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.

  • $\begingroup$ that's certainly the answer. It's an $R$ algebra by the rule $r \cdot r' \otimes a' = rr' \otimes a'$. $\endgroup$ – hunter Apr 1 '15 at 23:17
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    $\begingroup$ $G\left(A\right)$ is a $\mathbb Z$-algebra (being the tensor product of two $\mathbb Z$-algebras) and an $R$-module (due to $R$ acting on the left tensorand). It is easy to check that this makes it an $R$-algebra (the relevant axiom holds). Also, the left adjointness requires showing $\operatorname{Hom}_{R-\operatorname{Alg}}\left(G\left(A\right), B\right) \cong \operatorname{Hom}_{\operatorname{Ring}}\left(A, B\right)$ canonically. This isomorphism is what one would expect. ... $\endgroup$ – darij grinberg Apr 1 '15 at 23:17
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    $\begingroup$ ... In the backward direction, it sends a ring homomorphism $f : A \to B$ to the $R$-linear map $G\left(A\right)\to B, \ r \otimes a \mapsto rf\left(a\right)$, which is easily seen to be an $R$-algebra homomorphism. In the forward direction, it sends an $R$-algebra homomorphism $g : G\left(A\right) \to B$ to the ring homomorphism $A \to B, \ a \mapsto g\left(1\otimes a\right)$. This all assumes that your $R$-algebras are associative, unital and central and $R$ is commutative. Whether or not you require your $R$-algebras to be commutative does not matter. $\endgroup$ – darij grinberg Apr 1 '15 at 23:20
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    $\begingroup$ Oh of course I forgot that $\mathbb Z $-algebras are just rings in disguise, duh. $\endgroup$ – Matthew Levy Apr 1 '15 at 23:21

Intuitively, if a functor 'forgets structure' from one category of sets-with-algebraic-structure to another then it will have a left-adjoint; such adjunctions are called 'free-forgetful' adjunctions. Unfortunately $F$ is an unfortunate choice of letter because forgetful functors are usually denoted by $U$ and free functors by $F$, so for your $F$ I'll write $U$ and for your $G$ I'll write $F$.

The forgetful functor $U : \mathsf{Gp} \to \mathsf{Set}$ has a left-adjoint $F : \mathsf{Set} \to \mathsf{Gp}$, where given a set $X$, $FX$ is the free group on the set $X$ of generators.

In the case of $R$-algebras, the forgetful functor $U : R\text{-}\mathsf{Alg} \to \mathsf{Ring}$ will have a left-adjoint $F : \mathsf{Ring} \to R\text{-}\mathsf{Alg}$ sending a ring $A$ to the free $R$-algebra on $A$. That is, $FA$ is an $R$-algebra equipped with a ring map $A \to FA$ ('insertion of generators') such that, given any $R$-algebra $B$ and ring homomorphism $\theta : A \to B$, $\theta$ extends uniquely to an $R$-algebra homomorphism $\widehat{\theta} : FA \to B$.

...and if you check the details, this is precisely $R \otimes_{\mathbb{Z}} A$. The insertion-of-generators map $A \to R \otimes_{\mathbb{Z}} A$ is given by $a \mapsto 1_R \otimes a$, and the $R$-algebra structure is given by $r(s \otimes a) = (rs) \otimes a$.

  • $\begingroup$ There's clearly a problem though if $U:\mathsf {Field}\to \mathsf {Set}$ is the forgetful functor then there is no left adjoint even though Field a category of "sets-with-algebraic-structure". Could you explain this? $\endgroup$ – Matthew Levy Apr 2 '15 at 5:59
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    $\begingroup$ Personally, my conclusion is that fields are not sets with algebraic structure. And sure enough, in the definition of field, there is a non-algebraic axiom. (An algebraic axiom is a (universally quantified) equation.) $\endgroup$ – Zhen Lin Apr 2 '15 at 7:48
  • $\begingroup$ How are they not sets with algebraic structure? They are just a subcategory of $\mathsf {Ring}$ where each object contains all multiplicative inverses besides for $0$. $\endgroup$ – Matthew Levy Apr 3 '15 at 5:29
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    $\begingroup$ If I were being more precise I'd say that given any algebraic theory $\mathbb{T}$ the forgetful functor $U : \mathsf{Mod}_{\mathbb{T}} \to \mathsf{Set}$ from the category of models of $\mathbb{T}$ has a left-adjoint. The theory of fields is not algebraic, regardless of whether it's a subcategory of $\mathsf{Ring}$. (See, for example, Ch 3 in Borceux vol 2 for way more detail than I can fit into these comments.) $\endgroup$ – Clive Newstead Apr 3 '15 at 15:28

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.


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