J(A)$ is isomorphic to the product $\Pi F_i$ of a finite number (maybe equal to one!) of field extensions $K\to F_i$ (this follows from the Chinese Remainder Theorem).
Notice also that since $A$ is finite-dimensional over $K$, it is an artinian ring, so that the Jacobson radical coincides with the nilpotent radical : $J(A)=Nil(A)$.

The correct implication is (as you suggested) : $$\quad K/k \:\;\text {is purely inseparable} \Leftrightarrow A/Nil(A) =K$$

Edit
Since I find this equivalence rather difficult, I'll upgrade my preceding indications to a concise proof.

$\boxed {\Longleftarrow}$
Notice first that the hypothesis forces $A=K\oplus Nil(A)$.
Consider the separable part of your extension: $k\subset K_{sep }\subset K$.
Since $K\otimes _k K_{sep }\subset K\otimes _k K$ is reduced (separable extensions are universally reduced!) , we must have $K\otimes _k K_{sep }=K \subset K\oplus Nil(A)$ and thus $K_{sep }=k$, so that $k\subset K$ is indeed purely inseparable.

$\boxed {\Longrightarrow}$
It suffices to show that every element of $K\otimes_k K\setminus K\otimes_k k$ is nilpotent.
And for that it suffices to show that for any $b\in K$ the set $K\otimes_k k(b) \setminus K\otimes_k k$ is composed of nilpotents of $K\otimes_k k(b)$.
But $k(b) \simeq k[T]/(T^{p^r}-q) \;\; (q\in k, q=b^{p^r})$ by pure inseparability and so $K\otimes_k k(b) \simeq K[T]/(T^{p^r}-q) =K[T]/(T-b)^{p^r}$
In that last isomorphism the set $K\otimes_k k(b) \setminus K\otimes_k k$ corresponds to the set $K[T]/(T-b)^{p^r} \setminus K$, and this last set is indeed composed of nilpotents. More precisely, it is the nilpotent radical $(\bar T-b)$ of $K[T]/(T-b)^{p^r}$.

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Thank you very much for both parts of the answer! By the way, I meant that $A/J(A)$ is a product of a finite number of fields, not necessarily a product of 2 fields. This is true irregardless of whether $K/k$ is purely inseparable or not. In the purely inseparable case I assume that this finite number turns out to be 1 (this is just to help a future reader of this question and answer). –  Amy Jan 21 '12 at 8:47
Dear @Amy: if you allow products of just one factor, what you say is indeed completely correct . I'll edit my answer in order to suppress my unfair remark, for which I apologize. –  Georges Elencwajg Jan 21 '12 at 9:06
I have also edited my answer into a concise proof of the correct equivalence, since I feel that my preceding "hints" were a bit too cryptic . –  Georges Elencwajg Jan 21 '12 at 13:15
Dear Georges: Thanks! My (perhaps incorrect) argument goes as follows. Let $f$ be a $k$-algebra morphism from $K\otimes_kK$ to $\Omega$. Then there are $k$-algebra morphisms $g$ and $h$ from $K$ to $\Omega$ such that $f(x\otimes y)=g(x)h(y)$ for all $x$ and $y$ in $K$. But, $K/k$ being purely inseparable, we have $g(x)=x$ and $h(y)=y$, implying that the image of $f$ is $K$. –  Pierre-Yves Gaillard Jan 22 '12 at 4:13
Dear @Pierre-Yves, your argument is not only correct, but also very crisp and elegant: congratulations!(Oserais-je une question personnelle: est-ce que vous vous levez très tôt pour travailler ou êtes-vous à l'étranger?) –  Georges Elencwajg Jan 22 '12 at 12:55