6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.