$K_{1} \otimes K_{2}$ has no nonzero nilpotent elements.

Let $K_{1}$, $K_{2}$ be two algebraic extension of the field $F$ contained in the field $L$. Prove that the F-algebra $K_{1}\otimes K_{2}$ has no non-zero nilpotent elements? What if L has characterestic 0?

I don't have a clue. If $K_{1}, K_{2}$ are finite extensions of $F$ and $[K_{1}K_{2}:F]=[K_{1}:F][K_{2}:F]$, then $K_{1} \otimes K_{2} \cong K_{1}K_{2}$ as F-algebras. Hence $K_{1}\otimes K_{2}$ being a field will not contain any nilpotent elements except 0. But what about the other cases?

• @JefLaga beat me to it... Are you missing separability hypotheses? – peter a g Jun 1 '17 at 14:59
• You should probably indicate that you edited your question, I assume in response to the answer below - otherwise it makes the answer look a bit out of place... edit: Oh I see you did in your comment below – peter a g Jun 1 '17 at 15:21
• Yeah I actually did say in the comment below.. – Riju Jun 1 '17 at 15:22

(Edit: answer given before characteristic 0 assumption added)

This is not true. Let $F = \mathbb{F}_p(t)$, $K_1$ the algebraic closure of $F$ and $K_2 = F(\alpha) \subset K_1$ with $\alpha$ a root of the irreducible polynomial $X^p-t$. Then $K_2 = F[X]/(X^p-t)$ and $$K_1\otimes K_2 \cong K_1 \otimes F[X] /(X^p-t) \cong K_1[X]/(X^p-t) = K_1[X]/(X^p-\alpha^p) \cong K_1[X]/(X-\alpha)^p$$ And this last algebra contains the nilpotent $(X-\alpha)$.

(Edit: answer after characteristic 0 assumption added)

Suppose we would have an element $\sum x_i\otimes y_i$ of $K_1\otimes K_2$ which is nilpotent. Since it is a finite linear combination of elementary tensors, it can also be seen as an element of $L_1\otimes L_2$ where $L_i\subset K_i$ are finite extensions of $F$. Here we use the fact that the natural map $L_1\otimes L_2 \hookrightarrow K_1\otimes K_2$ is injective. So we only need to prove it when $L_1$ and $L_2$ are finite. This has already explained by another answer and can also be found for example here: http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/separable2.pdf

• Of course one can also take $L= K_1=K_2$. – peter a g Jun 1 '17 at 15:02
• @Jef Laga Sorry for the inconvenience.. I have edited the question.. but you can keep your answer .. Thanks!!! – Riju Jun 1 '17 at 15:19

The answer to your question is contained in the following general lemma (see e.g. Cassels-Fröhlich, chap. II, §9): keeping your notations $F, K_1, K_2$ and assuming that $K_2 /F$ is finite separable, one can show that $K_1 \otimes K_2$ is isomorphic to the direct sum of a finite numbers of fields $L_j$ which can be viewed as extensions of both $K_1 , K_2$. For details, see op. cit., but the principle of proof is simple enough: by hypothesis, $K_2$ is of the form $F(\alpha)$; let $f(X)\in F[X]$ be the irreducible polynomial of $\alpha$ over $F$, and decompose $f(X)$ into $\prod_j g_j (X)$, where the $g_j$'s are distinct irreducible in $K_1 [X]$; then $L_j \cong K_1 [X]/(g_j)$. The direct sum of the $L_j$'s may have divisors of zero, but it contains no non zero nilpotent element.

Note that this is not contradictory with the answer of @Jef Laga because of our additional separability hypothesis on $K_2$ .