I know that pullback is defined for commutative ring, but what about non-commutative case? Let's consider the following diagram, where $R_i,\bar{R}$ are rings and $R$ is the pullback:

enter image description here

Then $1\in R$ and $(R,+)$ is an additive group since it'a a pullback of additive groups. Also $(R-\{0\},\cdot)$ is closed since if $x^{1},x^{2}\in R$, $x^{i}=(x_{1}^{i},x_{2}^{i})$ with $\nu_{1}(x_{1}^{i})=\nu_{2}(x_{2}^{i})$, then $x^1x^2\in R$ too.

This does not means that $R$ is a ring too? Or maybe it's a ring but it's not the pullback of the diagram?

  • 5
    $\begingroup$ The pullback of (possibly) non-commutative rings is indeed perfectly well-defined. What made you think it wasn't? $\endgroup$ – Mees de Vries Jul 19 '18 at 10:27
  • $\begingroup$ @MeesdeVries: Every time I find "pullback of rings" just the commutative case is cited. Anyway I could not find a good reference for the commutative case no more $\endgroup$ – user84976 Jul 19 '18 at 10:32
  • 2
    $\begingroup$ It's probably because the rererences you read are specifically about commutative algebra. In fact the pullback of two morphisms always exists for algebraic structures $\endgroup$ – Max Jul 19 '18 at 11:56

Your Answer

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

Browse other questions tagged or ask your own question.