Graded rings: coproduct vs tensor product

In the category of commutative rings the coproduct and more generally pushouts can be neatly expressed by the tensor product. How about in the category of graded commutative rings (note: graded commutative, not commutative graded, ie. no signs to account for when switching factors), where the tensor product is graded in the usual way as $$(S\otimes S')_d = \bigoplus_{m+n=d}S_n\otimes S_m,$$ is this still the coproduct? Thanks!

• Yep, that's right. – Qiaochu Yuan Oct 30 '16 at 23:36

I guess that in that case you define multiplication on $S\otimes S'$ on generators by
$$(r\otimes r')\cdot (s \otimes s'):=(rs) \otimes (r's'), \;\; r,s \in S, \; r',s' \in S' \text{ homogeneous}.$$
In that case, I think it works: One has the maps $\alpha: S \rightarrow S \otimes S', \; \beta: S' \rightarrow S \otimes S'$ given on degrees by $$S_i \rightarrow S_i \otimes S'_0 \subseteq (S\otimes S')_i, \;\; s_i \mapsto s_i \otimes 1,$$ $$S'_i \rightarrow S_0 \otimes S'_i \subseteq (S\otimes S')_i, \;\; s'_i \mapsto 1 \otimes s'_i.$$
Now, given a pair of degree $0$ homogeneous homomorphisms $\gamma: S\rightarrow R,$ $\delta: S'\rightarrow R$, it should be easy to check that the map $\tau: S\otimes S' \rightarrow R,$ given by $s_i\otimes s'_j \mapsto \gamma(s_i)\cdot \delta(s'_j)$ is a correctly defined and unique degree $0$ homogeneous homomorphism with $\tau \alpha = \gamma, \; \tau \beta = \delta.$