Consider the endofunctor $$ \begin{align} (-)[x] \colon \operatorname{CRing} &\to \operatorname{CRing} \\ A &\mapsto A[x]. \end{align} $$
As explained here, this defines a monad over the category of commutative rings. In particular
- the unity $\eta_A \colon A \to A[x]$ is given by the canonical inclusion;
- the multiplication is given as $$ \begin{align} \mu_A \colon A[x][x] &\to A[x] \\ \sum_m \left (\sum_n a_{n,m} \, x^n \right) x^m &\mapsto \sum_{n, m} a_{n, m} \, x^{n+m}. \end{align} $$
I want to find out what is an algebra for this monad (in the linked post it is claimed that - in the case of rings - an algebra consists of a ring and an element of its center).
Following the definition, an algebra is a pair $(A, \xi)$ where
- A is a (commutative from now on) ring,
- $\xi \colon A[x] \to A$ is a ring morphism such that the square $\require{AMScd}$ \begin{CD} A[x][x] @> T \, \xi >> A[x]\\ @V \mu_A VV @VV \xi V\\ A[x] @>> \xi > A \end{CD} commutes (!!! see the first comment below).
The morphism $\xi$ is determined by a ring morphism $\xi_0 \colon A \to A$ and and elment $\xi(x) \in A$ from $$ \xi \left(\sum_n a_n x^n \right) = \sum_n \xi_0(a_n) \, \xi(x)^n. $$
Chasing the element $(a \, x^0) \, x^0$ in the commutative square gives the identity $$ \xi_0(a) = \xi_0(\xi_0(a)) $$ for every $a \in A$, hence $\xi_0$ is a projection (idempotent endomorphism of $A$).
Using this fact and chasing the generic element $$ \sum_m \left (\sum_n a_{n,m} \, x^n \right) x^m $$ in the commutative square we have
- passing through the left lower corner, the element $$ \xi \left(\sum_{n, m} a_{n, m} \, x^{n+m} \right) = \sum_{n, m} \xi_0(a_{n, m}) \, \xi(x)^{n+m}; $$
- passing through the right upper corner, the element $$ \xi \left(\sum_m \left (\sum_n \xi_0(a_{n,m}) \, \xi(x)^n \right) x^m \right) = \sum_{n, m} \underbrace{\xi_0(\xi_0(a_{n, m}))}_{= \xi_0(a_{n, m})} \, \xi(\xi(x))^{n} \xi(x)^m. $$
Choosing $a_{n, m} = 1$ and $n_{\text{max}} = 1$, $m_{\text{max}} = 0$, this gives $$\xi(x) = \xi(\xi(x))$$ ie. $\xi(x)$ is in the image of $\xi_0$. Clearly this condition is also sufficient for the commutativity of the square.
Conclusion: an algebra for this monad consists of
- a (commutative) ring $A$,
- a morphism $\xi_0 \colon A \to A$ such that $\xi_0^2 = \xi_0$,
- an element $\xi(x) \in \operatorname{im}(\xi_0)$.
Question 1: is the above discussion correct?
Question 2: is the claim in the linked post correct? I must have used commutativity but I don't realize where.