This is from Rotman's Advanced Modern Algebra, Part I.

A positively graded algebra $A = \bigoplus_{p \geq 0} A_p$ over a commutative ring $R$ is

  • anticommutative if $ab = (-1)^{pq}ba$ for $a \in A_p, b \in A_q$,

  • alternating if $a^2 = 0$ for all $a \in A_p$ with $p$ odd.

Let $\mathsf{C}$ be the category of anticommutative alternating graded $R$-algebras with morphisms being graded $R$-algebra homomorphisms of degree $0$, that is, the ones which preserve grading: $\phi(A_p) \subseteq B_p$.

Rotman claims that a tensor product of such algebras is a coproduct in $\mathsf{C}$.

Let $A = \bigoplus_{p \geq 0} A_p$ and $B = \bigoplus_{p \geq 0} B_p$ be alternating anticommutative graded $R$-algebras. My problem is determining the coproduct maps $\rho\colon A\to A\otimes_R B$ and $\sigma\colon B\to A\otimes_R B$. The maps inherited from the category of commutative rings, that is, $a \mapsto a\otimes 1$ and $b\mapsto 1\otimes b$ don't seems to work here. My another idea was to use $\rho\colon\sum_{p \geq 0} a_p \mapsto \sum_{p \geq 0} (-1)^p a_p\otimes 1$ and $\sigma\colon\sum_{p \geq 0} b_p\mapsto \sum_{p \geq 0} (-1)^p 1\otimes b_p$, but this appers to don't lead anywhere.

So what are those maps? Also, I'm not sure if the assumption of algebras being alternating is necessary. Rotman advises to use anticommutativity and says nothing being another condition.

  • $\begingroup$ Does this question have an answer here: math.stackexchange.com/questions/984151/… ? $\endgroup$ – Matt Aug 27 at 21:19
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    $\begingroup$ Why doesn't your first attempt work ? Recall that with the Koszul sign rule multiplication in $A\otimes B$ is $(a\otimes b)(a'\otimes b') := (-1)^{|a'||b|}(aa')\otimes (bb')$ (with this your first guess seems to work out fine, and indeed anticommutativity - usually called graded commutativity - in $C$ (the codomain) is crucial) $\endgroup$ – Max Aug 27 at 22:01
  • $\begingroup$ @Max Oh, it appears that my misunderstanding stemmed from not understanding what is a tensor product of two anticommutative graded algebras is. Could you tell what the grading is? Is it $(\bigoplus_{i + j = n} M_i\otimes_R N_i)$? $\endgroup$ – Jxt921 Aug 28 at 7:27
  • $\begingroup$ Yes, that would be the natural grading. Then you can check the following equalities mod 2 : $pq' + pp' + qq' = q' (p+q) + p'(p+q) + p'q = (p+q)(p'+q') + p'q$ which tell you that with the Koszul sign rule and your proposed grading, $A\otimes B$ is anticommutative as well $\endgroup$ – Max Aug 28 at 8:48
  • $\begingroup$ Since you used "anticommutative" in your question I'm going to assume that the right parentheses here are " graded (commutative algebras)" and not "(graded commutative) algebras". In this case if you have the Koszul sign rule for the multiplication on the tensor product you don't get commutativity on said tensor product, so you have to use the usual multiplication (that is, the one that you would use for commutative algebras, gradedness put aside) and then it works as well as for usual commutative algebras $\endgroup$ – Max Aug 28 at 11:24

The first "injections" work for the following reason : the multiplication on $A\otimes B $ is (in this context) defined with the Koszul sign rule : each time something of degree $p$ goes through something of degree $q$, you add a sign $(-1)^{pq}$.

Concretely here this means $(a\otimes b) (a'\otimes b') := (-1)^{|a'||b|} (aa')\otimes (bb')$ for homogeneous elements. (as explained in the chat, this is well defined as we may define it on each factor $A_p\otimes B_q\otimes A_r\otimes B_s$ independently, factors where it's clearly $4$-linear).

Of course as tensor products distribute over direct sums, the following is a decomposition of $A\otimes B$: $\bigoplus_n \bigoplus_{p+q=n} A_p\otimes B_q$, which we take to be our decomposition.

You can then check (an easy computation mod $2$, as I mentioned in the comments) that this makes $A\otimes B$ into an anticommutative (usually called "graded commutative") graded $R$-algebra.

In particular if we have an anticommutative graded $R$-algebra $C$ with maps $f : A\to C, g: B\to C$, then we have an obvious map $f\otimes g: A\otimes B\to C$ defined by $(f\otimes g)(a\otimes b) = f(a)g(b)$ (note that the Koszul sign rule is respected here as well : $g$ goes through $a$, but $g$ has degree $0$ so no sign is added).

As it is defined by the universal property of the tensor product this is clearly an $R$-module map; so we only need to check that it is multiplicative.

