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Stream: learning: questions

Topic: Exponentiation

Mike Stay (Oct 18 2023 at 05:02):

Has anyone considered weakening the idea of an [[exponential ring]] so that the ring $R$ of exponents is different than the ring $R'$ of numbers? That would allow finite fields to be nontrivial examples, e.g. $R' = \mathbb{Z}_p$ and $R = \mathbb{Z}_{p-1}.$ Are exponential rings and finite fields the only nontrivial examples of such a structure?

Given a unital commutative ring $R$, how would I go about enumerating those rings $R'$ such that there exists a surjective monoid homomorphism $E:R\to R'^*$ such that for all $x, y \in R$, $E(x+y) = E(x)E(y)?$ Maybe Sage or something?

Jean-Baptiste Vienney (Oct 18 2023 at 14:16):

Given any two rings $R,R'$ such that $R$ is an exponential ring, and given a monoid homomorphism $\rho:(R',+) \rightarrow (R,+)$, define $r^{r'}=r^{\rho(r')}$. Then $r^0=r^{\rho(0)}=r^0=1$ and $r^{r'_1+r'_2}=r^{\rho(r'_1+r'_2)}=r^{\rho(r'_1)+\rho(r'2)}=r^{\rho(r'_1)}.r^{\rho(r'_2)}$ so that you get the kind of structure you're interested by.

Jean-Baptiste Vienney (Oct 18 2023 at 14:22):

Probably, it provides a functor between two appropriate categories.

fosco (Oct 18 2023 at 14:22):

Interesting, this looks like a change of base functor

fosco (Oct 18 2023 at 14:22):

ah, I was about to write that as well ;-)

Jean-Baptiste Vienney (Oct 18 2023 at 14:22):

Yes, I was thinking to that indeed :) (the change of base)

fosco (Oct 18 2023 at 14:25):

adjoint to the forgetful from expoRing to Ring?

fosco (Oct 18 2023 at 14:25):

but you need a homomorphism.. hm

Jean-Baptiste Vienney (Oct 18 2023 at 16:17):

It makes me think to a more categorified concept. Consider a (finitary) distributive monoidal category $(\mathcal{C},\otimes,I,\times,\top)$ (it is a kind of categorified semi-ring). Then an exponential functor $!:\mathcal{C} \rightarrow \mathcal{C}$ is a strong monoidal functor $(\mathcal{C},\times,\top) \rightarrow (\mathcal{C},\otimes,I)$

Jean-Baptiste Vienney (Oct 18 2023 at 16:18):

This little guys appear in math and in logic

Jean-Baptiste Vienney (Oct 18 2023 at 16:21):

In linear logic, you ask for $!$ to be a comonad and more. There is also the notion of relative monad when $!$ would be between different categories. And there is still to invent the notion of relative linear logic. It makes me think to that because it is somehow the same move than going from exponential ring to relative exponential ring (let me call your concept by this name).

Mike Stay (Oct 19 2023 at 04:15):

Jean-Baptiste Vienney said:

Given any two rings $R,R'$ such that $R$ is an exponential ring, and given a monoid homomorphism $\rho:(R',+) \rightarrow (R,+)$, define $r^{r'}=r^{\rho(r')}$. Then $r^0=r^{\rho(0)}=r^0=1$ and $r^{r'_1+r'_2}=r^{\rho(r'_1+r'_2)}=r^{\rho(r'_1)+\rho(r'2)}=r^{\rho(r'_1)}.r^{\rho(r'_2)}$ so that you get the kind of structure you're interested by.

Thanks! Hmm, it looks like you swapped which ring has the prime on it. I'll restate using $B$ for base ring, $E$ for exponents ring, and $\mu$ for monoid homomorphism:

Given any two rings $E, B$ such that $B$ is an exponential ring, and given a monoid homomorphism $\mu:(E,+) \rightarrow (B,+)$, define $b^{e}=b^{\mu(e)}$. Then $b^{0_e}=b^{\mu(0_e)}=b^{0_b}=1$ and $b^{e_1+e_2}=b^{\mu(e_1+e_2)}=b^{\mu(e_1)+\mu(e_2)}=b^{\mu(e_1)}\cdot b^{\mu(e_2)}$ so that you get the kind of structure you're interested by.

