Tensor product of field extensions Let $K$ be a field of characteristic 0 and $L$ a finite extension of $K$. Denote by $m$ the natural multiplication map from $L \otimes_K L$ to $L$. Denote by $I$ the kernel of the morphism $m$. Is $I$ nilpotent?
 A: In general, if $A$ is a field of characteristic $0$ and $B/A, C/A$ are two finite extensions of $A$, then the tensor product $B \otimes_A C$ is isomorphic to the product (i.e. direct product of $K$-algebras) of all "composita" (plural of "compositum") of $B/A$ and $C/A$. Here a "compositum" of the two extensions is a finite extension $D/A$ together with injections $B \hookrightarrow D$ and $C \hookrightarrow D$, which make an obvious commutative diagram, and such that $D$ is generated by the images of $B$ and $C$.
In your case, there is a trivial compositum of $L$ and itself, namely $L$. Let $d$ denote the degree $[L:K]$. The algebra $L \otimes_K L$ has dimension $d^2$ over $K$, and the trivial compositum has dimension $d$ over $K$. Therefore, if $L$ is not equal to $K$, then there exist necessarily composita other than the trivial one.
If one writes $L \otimes_K L = L \times (\textit{product of other composita})$, then the morphism $m: L \otimes_K L \rightarrow L$ coincides with the "projection to the first summand".
From this, it is clear that the kernel $I$ is not nilpotent, whenever $L$ is not equal to $K$. In fact, it is a product of some field extensions of $K$.

A simple example: take $K = \mathbb{R}$ and $L = \mathbb{C}$. Then we have the isomorphism: $$ \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathbb{C} \times \mathbb{C},$$ the map being given by $a \otimes b \mapsto (ab, a\overline{b})$. The morphism $m$ is then the projection to the first summand $\mathbb{C}$, and the kernel is the whole second summand $\mathbb{C}$, which is certainly not nilpotent.

UPDATE: I attach here a proof of the claim in the first paragraph.
By primitive element theorem, we may choose a primitive element $b\in B$ so that the ring homomorphism $A[X]\rightarrow B$ sending $X$ to $b$ induces an isomorphism $A[X]/f \simeq B$, where $f\in A[X]$ is the minimal polynomial of $x$ over $A$.
Note that $f$ only has simple roots, as the extension $B/A$ is separable. This means that, in the ring $C[X]$, $f$ decomposes as $g_1\dots g_r$ with distinct irreducible polynomials $g_1, \dots, g_r \in C[X]$.
Therefore we have $B\otimes_A C\simeq \prod_{i = 1}^r D_i$, where $D_i = C[X]/g_i$ is a field. Note that each $D_i$ is a compositum of $B$ and $C$: the embedding $C\hookrightarrow D_i$ is clear, and the embedding $B\hookrightarrow D_i$ is induced by the natural map $A[X] \rightarrow C[X]$. It is clear that the images of $B$ and $C$ generate $D_i$.
It remains to show that:

*

*Every compositum $D$ of $B$ and $C$ is isomorphic to one such $D_i$ (isomorphic in the sense of composita, i.e. $D$ isomorphic to $D'$ if there exists a field isomorphism $D \simeq D'$ which makes a big commutative diagram);

*For different indices $i, j$, the composita $D_i$ and $D_j$ are not isomorphic.

For 1: Take a compositum $D$ and look at the image of $b$ (the primitive element that we have chosen) in $D$. Let's call the image $d$ and look at the minimal polynomial $g$ of $d$ over $C$.
Since $f(d) = 0$, we know that $g$ must be equal to some $g_i$. Hence the homomorphism $C[X] \rightarrow D$ sending $X$ to $d$ induces a homomorphism $D_i \rightarrow D$, which (as both sides are fields) must be injective. It is also surjective, because $D$ is generated over $A$ by $d$ and $C$, by definition of compositum.
Therefore we get an isomorphism $D_i \simeq D$ and it is easy to verify the commutative diagram.
For 2: If we have an isomorphism $D_i \simeq D_j$ which makes a commutative diagram, then the image of $b$ in $D_i$ (call it $d_i$) is mapped via this isomorphism to the image of $b$ in $D_j$ (call it $d_j$).
This means that the minimal polynomials of $d_i$ and $d_j$ over $C$ should be the same. However, by definition, these minimal polynomials are $g_i$ and $g_j$, respectively, which are different.

Remark: Although we have made the choice of a primitive element in the proof, the isomorphism $B\otimes_A C\rightarrow \prod D$ is canonical and does not depend on such a choice. In fact, for each $D$, the canonical injections $B\hookrightarrow D$ and $C\hookrightarrow D$ induce a homomorphism $B\otimes_A C \rightarrow D$ and these pack together to give a homomorphism $B \otimes_A C \rightarrow \prod D$.