This is a straightforward calculation, indeed $$(f\otimes g)((a\otimes b)(a'\otimes b')) = (f\otimes g) ((-1)^{|a'||b|}(aa')\otimes (bb')) = (-1)^{|a'||b|} f(aa')g(bb') =(-1)^{|a'||b|} f(a)f(a')g(b)g(b') = f(a)g(b)f(a')g(b') = (f\otimes g)(a\otimes b)(f\otimes g)(a'\otimes b')$$, where the first equality follows from definition of multiplication, the second is the definition of $f\otimes g$, the third is multiplicativity of $f,g$, the fourth is anticommutativity of $C$ and the last is again the definition of $f\otimes g$.

You can extend this for finite coproducts, as in any category $C$ with binary coproducts $\coprod$, and any objects $A_1,...,A_n$, $(...(A_1\coprod A_2)\coprod ... )\coprod A_n$ is a coproduct of $A_1,...,A_n$. This tells you how to get the correct multiplication on $A_1\otimes... \otimes A_n$, although again one can check (by induction) that we just end up with the Koszul convention.

(We note here that alternativity plays no role here. We could of course check that if $A,B$ are alternating, so is $A\otimes B$; warning : one has to prove $a^2=0$ not only for pure tensors, but it's not a problem)

ADDED: (proof that $A\otimes B$ is alternating if $A$ and $B$ are) I won't do a full proof, only a sufficiently general example. Assume $a\otimes b$ and $a'\otimes b'$ are of odd degree, and consider $(a\otimes b + a'\otimes b')^2 = (a\otimes b)^2 + (a'\otimes b')^2 + (a\otimes b) (a'\otimes b') + (a'\otimes b')(a\otimes b)$

$(a\otimes b)^2 = (-1)^{|a||b|}a^2\otimes b^2$. Now if $|a\otimes b | $ is odd, it means $|a|$ is odd or $|b|$ is odd, so $a^2=0$ or $b^2=0$ (as $A,B$ are alternating), so $(a\otimes b)^2 = 0$, and same for $(a'\otimes b')^2$.

Then $(a'\otimes b')(a\otimes b) = (-1)^{|a\otimes b||a'\otimes b'|}(a\otimes b)(a'\otimes b')$ by anticommutativity of $A\otimes B$, and the degrees are both odd, so $(a'\otimes b')(a\otimes b) = -(a\otimes b)(a'\otimes b')$

So the square is $0$. Of course the same proof holds for more than two pure tensors.

If you got this result without hypotheses on $A,B$, I suspect you only dealt with the "double products" and not the "inner squares" $(a\otimes b)^2$. An example where this doesn't work is the following : take $\mathbb F_2 [x]$ with $x$ in degree $1$, which is commutative, hence as $1=-1$ anticommutative. However it is not alternating. If you take its tensor square over $\mathbb F_2$ you get $\mathbb F_2[x,y]$ which is still not alternating. (note that an example "has to be in characteristic $2$" because of course if $2$ is regular in $A\otimes B$ then anticommutative implies alternating - so not necessarily characteristic $2$, but $2$ has to be non invertible for instance)

  • $\begingroup$ Dear Max, could you elaborate on proving $a^2 = 0$ for elements of odd degree in $A\otimes B$? When I'm trying to prove that I get that it is true regardless of $A$ and $B$ being alternating, so I suspect there is a mistake. $\endgroup$ – Jxt921 Aug 30 at 9:00
  • $\begingroup$ That is, if $r + s = n = p + q$ is odd, then for $a \in A_p, a' \in A_r, b \in B_q, b' \in B_s$, then $(a\otimes b)(a'\otimes b') = (-1)^{n^2} (a'\otimes b')(a\otimes b) = - (a'\otimes b')(a\otimes b)$, so, even for not pure tensors pure, tensor components of that sum that is $x^2$ for $x \in \bigoplus_{p + q = n} A_p\otimes B_q$ cancel each other out. $\endgroup$ – Jxt921 Aug 30 at 9:03
  • $\begingroup$ I added a few words. Note two things : 1- your "proof" doesn't rely on $A\otimes B$, just on anticommutativity of an algebra; 2- $x=-x$ does not imply $x=0$ over a general algebra ! That's the issue here $\endgroup$ – Max Aug 30 at 9:14
  • $\begingroup$ Which is why for instance $\mathbb Z [x]$ as a graded commutative algebra with $x$ in degree $1$ is not free in any reasonable sense : $2x^2 = 0$ so $x^2=0$ (over $\mathbb Z$), but $\mathbb F_2[x]$ has $x^2\neq 0$, so there's no map of graded algebras $\mathbb Z [x] \to \mathbb F_2 [x]$ that sends $x\mapsto x$ $\endgroup$ – Max Aug 30 at 9:16
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    $\begingroup$ Not exactly, but you can just write $\sum_i a_i\otimes b_i$ with each $a_i, b_i$ homogeneous and $|a_i| + |b_i| = n$ odd (you might get unpure tensors of bidegree $(p,q)$) $\endgroup$ – Max Aug 30 at 9:59

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