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Stream: deprecated: mathematics

Topic: polynomial ring

Hugo Jenkins (Mar 23 2023 at 22:06):

If $\mathcal{C}$ is the category of unital commutative rings, we can consider the endofunctor $A\mapsto A[X]$. It outputs the ``free $A$-algebra''.

How should I think about the interactions of this endofunctor with tensor product? Such as:

• $A[X]\otimes_A A[X]\cong A[X,Y]$
• If $A\overset{\iota}{\to} B$, then $A[X]/I \otimes_A B \cong B[X]/B\iota(I)$.

Jean-Baptiste Vienney (Mar 23 2023 at 22:12):

Good question, polynomial algebras give a codifferential category. I can explain this a bit later, but maybe @JS PL (he/him) wants to do it also.

JS PL (he/him) (Mar 23 2023 at 23:52):

Jean-Baptiste Vienney said:

Good question, polynomial algebras give a codifferential category. I can explain this a bit later, but maybe JS PL (he/him) wants to do it also.

No that's not what @Hugo Jenkins is talking about here... This question is not related to differential categories in the way you are thinking. (While I do love differential categories, not everything can be answered by differential categories!)

JS PL (he/him) (Mar 23 2023 at 23:56):

But here's an attempt at an answer:
Let $\mathsf{CRING}$ be the category of commutative (unital) rings. And let $P: \mathsf{CRING} \to \mathsf{CRING}$ be the functor which maps $P(A) = A[X]$.

JS PL (he/him) (Mar 23 2023 at 23:57):

Alternatively, we may write this as $P(A) = A \otimes \mathbb{Z}[X]$ -- where the tensor product of $\mathbb{Z}$-modules, which is the coproduct in $\mathsf{CRING}$

JS PL (he/him) (Mar 24 2023 at 00:07):

It's neat to note that $P$ is both a monad and comonad

John Baez (Mar 24 2023 at 00:11):

I guess the multiplication $PP \Rightarrow P$ gives the ring homomorphism $A[X_1,X_2] \to A[X]$ sending both $X_1$ and $X_2$ to $X$. What does the comultiplication do? Send $X$ to $X_1 + X_2$?

(In other notation $X_1 + X_2$ is $X \otimes 1 + 1 \otimes X \in A[X] \otimes_A A[X]$, which is a respectable-looking thing for a comultiplication to do.)

JS PL (he/him) (Mar 24 2023 at 00:13):

John Baez said:

What does the comultiplication do? Send $X$ to $X_1 + X_2$?

Yup that's correct! The comonad structure comes from the fact that $\mathbb{Z}[X]$ is a comonoid in $\mathsf{CRING}$ with respect to $\otimes$
(which is just a fancy way of saying that $\mathbb{Z}[X]$ is a bialgebra)

JS PL (he/him) (Mar 24 2023 at 00:14):

John Baez said:

(In other notation $X_1 + X_2$ is $X \otimes 1 + 1 \otimes X \in A[X] \otimes_A A[X]$, which is a respectable-looking thing for a comultiplication to do.)

Yes this is the canonical comultiplication of the polynomial ring, which makes it a bialgebra (and also a Hopf algebra)

JS PL (he/him) (Mar 24 2023 at 00:16):

Hugo Jenkins said:

• $A[X]\otimes_A A[X]\cong A[X,Y]$

So if $\mathsf{CALG}_A$ is the category of commutative $A$-algebras. In this category, $\otimes_A$ is the coproduct. What this identity is saying is that $PP(A)$ is the coproduct of $P(A)$ and $P(A)$ in $\mathsf{CALG}_A$.
But if you want to stay within $\mathsf{CRING}$ then $\otimes_A$ is a coequalizer. So this identity instead says that $PP(A)$ is a coequalizer of the parallel maps $A \otimes P(A) \otimes P(A) \to P(A) \otimes P(A)$ where you multiply $A$ with either the first $P(A)$ or the second.

Jean-Baptiste Vienney (Mar 24 2023 at 02:27):

Hugo Jenkins said:

If $\mathcal{C}$ is the category of unital commutative rings, we can consider the endofunctor $A\mapsto A[X]$. It outputs the ``free $A$-algebra''.

