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The construction of polynomials $R \mapsto R[x]$ gives a functor $P: \mathbf{Ring} \to \mathbf{Ring}$ on the category of possibly noncommutative rings. Choosing a ring $R$ for the moment, there is a nice homomorphism $R \to P(R)$ which embeds in the obvious way, taking $r$ to the constant polynomial $r \cdot 1$. There also seems to be a homomorphism $P(P(R)) \to P(R)$: given a polynomial in $x$ with coefficients in $R[x]$, just do the multiplication and addition to get a polynomial with coefficients in $R$.

This sounds suspiciously like a monad on $\mathbf{Ring}$. I think that the above maps are in fact natural transformations $\eta: 1_{\mathbf{Ring}} \to P$ and $\mu: P^2 \to P$, and that $\mu \circ P \eta = \mu \circ \eta P = 1_P$ and $\mu \circ P \mu = \mu \circ \mu P$.

Is this right? And if so, what have people done with this idea? For example, I see that there are things called polynomial monads, but it's not clear how they might be related.

Hew Wolff
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  • The choice of evaluation $P(P(R))\to P(R)$ is the key question - there are many such evaluation homomorphisms, and it requires you to distinguish $x$ in $P(R)$ to know which evaluation to take. But $R[x]\cong R[x-1]\cong R[x-r]$. So it is not clear that the map $P(P(R))\to P(R)$ is natural. – Thomas Andrews Oct 06 '15 at 18:51
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    @ThomasAndrews: The evaluation $P(P(R))=R[x,y]\to P(R)=R[x]$ sending $y$ to $x$ is natural. It will not give a commutative diagram involving arbitrary maps out of the object $P(R)$, but it will whenever those maps are obtained from a map out of $R$ by applying the functor $P$, which is all that matters. – Eric Wofsey Oct 06 '15 at 19:01
  • Slightly OT: It's past 3am and my half-asleep brain somehow came to this question. Thank you @Hew Wolff for asking this before, you probably saved me an hour of thinking about it instead of sleeping. ;) – Lukas Juhrich May 03 '19 at 01:28

1 Answers1

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Yes, this is a monad. Much more generally, if $\mathcal{C}$ is a monoidal category and $M$ is a monoid object in $\mathcal{C}$, then the functor $P(R)=M\otimes R$ is a monad using the monoid structure of $M$. In this case, $\mathcal{C}=\mathbf{Ring}$, the monoidal structure is $\otimes_\mathbb{Z}$, and $M=\mathbb{Z}[x]$ (note that a monoid object in $\mathbf{Ring}$ is the same thing as a commutative ring).

I don't know of any particular applications of this monad. An algebra over this monad is just a $\mathbb{Z}[x]$-algebra, or equivalently a ring $R$ together with a chosen central element $x\in Z(R)$ (the image of $x$ under the structure map $R[x]\to R$). I'm not familiar with polynomial monads, but from reading a little on nlab they seem to be totally unrelated.

Eric Wofsey
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