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Does the forgetful functor $U: R-\mathbf{Alg} \to \mathbf{CRing}$ have a right-adjoint? I checked that it commutes with finite colimits but I couldn't guess any other candidate than the tensor product, which is left-adjoint to $U$.

Edit: I checked that it commutes with fibered coproducts as in $$U(A) \otimes_{U(C)} U(B) \simeq U(A \otimes_C B)$$ but this does of course not imply that it commutes with coproducts in general, as pointed out in Hannos excellent answer.

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    $\begingroup$ Forgetful functors between algebraic categories usually have left adjoints. In this case the left adjoint is tensoring with $R$. $\endgroup$
    – Qiaochu Yuan
    Nov 8, 2014 at 19:51
  • $\begingroup$ And note that the reason why Hanno's counterexample and your check don't contradict each other is that $U$ does even not preserve the initial object. $\endgroup$
    – Qiaochu Yuan
    Nov 8, 2014 at 23:48

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As a necessary condition for $U$ to be a left adjoint, you need to check that $U$ preserves colimits. For example, the coproduct in $R\textbf{-Alg}$ is given by $-\otimes_R -$, while in $\textbf{CRing}\cong{\mathbb Z}\textbf{-Alg}$ it is given by $-\otimes_{\mathbb Z}-$. Therefore the right adjoint will not exist in general: e.g., you can take $R = {\mathbb Z}[X]$, and then $${\mathbb Z}[X,Y]\cong U(R)\otimes_{\mathbb Z} U(R)\xrightarrow{X,Y\mapsto X} U(R\otimes_R R)\cong {\mathbb Z}[X]$$ is not bijective.

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