How should I think about the interactions of this endofunctor with tensor product? Such as:

• $A[X]\otimes_A A[X]\cong A[X,Y]$
• If $A\overset{\iota}{\to} B$, then $A[X]/I \otimes_A B \cong B[X]/B\iota(I)$.

The first isomorphism is an example of the Seely isomorphism of linear logic/differential categories. But maybe the answer without differential categories is simpler depending of your definition of simple.

JS PL (he/him) (Mar 24 2023 at 02:41):

Jean-Baptiste Vienney said:

The first isomorphism is an example of the Seely isomorphism of linear logic/differential categories.

The question was about the question the endofunctor on the category of $\mathsf{CRING}$. So it is not the so called Seely isomorphism in this case -- at least not directly.

JS PL (he/him) (Mar 24 2023 at 02:43):

To @Jean-Baptiste Vienney point, if you consider the endofunctor on $\mathsf{MOD}_A$, the category of $A$-modules, $S_A: \mathsf{MOD}_A \to \mathsf{MOD}_A$ which maps an $A$-module $M$ to the free commutative $A$-algebra over $M$, $S_A(M)$, then yes that is that first identity is the Seely isomorphism, which in general is $S_A(M \times N) \cong S_A(M) \otimes_A S_A(N)$. And for the particular case you are thinking about:
$S_A(A) = A[X]$ and $S_A(A \times A) \cong A[X,Y]$
so $S_A(A) \otimes_A S_A(A) \cong S_A(A \times A)$

JS PL (he/him) (Mar 24 2023 at 02:45):

But @Hugo Jenkins question was not about this. It was about the endofunctor $P: \mathsf{CRING} \to \mathsf{CRING}$, mapping a commutative ring $A$ to the polynomial ring in one variable $A[X]$, so $P(A) = S_A(A)$. Which is different than the functor $S_A: \mathsf{MOD}_A \to \mathsf{MOD}_A$.

JS PL (he/him) (Mar 24 2023 at 02:52):

(and yes, my original comments about me disagreeing on using differential categories here were too confrontational than needed to be -- so apologies)

JS PL (he/him) (Mar 24 2023 at 02:56):

The point I attempted to make (badly...): was I disagree that differential categories/Seely isomorphisms were the appropriate approach/tools for answering this question.

John Baez (Mar 24 2023 at 23:20):

Be nice.

Hugo Jenkins (Apr 07 2023 at 13:55):

JS PL (he/him) said:

Yup that's correct! The comonad structure comes from the fact that $\mathbb{Z}[X]$ is a comonoid in $\mathsf{CRING}$ with respect to $\otimes$
(which is just a fancy way of saying that $\mathbb{Z}[X]$ is a bialgebra)

Perhaps the fact I was looking for is just that the functor $P$ is equal to $-\otimes \mathbb{Z[x]}$. What is the universal property of $\mathbb{Z}[x]$ with respect to $\mathsf{CRING}, \otimes$?

Does $P$ being $-\otimes \mathbb{Z}[x]$ explain the evaluation map/universal property of $A[x]$? (I.e. that a map $A\to B$ extends uniquely to $A[x]\to B$ sending $x$ to any prescribed element)?

Josselin Poiret (Apr 08 2023 at 09:48):

for $\mathbf{CRing}$, $\otimes$ is the coproduct, and the free ring on a set of elements (the corresponding polynomial ring $\mathbb{Z}[S]$) is a left adjoint, so we get coproduct preservation. You can probably derive everything from just this.

Josselin Poiret (Apr 08 2023 at 09:50):

the universal property of $\mathbb{Z}[x]$ is really just the commutative ring generated from one element, no need to put $\otimes$ in the picture yet

JS PL (he/him) (Apr 09 2023 at 23:07):

Hugo Jenkins said:

What is the universal property of $\mathbb{Z}[x]$ with respect to $\mathsf{CRING}, \otimes$?

