Let $C$ be the category of associative commutative rings with 1 and let $F:C\to C$ be a functor which commutes with the forgetful functor to abelian groups (i.e. $F$ is a functorial way to define another multiplication on every associative commutative ring with 1). Assume also that on $\mathbb Z$ the new multiplication is equal to the old one. Is it true that $F$ is the identity functor?
$\begingroup$
$\endgroup$
2
asked Dec 19, 2017 at 12:43
-
8$\begingroup$ Given a ring define the bijection $\varphi:R\to R$ on the underlying set by imposing $\varphi(x)=-x$. Define a new ring structure $R^-$ on the underlying set $R$ by transfer of structure along $\varphi$. Addition is the same, but multiplication is given by $x\times^-x'=-xx'$. This should define a functor $C\to C$ where any morphism of commutative rings is sent to the same underlying set map. Since addition is the same, this functor commutes with the forgetful functor to abelian groups. Does this work ? $\endgroup$– Olivier BégassatCommented Dec 19, 2017 at 13:00
-
1$\begingroup$ I forgot to say that unit remain a unit $\endgroup$– Alexander BravermanCommented Dec 19, 2017 at 14:14
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Yes, $F$ is the identity functor.
The multiplication on $R$ would come in the form of a natural transformation $R \times R \to R$ which is bilinear in each variable. In commutative rings, the functor $R \mapsto R \times R$ is represented by the polynomial algebra $\Bbb Z[x,y]$, and so by the Yoneda lemma natural transformations $R \times R \to R$ are represented by polynomials $f(x,y) \in \Bbb Z[x,y]$. In short, your new multiplication on $F(R)$ must be of the form $x * y = f(x,y)$ for $f$ some polynomial with integer coefficients.
In order for $x * y$ to be distributive, we need $f(x + x',y) \equiv f(x,y) + f(x',y)$, so that $f$ is linear in each variable. This forces $x * y = axy$ for some $a \in \Bbb Z$. However, the only choices of $a$ that ensure $F(R)$ is always unital are $a = \pm 1$, and the only one that preserves the unit of $\Bbb Z$ is $a=1$.
answered Dec 19, 2017 at 16:16
-
$\begingroup$ Thanks. I wonder if this argument can be extended to algebras in more general monoidal categories. For example, rings with $G$-action, where $G$ is some algebraic group. $\endgroup$ Commented Dec 20, 2017 at 9:19
-
$\begingroup$ @AlexanderBraverman That's a fair question. The argument above does not quite prove uniqueness, e.g. in the case where $G$ is the discrete group $\Bbb Z/2$, but we also have not used the constraint that $f$ should define an associative multiplication yet. There should be some mileage yet to get out of this. $\endgroup$ Commented Dec 20, 2017 at 17:45