Any exponential ring $B$ comes equipped with a monoid homomorphism $\exp (-): (B, +) \to (B, \cdot)$; it's usually of the form $b^-$ for some $b$ in $B.$ And you're pointing out that you can extend that with any other monoid homomorphism $\mu: (E, +) \to (B, +)$ to get a relative exponential ring $b^{\mu(-)}$. Nice!

I suppose we can do something similar on the other side: given a monoid homomorphism $\nu: (B, \cdot) \to (C, \cdot)$, we get a relative exponential ring $\nu(b^-): (B, +) \to (C, \cdot)$ so $\nu(b^0) = \nu(1_b) = 1_c$ and

$\nu(b^{b_1 +_b b_2}) = \nu(b^{b_1}\cdot_b b^{b_2}) = \nu(b^{b_1}) \cdot_c \nu(b^{b_2}).$

Similar pre- and postcomposition lets us create relative exponential rings from other relative exponential rings.

Mike Stay (Oct 19 2023 at 04:17):

Do all nontrivial relative exponential rings factor through either a nontrivial exponential ring or a finite field?

My definition of relative exponential ring above also included the requirement that the $\exp$ homomorphism should be onto $(B, \cdot)$ so that every nonzero element has a logarithm, but I guess that's not standard for exponential rings.

Jean-Baptiste Vienney (Oct 19 2023 at 18:04):

Mike Stay said:

Do all relative exponential rings factor through either an exponential ring or a finite field?

I don't know... For now I think we should organize all the structure, it looks like there are a few functors in sight! I'll don't include the onto requirement first.

I will consider that ring means "nontrivial ring"

Jean-Baptiste Vienney (Oct 19 2023 at 18:14):

So:

• Let's denote $U_{+}:Ring \rightarrow Mon$ the functor which sends a ring $R$ to the underlying additive monoid.
• Denote $U_{\times}:Ring \rightarrow Mon$ the functor which sends a ring $R$ to the underlying multiplicative monoid $R \backslash \{0\}$ of $R$.

Jean-Baptiste Vienney (Oct 19 2023 at 18:16):

• An exponential ring is a tuple $(R,exp)$ where $R$ is a ring and $exp:U_{+}(R) \rightarrow U_{\times}(R)$ a monoid homomorphism.

Jean-Baptiste Vienney (Oct 19 2023 at 18:18):

• A relative exponential ring is a tuple $(B,R,\mu)$ where $B$ and $R$ are rings and $\mu:U_+(B) \rightarrow U_{\times}(E)$ is a monoid homomorphism.

Jean-Baptiste Vienney (Oct 19 2023 at 18:27):

Now, what we said before looks easy... It's just a matter of composing monoid homomorphisms.

Jean-Baptiste Vienney (Oct 19 2023 at 18:27):

We can still define morphisms of relative exponential rings:

Jean-Baptiste Vienney (Oct 19 2023 at 18:29):

A morphism from $(B_1,R_1,\mu_1)$ to $(B_2,R_2,\mu_2)$ is given by a pair $(f,g)$ of monoid homomorphisms such as below:

Jean-Baptiste Vienney (Oct 19 2023 at 18:30):

Screenshot-2023-10-19-at-2.29.59-PM.png

Jean-Baptiste Vienney (Oct 19 2023 at 18:33):

Of course there is the morphism $(id,id)$ from $(B,R,\mu)$ to itself and we can also compose morphisms:

Jean-Baptiste Vienney (Oct 19 2023 at 18:34):

Screenshot-2023-10-19-at-2.34.37-PM.png

Jean-Baptiste Vienney (Oct 19 2023 at 18:35):

So that we get a category $RelExpRing$ (whether we include the epi condition or not)

Jean-Baptiste Vienney (Oct 19 2023 at 18:43):

