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!

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

If I understand your comment, you want to work with commutative rings that are graded.

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 $\tau \alpha = \gamma, \; \tau \beta = \delta.$


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