As mentioned by @Josselin Poiret , $\mathbb{Z}[x]$ is the free ring over the singleton set, which means that for commutative ring $R$ and every function $f: \lbrace x \rbrace \to R$$, there exists a unique ring morphism $f^\sharp: \mathbb{Z}[x] \to R$ such that $f^\sharp (x) = f(x)$

JS PL (he/him) (Apr 09 2023 at 23:07):

but this steps outside of $\mathsf{CRING}$ -- and I don't know if $\mathbb{Z}[x]$ has a universal property apart from that

JS PL (he/him) (Apr 09 2023 at 23:14):

You could say that $\mathbb{Z}[x]$ is the initial ring with a chosen element. That is, for every commutative ring $R$ and a chosen element $r \in R$, so a pointed ring $(R,r)$, there exists a unique ring morphism $e_r: \mathbb{Z}[x] \to R$ such that $e_r(x) = r$

JS PL (he/him) (Apr 09 2023 at 23:16):

Hugo Jenkins said:

JS PL (he/him) said:

Yup that's correct! The comonad structure comes from the fact that $\mathbb{Z}[X]$ is a comonoid in $\mathsf{CRING}$ with respect to $\otimes$
(which is just a fancy way of saying that $\mathbb{Z}[X]$ is a bialgebra)

Does $P$ being $-\otimes \mathbb{Z}[x]$ explain the evaluation map/universal property of $A[x]$? (I.e. that a map $A\to B$ extends uniquely to $A[x]\to B$ sending $x$ to any prescribed element)?

So then given a ring morphism $f: A \to B$ and a chosen element $b \in B$, your unique map is $f \otimes e_b: A \otimes \mathbb{Z}[x] \to B$

John Baez (Apr 09 2023 at 23:50):

JS PL (he/him) said:

You could say that $\mathbb{Z}[x]$ is the initial ring with a chosen element. That is, for every commutative ring $R$ and a chosen element $r \in R$, so a pointed ring $(R,r)$, there exists a unique ring morphism $e_r: \mathbb{Z}[x] \to R$ such that $e_r(x) = r$

That's a sneakier way of "stepping outside of $\mathsf{CommRing}$, since here a "chosen element" of a ring is really a chosen element of the underlying set of the commutative ring.

John Baez (Apr 09 2023 at 23:53):

So if you want to describe the universal property of $\mathbb{Z}[x]$ as something like "the free ring on one element", I think you should use the adjoint functors $U : \mathsf{CommRing} \to \mathsf{Set}$, $F: \mathsf{Set} \to \mathsf{CommRing}$ (either one determines the other up to natural isomorphism).

JS PL (he/him) (Apr 09 2023 at 23:54):

John Baez said:

That's a sneakier way of "stepping outside of $\mathsf{CRING}$, since here a "chosen element" of a ring is really a chosen element of the underlying set of the commutative ring.

Agreed, and if you want to stay inside $\mathsf{CRING}$, the way to pick out an element of $R$ corresponds to ring morphisms of type $\mathbb{Z}[x] \to R$. So a pointed ring is a commutative ring $R$ with a chosen ring morphism $\mathbb{Z}[x] \to R$, and $\mathbb{Z}[x]$ is the initial one with the canonical choice being the identity morphism $\mathbb{Z}[x] \to \mathbb{Z}[x]$

John Baez (Apr 09 2023 at 23:55):

Right! Another way to say it: once someone whispers in your ear which object of $\mathsf{CommRing}$ is $\mathbb{Z}[x]$, you can define $U : \mathsf{CommRing} \to \mathsf{Set}$ by

$U = \mathsf{CommRing}(\mathbb{Z}[x], -)$

John Baez (Apr 09 2023 at 23:57):

Then $U$ has a left adjoint $F$ and you can check $\mathbb{Z}[x] \cong F(1)$.

JS PL (he/him) (Apr 10 2023 at 00:00):

John Baez said:

$U = \mathsf{CommRing}(\mathbb{Z}[x], -)$

From this point of view, you can get the $\otimes$ nicely in the story of $P$ since its a coproduct we get:
$\mathsf{CommRing}(A \otimes \mathbb{Z}[x], -) \cong \mathsf{CommRing}(A, -) \times \mathsf{CommRing}(\mathbb{Z}[x], -)$