• You ask whether each relative exponential ring (with the epi condition) $\mu:U_+(B) \twoheadrightarrow U_{\times}(R)$ is of the form:
  (1) $U_+(B) \overset{exp}{\twoheadrightarrow} U_{\times}(B) \overset{m}{\longrightarrow} U_{\times}(R)$
  or
  (2) $U_+(B) \overset{m}{\longrightarrow} U_+(R) \overset{exp}{\twoheadrightarrow} U_{\times}(R)$

Jean-Baptiste Vienney (Oct 19 2023 at 18:46):

for some exponential ring $exp$ and monoid homomorphism $m$

Jean-Baptiste Vienney (Oct 19 2023 at 18:53):

Well, well, well, I think we should introduce "logarithm rings"

Jean-Baptiste Vienney (Oct 19 2023 at 18:56):

• Let's say that a logarithm ring is a tuple $(R,log)$ where $R$ is a ring and $log:U_\times(R) \rightarrow U_+(R)$ is a monoid homomorphism

Jean-Baptiste Vienney (Oct 19 2023 at 18:57):

• (We could as well introduce relative logarithm rings)

Jean-Baptiste Vienney (Oct 19 2023 at 19:01):

• Let's say that an exp/log ring is a tuple $(R,exp,log)$ where $R$ is a ring, $exp:U_+(R) \rightarrow U_\times(R)$ and $log:U_\times(R) \rightarrow U_+(R)$ are monoid homomorphisms such that
  $log;exp=1_{U_{+}(R)}$ and $exp;log=1_{U_{\times}(R)}$.

Jean-Baptiste Vienney (Oct 19 2023 at 19:02):

• (We could as well introduce relative exp/log rings)

Jean-Baptiste Vienney (Oct 19 2023 at 19:12):

Proposition: If $(R,exp,log)$ is an exp/log ring then any relative exponential ring $\mu:U_{+}(B) \rightarrow U_{\times}(R)$ can be decomposed under the form:

Jean-Baptiste Vienney (Oct 19 2023 at 19:21):

Screenshot-2023-10-19-at-3.20.58-PM.png

Jean-Baptiste Vienney (Oct 19 2023 at 19:21):

so that it comes from the exponential ring $(R,exp)$

Jean-Baptiste Vienney (Oct 19 2023 at 19:22):

Proposition: If $(B,exp,log)$ is an exp/log ring then any relative exponential ring $\mu:U_{+}(B) \rightarrow U_{\times}(R)$ can be decomposed under the form:

Jean-Baptiste Vienney (Oct 19 2023 at 19:26):

Screenshot-2023-10-19-at-3.26.06-PM.png

Jean-Baptiste Vienney (Oct 19 2023 at 19:26):

so that it comes from the exponential ring $(B,exp)$

Jean-Baptiste Vienney (Oct 19 2023 at 19:27):

It doesn't answer your question, but I think we've done some good progress into structuring the framework. I'll stop now for a bit

Jean-Baptiste Vienney (Oct 20 2023 at 12:59):

I'm still excited by this question. I can refine a bit what I wrote above.

Jean-Baptiste Vienney (Oct 20 2023 at 13:02):

• Define an $exp;log$ ring to be a tuple $(R,exp,log)$ where $exp:U_+(B)\rightarrow U_\times(B)$ and $log:U_\times(R)\rightarrow U_+(R)$ are two monoid homomorphisms such that $exp;log=1_{U_+(B)}$

Jean-Baptiste Vienney (Oct 20 2023 at 13:02):

• Define a $log;exp$ ring to be a tuple $(R,exp,log)$ where $exp:U_+(B)\rightarrow U_\times(B)$ and $log:U_\times(R)\rightarrow U_+(R)$ are two monoid homomorphisms such that $log;exp=1_{U_\times(B)}$

Jean-Baptiste Vienney (Oct 20 2023 at 13:03):

Then, the above propositions become:

Jean-Baptiste Vienney (Oct 20 2023 at 13:06):

