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:
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?