Let L and N be field extension of K.
Tensor product of L and N over K is always field？
If there are counterexamples, I would be appreciated if you could give me an example and explanation.
Tensor product of fields may not be a field
field-theorytensor-products
Related Solutions
As you seem to have realized the concept of a compositum is a bit troubling if we are not working inside a bigger field. Anyway, here come a few quick and dirty answers:
1+2) Yes (up to identification). The tensor product $E\otimes_k F$ is a commutative $k$-algebra. If $I$ is a maximal ideal in there (call upon Zorn's lemma to get the existence), then $M=E\otimes_k F/I$ is a field. The mappings $e\mapsto e\otimes 1 +I$ and $f\mapsto 1\otimes f +I$ are then homomorphisms of $k$-algebras, and thus injective (scores of details to check here). Therefore we can identify their images with $E$ and $F$ respectively. However, the choice of $I$ may make quite a difference. For example, if $E$ and $F$ are isomorphic (think: $\mathbf{Q}(\root 3\of 2)$ and $\mathbf{Q}(\omega\root3\of2)$, $\omega=(-1+i\sqrt3)/2$), then their images are equal for an appropriate choice of $I$, but may intersect trivially for another one.
[Edit: It is possible that the images of $E$ and $F$ never intersect trivially. A trivial intersection occurs in my example case, but clearly not always. Sorry about any possible confusion this error in the original version may have created.]
3) No. This never happens. The tensor product of two non-trivial vector spaces is never zero. You get a basis for the tensor product from pairwise elementary tensors of basis elements of the factor spaces. When we mod out a maximal ideal as above, some linear dependencies may or may not be introduced (the ideal may be zero).
Yes. Take a field $k$. Then $k[[t]]$ is (isomorphic to) the product of infinitely many copies of $k$. If $V = k^\infty$ is an infinite dimensional space, then the canonical map $k[[t]] \otimes_k V \to V[[t]] \simeq \prod_{n=1}^\infty k \otimes V$is not an isomorphism: its image is the subspace of elements of the product such that the coefficients span a finite dimensional subspace of $V$.
Indeed, elements of its image are of the form $x(t) = \sum_{k=1}^d R_k(t) \otimes v_k$ where $R_k$ are power series and $v_k \in V$, and the coefficients of $x$ are in the vector space $\mathrm{Span}(v_1, \dots, v_d)$ which is finite dimensional. On the other hand, if $(e_i)$ is a basis of $k^\infty$, then $\sum_{i=1}^\infty e_i t^i$ is not in the image.
This example is the motivation for the definition of the completed tensor product, by the way.
Best Answer
Given any proper field extension $K \subset L$, the tensor product $L \otimes_K L$ is never a field, because the kernel of the multiplication map $L \otimes_K L \to L$ is a nonzero proper ideal.