Proposition: If $(B,exp,log)$ is an $exp;log$ ring then any relative exponential ring $\mu:U_{+}(B) \rightarrow U_{\times}(R)$ comes from the exponential ring $(B,exp)$

Jean-Baptiste Vienney (Oct 20 2023 at 13:06):

Proposition: If $(R,exp,log)$ is a $log;exp$ ring then any relative exponential ring $\mu:U_{+}(B) \rightarrow U_{\times}(R)$ comes from the exponential ring $(R,exp)$

Jean-Baptiste Vienney (Oct 20 2023 at 13:08):

So if we want to find a relative exponential ring $\mu:U_{+}(B) \rightarrow U_{\times}(R)$ which doesn't come from an exponential ring, we need at least that:
(1) $B$ can't be equipped with any structure of an $exp;log$ ring
(2) $R$ can't be equipped with any structure of a $log;exp$ ring

Jean-Baptiste Vienney (Oct 20 2023 at 13:19):

Example: for every ring $R$, if we denote $\epsilon_1:R \rightarrow R$ the constant function equal to $1$, then $(R,\epsilon_1)$ is an exponential ring.

Jean-Baptiste Vienney (Oct 20 2023 at 13:20):

Proof: $\epsilon_1(0)=1$ and $\epsilon_1(r+r')=1=1.1=\epsilon_1(r).\epsilon_1(r')$ so that $\epsilon_1:U_+(R)\rightarrow U_\times(R)$ is a monoid homomorphism

Jean-Baptiste Vienney (Oct 20 2023 at 13:29):

Example: for every ring $R$, if we denote $\epsilon_0:R \rightarrow R$ the constant function equal to $0$, then $(R,\epsilon_0)$ is a logarithm ring.

Jean-Baptiste Vienney (Oct 20 2023 at 13:30):

Proof: $\epsilon_0(1)=0$ and $\epsilon_0(r.r')=0=0+0=\epsilon_0(r)+\epsilon_0(r')$ so that $\epsilon_0:U_\times(R)\rightarrow U_+(R)$ is a monoid homomorphism

Jean-Baptiste Vienney (Oct 20 2023 at 13:40):

But $(R,\epsilon_1)$ can never be made into a $log;exp$ ring or an $exp;log$ ring because, then we would have that $R$ if of cardinal $1$ and thus trivial. Same for $(R,\epsilon_0)$.

Jean-Baptiste Vienney (Oct 20 2023 at 16:10):

Example
If $q=p^{k}$ where $k \ge 1$ and $p$ is a prime number, then we have the finite field $GF(q)$. There always exists what we call a primitive element: an element $m \in GF(q)$ which is a generator of the group $(GF(q)\backslash \{0\}, \times)$. Let $m$ be a primitive element of $GF(q)$. We then have a group homomorphism
$log_m:(GF(q) \backslash \{0\},\times) \rightarrow (\mathbb{Z}/(q-1)\mathbb{Z},+)$
such that if we define the group homomorphism
$log_m:(\mathbb{Z}/(q-1)\mathbb{Z},+) \rightarrow (GF(q) \backslash \{0\},\times)$ by
$exp_m(k)=k^m$
then $(GF(q),\mathbb{Z}/(q-1)\mathbb{Z},exp_m,log_m)$ is a relative exp/log ring.

Jean-Baptiste Vienney (Oct 20 2023 at 16:31):

There is also the example of the matrices

Jean-Baptiste Vienney (Oct 20 2023 at 16:32):

Maybe matrices over $\mathbb{Z}/n\mathbb{Z}$ or something similar would provide a relative exponential ring which doesn't come from an exponential ring or a finite field

Mike Stay (Oct 23 2023 at 03:57):

Thanks!

JS PL (he/him) (Oct 23 2023 at 05:56):

I just wanted to say: @Mike Stay this a very interesting idea, and @Jean-Baptiste Vienney fantastic, well done: very cool observations about $exp$ and $log$ rings here.

