TITLE: Adjoint functor to an R-algebra only "remembering" itself as a ring
QUESTION [4 upvotes]: I have been wondering this question while trying to comprehend adjoint functors and the various definitions. If you let 
$$F:\mathbf {R\text - Alg}\to  \mathbf {Ring}$$
be the functor that sends each R-algebra to itself viewed as a ring in $\mathbf {Ring}$ and each R-algebra homomorphism to the same map viewed as a ring homomorphism. Does this functor have a right adjoint? What about a left adjoint? (Does the forgetful functor from $\mathbf {Grp}$ to $\mathbf {Set}$ have a left adjoint?)
I have an intuition for what the functor should do if it did exist. It should take any ring and turn it into the most general R-algebra possible from this. I assume this most general will come as a universal property, but again I'm not sure what this would be.
I asked my professor at the end of the class the other day and he didn't have much time to think about it but suggested it may be 
$$G(A):=R\otimes_{\mathbb Z}A\in Ob_{\mathbf {R\text - Alg}}$$
the functor that sends every ring $A$ to this tensor product viewed as an R-algebra. I don't really see how this is an R-algebra even. Can somebody please help me understand this abstract non-sense. Thanks.

REPLY [2 votes]: Of course, the algebraic intuition should be the one given by Clive Newstead. However, it might be worth noting that your problem is in fact an instance of something more general.
Take $\mathcal C$ to be a category with pushouts. Denotes ${}_{x \backslash}\!\mathcal C$ the category whose objects are the morphisms $x \to a$ of $\mathcal C$ and whose morphisms are the commutative triangles (see coslice category at nlab for more detail). Then any morphism $f \colon x \to y$ of $\mathcal C$ induces a functor
$$ \begin{aligned} f^\ast \colon {}_{y \backslash}\!\mathcal C &\to {}_{x \backslash}\!\mathcal C, \\(y\to a) &\mapsto (x\stackrel f \to y \to a). \end{aligned} $$
This functor $f^\ast$ has a left adjoint: indeed, the very definition of the pushout gives for any $g \colon x\to a$ and $h\colon y \to b$ a natural isomorphism (draw the diagrams!)
$$ \{ j \colon a \to b \in \mathcal C: h\circ f = j\circ g\} \simeq \{k \colon y\sqcup_x a \to b \in \mathcal C : h = (y \to y\sqcup_x a \stackrel k \to b) \}. $$
But the left-hand side is precisely $\hom_{{}_{x \backslash}\!\mathcal C}(g,f^\ast h)$ and the right-hand side $\hom_{{}_{y \backslash}\!\mathcal C}(y\to y\sqcup_x a,h)$. This means that the functor ${}_{x \backslash}\!\mathcal C \to {}_{y \backslash}\!\mathcal C$ mapping every $x\to a$ to its pushout along $f$ is a left adjoint to $f^\ast$.
Now, apply it to $\mathcal C = \mathsf{Ring}$ with $x=\mathbb Z$ and $y=R$ to obtain the result of your problem.
{"set_name": "stack_exchange", "score": 4, "question_id": 1216535}