It is clear that if $A$ and $B$ are unital algebras (over a field), then the tensor product $A \otimes B$ is also unital, with the unit being $1_A \otimes 1_B$. I came across an exercise that questions about the converse statement. That is, if $A \otimes B$ is a unital non-zero algebra then $A$ and $B$ must also be unital.

I started by denoting $e$ the unit of $A \otimes B$. We can write $e = \sum_{i=1}^{n} a_i \otimes b_i$, with $n$ being minimal. This minimality implies that $a_1, \cdots, a_n$ and $b_1, \cdots, b_n$ are linearly independent. If we can prove that $n = 1$, then $e = a \otimes b$ is a pure tensor. These elements $a \in A$ and $b \in B$ are the ideal candidates for units in $A$ and $B$, respectively. However, I have not been able to arrive to a contradiction if $n > 1$ using only the basic tools and computations of tensors. Since no properties from $A$ or $B$ are assumed, I do not know what other tools can be used in this generality.

Note: Said exercise can be found in Introduction to Noncommutative Algebra by M. Bresar, chapter 4, page 104.

  • $\begingroup$ I forgot to mention, but this is my first question in M.SE, the first of many I hope. $\endgroup$ – solid Feb 12 '16 at 3:43
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    $\begingroup$ False. $0 \times A \cong 0$ is always unital, no matter what $A $ is. Maybe you want $A $ and $B $ to be nonzero? $\endgroup$ – darij grinberg Feb 12 '16 at 4:27
  • $\begingroup$ I think you mean $\otimes$ instead of $\times$, but the point is $A \otimes B$ being non-zero. Otherwise it works just as you say and it is not very interesting. I will edit the question. $\endgroup$ – solid Feb 12 '16 at 12:09
  • $\begingroup$ Oops, yes, I meant the tensor product (I think this was an autocorrect I failed to spot). Given the context of this exercise in Bresar's book, I suspect a mistake by the author (since the surrounding exercises are rather simple). $\endgroup$ – darij grinberg Feb 12 '16 at 21:16

I was able to find a positive answer thanks to my lecturer who, as far as I know, is not present on M.SE. I will write a proof based on the idea he gave me, which is his credit.

Let $A$, $B$ be algebras over a field $F$ and suppose $A \otimes B$ is nonzero and unital. Since $A \otimes B$ is nonzero, then both $A$ and $B$ are nonzero. Take $0 \neq x \in A$ and $y \in B$ arbitrary elements. Write $e = \sum_{i=1}^{n} a_i \otimes b_i$ (as I did on the question). Since $e$ is the unit in $A \otimes B$, we have

\begin{align} x \otimes y = (x\otimes y)e = \sum_{i=1}^{n} xa_i \otimes yb_i \tag{1}. \end{align}

Furthermore, since $x \neq 0$, we can expand $x$ to a basis of $A$ and hence, write $xa_i = \lambda_ix + v_i$, where $\lambda_i \in F$ and $v_i$ is linearly independent with $x$. Plugging this to $(1)$, we get

\begin{align} x \otimes y = \sum_{i=1}^{n} (\lambda_i x +v_i)\otimes yb_i = x \otimes\sum_{i=1}^{n} \lambda_i y b_i + \sum_{i=1}^{n} v_i \otimes yb_i \tag{2}\end{align}

or equivalently, after arranging terms

\begin{align} x \otimes (y - \sum_{i=1}^{n}\lambda_i y b_i)= \sum_{i=1}^{n} v_i \otimes yb_i \tag{3}.\end{align}

Since $x$ is linearly independent with each $v_i$, we conclude that

\begin{align} y = \sum_{i=1}^{n}\lambda_i y b_i = y (\sum_{i=1}^{n} \lambda_ib_i) \tag{4}\end{align}

(via a result on linear independence of tensor products, for instance, Lemma 4.8 in Bresar).

Analogously, using $x \otimes y = e(x \otimes y)$, we can conclude that $y = (\sum_{i=1}^{n} \lambda_ib_i)y$ and since $y \in B$ is arbitrary, this means that $\sum_{i=1}^{n} \lambda_ib_i$ is the unit of $B$.

The same argument works for proving that $A$ has unit as well.

  • $\begingroup$ The argument is very elementary and computation-based. The thought that eluded me was writing each $xa_i$ as linear combination of $x$ and something else. It goes straight-forward onwards. This proof works but I wonder if there are other ways, more conceptual and less about playing with tensors. $\endgroup$ – solid Feb 13 '16 at 17:26
  • $\begingroup$ I don't think you want to say "$v_i$ is linearly independent with $x$"; your $v_i$ could be $0$ for all we know. I guess the correct wording is "$v_i \in Y$", where $Y$ is some (chosen for once) complement to the $F$-linear span of $x$ in $A$. Nice proof! $\endgroup$ – darij grinberg Mar 17 '16 at 4:04
  • $\begingroup$ Once again, you are right. The philosophy is to choose those elements $v_i$ in a fixed complement of the span of $x$. Then, each $v_i$ is linearly independent with $x$ if and only if $v_i$ is non-zero. $\endgroup$ – solid Mar 22 '16 at 14:34

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