JS PL (he/him) (Oct 23 2023 at 06:10):

I have done work on exponential rings before, but also with the added assumption of working with a differential exponential ring, so an exponential ring $(R,exp)$ with a derivation $D: R \to R$ such that $D(exp(x)) = exp(x)D(x)$ (think of this as the chain rule for $e^{f(x)}$

JS PL (he/him) (Oct 23 2023 at 06:14):

Now won't the function $exp: \mathbb{R} \to \mathbb{R} [[x ]]$, where the codomain is the ring of formal powers series, where $exp(x) = \sum \frac{x^n}{n!}$ be a relative exponential ring?

Now of course, $\mathbb{R}$ is an exponential ring -- so maybe not the example you are looking for. But this should work for any commutative ring $R$ where you can divide by $\frac{1}{n!}$

JS PL (he/him) (Oct 23 2023 at 06:18):

In fact, if you take the cofree diferential ring over any ring $R$, which is called the Hurwitz ring $H(R)$ (and very similar to power series), I belive you always get a relative exponential ring $R \to H(R)$ as described in https://www.sciencedirect.com/science/article/pii/S0022404998000991

(I am absolutely love Hurwitz rings! so happy to talk more about them if this seems like a relevant example)

Mike Stay (Oct 23 2023 at 16:14):

JS PL (he/him) said:

Now won't the function $exp: \mathbb{R} \to \mathbb{R} [[x ]]$, where the codomain is the ring of formal powers series, where $exp(x) = \sum \frac{x^n}{n!}$ be a relative exponential ring?

I'm confused: the formal variable $x$ isn't an element of $\mathbb{R}$.

JS PL (he/him) said:

In fact, if you take the cofree diferential ring over any ring $R$, which is called the Hurwitz ring $H(R)$ (and very similar to power series), I belive you always get a relative exponential ring $R \to H(R)$ as described in https://www.sciencedirect.com/science/article/pii/S0022404998000991

(I am absolutely love Hurwitz rings! so happy to talk more about them if this seems like a relevant example)

OK, the Hurwitz ring $H(R)$ consists of countable sequences of elements of $R,$ and the derivation is a shift operator. The elements are very similar to the list of coefficients from a formal power series. And the map $\mathbb{R} \to H(\mathbb{R})$ taking 0 to $(1, 0, 0, ...)$ and $r \ne 0$ to the series $a_n = r^n$ (so that the resulting ring element is basically $\exp(rx)$) gives a relative exponential ring, since $\exp((r+s)x) = \exp(rx) \exp(sx)$ in $H(\mathbb{R}).$ Nice!

I think Faà di Bruno's formula should make $H(\mathbb{R})$ into an exponential ring in its own right.

Jean-Baptiste Vienney (Oct 23 2023 at 16:33):

But well in any case, as $\mathbb{R}$ is an exp;log ring with the above terminology, a relative exponential ring $exp:(\mathbb{R},+) \rightarrow (\mathbb{R}[[x]]\backslash \{0\}, \times)$ comes necessarily from the exponential ring $\mathbb{R}$ followed by a morphism of multiplicative monoids through the factorization $(\mathbb{R},+) \overset{exp_{\mathbb{R}}}{\rightarrow} (\mathbb{R} \backslash \{0\}, \times)$ $\overset{log_{\mathbb{R}}}{\rightarrow} (\mathbb{R},+) \overset{exp}{\rightarrow}(\mathbb{R}[[x]] \backslash \{0\},\times)$

Jean-Baptiste Vienney (Oct 23 2023 at 16:41):

However, with a more exotic ring $R$, it could gives a solution to your problem

Jean-Baptiste Vienney (Oct 23 2023 at 16:43):

In the same way, taking $R=GF(q)$, we will get the same kind of factorization

Jean-Baptiste Vienney (Oct 23 2023 at 16:44):

Maybe a morphism of monoids $exp:(\mathbb{Z},+) \rightarrow (H(\mathbb{Z}) \backslash \{0\}, \times)$ would work

JS PL (he/him) (Oct 23 2023 at 20:22):

Mike Stay said:

JS PL (he/him) said:

Now won't the function $exp: \mathbb{R} \to \mathbb{R} [[x ]]$, where the codomain is the ring of formal powers series, where $exp(x) = \sum \frac{x^n}{n!}$ be a relative exponential ring?

