# Representations of tensor products of algebras

For two associative unital algebras $$A$$ and $$B$$, defined over $$\mathbb{K} = \mathbb{R}, \mathbb{C}$$, is it possible to have an irreducible representation of $$A \otimes_{\mathbb{K}}B$$ which is not of the form $$V \otimes W$$, where $$V$$ is a representation of $$A$$ and $$W$$ is a representation of $$B$$?

• Yes, very easily. For example, $(V \otimes W) \oplus (V \otimes W)$ does not in general factorize as $V' \otimes W'$. Did you want some irreducibility condition? Jan 15 '20 at 14:39
• Yes, I want irreducibility. I have now written this. Jan 15 '20 at 14:42
• K=R, A=B=C is a standard example where C/R is any nontrivial field extension. Jan 15 '20 at 14:51
• @PeterMcNamara Why writing answers in the comment section? (I ask this on MO since 10 years ...) Jan 16 '20 at 12:56
• @MartinBrandenburg, when I do something like that, it's because I suspect that the question might change. For example, probably the question should have been not just about irreducible representations (as @‌MarkWildon suggested) but about absolutely irreducible representations (as indicated in @‌Mare's answer). Jan 16 '20 at 18:13

In case your two algebras $$A,B$$ are finite dimensional and the field is algebraically closed (or more generally the two algebras are split over the field), then all simple modules over $$A \otimes_K B$$ are indeed of the form $$V \otimes_K W$$ for a simple $$A$$-module $$V$$ and a simple $$B$$-module $$W$$.

This is not true when the algebras are not split: Let $$K= \mathbb{R}$$ and $$A=B=\mathbb{C}= \mathbb{R}[x]/(x^2+1)$$.

Then $$A \otimes_K B= \mathbb{C}[x]/(x^2+1)=\mathbb{C}[x]/(x+i) \times \mathbb{C}[x]/(x-i)=\mathbb{C} \times \mathbb{C}$$.

Thus $$A \otimes_K B$$ has a simple modules of $$K$$-dimension two, while all non-zero $$A \otimes_K B$$-modules of the form $$V \otimes_K W$$ have $$K$$-dimension at least 4.

• What does it mean for an algebra to be split? Isomorphic to a direct power $K^{\oplus n}$? Jan 16 '20 at 18:14
• Also, note that @PeterMcNamara gave a variant of your counterexample in the comments. Jan 16 '20 at 18:15
• @LSpice It means that the algebra modulo its jacobson radical is a products of matrix rings over the field $K$. What you say would be split and basic. Being split is also often called elementary.
– Mare
Jan 16 '20 at 18:28
• What happens in the infinite dimensional case? Jan 19 '20 at 15:28