I'm confused: the formal variable $x$ isn't an element of $\mathbb{R}$.

Yes quite right, it should be instead $exp(r) = \sum \frac{r^nx^n}{n!}$

JS PL (he/him) (Oct 23 2023 at 20:30):

Jean-Baptiste Vienney said:

Maybe a morphism of monoids $exp:(\mathbb{Z},+) \rightarrow (H(\mathbb{Z}) \backslash \{0\}, \times)$ would work

I don't think you need to remove $0$. (unless you really want your $log$ factorization

Jean-Baptiste Vienney (Oct 23 2023 at 21:37):

Maybe it can help: the only exponential morphisms on $\mathbb{Z}$ are $f(n)=1$ and $f(n)=(-1)^n$. Indeed, if $f$ is an exponential morphism, then $f(0)=f(-1).f(1)=1$, therefore $f(1)$ is a unit, ie $f(1)=\pm 1$ and moreover $f(n)=f(1)^n$ if $n \ge 1$ and $f(-n)=f(-1)^n=f(1)^{-n}$ if $n \le -1$.

Jean-Baptiste Vienney (Oct 23 2023 at 21:38):

More generally if $R$ is any exponential ring, then $f(1)$ is a unit and $f(n)=f(1)^n$ for every $n \in \mathbb{Z}$.

JS PL (he/him) (Oct 23 2023 at 21:39):

Are you not assuming that $exp(0) = 1$?

Jean-Baptiste Vienney (Oct 23 2023 at 21:40):

Thanks, corrected

Jean-Baptiste Vienney (Oct 23 2023 at 21:41):

Yes, we assume this

JS PL (he/him) (Oct 23 2023 at 21:50):

Right so just as a sanity check for myself, what are the exponentials $e: \mathbb{Z} \to \mathbb{Z}$ with $e(n+m) = e(n) + e(m)$ and $e(0) = 1$

Well for $n \in \mathbb{N}$, we get that $e(n) = e(1+ ... + 1) = e(1) ... e(1) = e(1)^n$, so $e(n) = e(1)^n$
and similarly $e(-n) = e(-1 + ... + -1) = e(-1) ... e(-1) = e(-1)^n$, so $e(-n) = e(-1)^n$

So $e$ is completely determined by $e(1)$ and $e(-1)$.

Now we also have that $1=e(0) = e(1 - 1) = e(1) e(-1)$, but in $\mathbb{Z}$ the only possibilities of $xy =1$ are $x=y=1$ or $x=y=-1$

So we must have $e(1) = e(-1) = 1$, which implies that $e(x) = 1$ for all $x \in \mathbb{Z}$
Or we must have $e(1) = e(-1) = -1$, which implies that $e(x)^n = (-1)^{\vert x \vert}$ for all $x \in \mathbb{Z}$

JS PL (he/him) (Oct 23 2023 at 21:52):

But the relative exponential $\mathbb{Z} \to \mathbb{Z}[[x]]$ mapping $a \to \sum \frac{a^n x^n}{n!}$ does not factor through the exponentials on $\mathbb{Z}$

Jean-Baptiste Vienney (Oct 23 2023 at 21:53):

How do you prove this?

Jean-Baptiste Vienney (Oct 23 2023 at 21:54):

Jean-Baptiste Vienney said:

More generally if $R$ is any exponential ring, then $f(1)$ is a unit and $f(n)=f(1)^n$ for every $n \in \mathbb{Z}$.

In the same vein, I find it interesting that any unit $u \in R^\times$ generates a relative exponential ring $\mathbb{Z} \rightarrow R$ by $e(n)=u^{n}$.

Jean-Baptiste Vienney (Oct 23 2023 at 22:08):

Ok, I got it, if you had a factorization, then the relative exponential would take a maximum of two different values

JS PL (he/him) (Oct 23 2023 at 22:08):

Jean-Baptiste Vienney said:

How do you prove this?

Well clearly both $e(x) = 1$ can't have a $l$ such that $l(e(x)) =x$. Similarly for $e(x) = (-1)^{\vert x \vert}$ I believe

Jean-Baptiste Vienney (Oct 23 2023 at 22:09):

Hmm, yes, but you could maybe have a factorization without having a logarithm for your exponential on $\mathbb{Z}$

Jean-Baptiste Vienney (Oct 23 2023 at 22:10):

But this is true that we don't have a factorization through the exponential on $\mathbb{Z}$ followed by a monoid homomorphism $m:(\mathbb{Z},\times) \rightarrow (\mathbb{Z}[[x]],\times)$. If we had one then the relative exponential would take only two values $m(1)$ and $m(-1)$

Jean-Baptiste Vienney (Oct 23 2023 at 23:25):

Mhh, I've been thinking to something. I think it is worth mentionning that for every exp ring $R$, every element $e(x)$ of the image of the exponential is a unit since $e(-x+x)=e(0)=1=e(x)e(-x)$. And so we always get a factorization $(R,+) \rightarrow (R^{\times},\times) \rightarrow (R,\times)$ of the exponential as a composition of two monoid homomorphism, through the group of units.

Jean-Baptiste Vienney (Oct 23 2023 at 23:43):

I think it is better to define an exponential and a relative exponential by writing the domain as $(R \backslash \{0\}, \times)$ if we want the above factorization to be a factorization through an exponential ring. Because here $R^{\times}$ is never a subring of $R$, unless when $R$ is trivial (because it needs to contain $0$).

However, it could happen that $S=R^{\times} \cup \{0\}$ is a subring of $R$. For instance if $R$ is a field (but in this case the above factorization is trivial). But also, if we consider $R=k[x]$ where $k$ is a field, then $S=k$ which is a subfield of $R$.

Jean-Baptiste Vienney (Oct 23 2023 at 23:47):

In that case the factorization is a factorization through a relative exponential ring: $(R,+) \rightarrow (S \backslash \{0\}, \times) \rightarrow (R \backslash \{0\}, \times)$

Jean-Baptiste Vienney (Oct 23 2023 at 23:48):

If we consider $R \neq 0$, then a ring such that $S$ is a subring is exactly "a nontrivial ring such that the sum of two units is either zero or a unit"

Jean-Baptiste Vienney (Oct 23 2023 at 23:51):

It sounds a lot like the definition of a local ring which is "a nontrivial ring such that the sum of two non-units is a non-unit" but this is not the same

Jean-Baptiste Vienney (Oct 23 2023 at 23:58):

On mathsatckexchange, someone asked the question how to name such a ring and someone else answered that we could name such a ring "a good ring" but the guy who answered asks the question whether good rings are useful or not.

Jean-Baptiste Vienney (Oct 24 2023 at 00:05):

It looks like a condition a bit stronger than asking $R$ to be an algebra over a field (in the case that $R$ is commutative, since it then makes $S$ a field )

JS PL (he/him) (Oct 24 2023 at 00:53):

Jean-Baptiste Vienney said:

I think it is better to define an exponential and a relative exponential by writing the domain as $(R \backslash \{0\}, +)$

$(R \backslash \{0\}, +)$ does not make sense. What do you with $x + (-x)$?

Jean-Baptiste Vienney (Oct 24 2023 at 01:05):

What matters is that it is a monoid (if $R \neq 0$)

JS PL (he/him) (Oct 24 2023 at 01:14):

But again its not a monoid because $x + (-x)$ is undefined...

Jean-Baptiste Vienney (Oct 24 2023 at 01:17):

Oh yes

Jean-Baptiste Vienney (Oct 24 2023 at 01:18):

It's annoying. How do you define a log ring so?

JS PL (he/him) (Oct 24 2023 at 01:20):

The way you did it above. If $R$ is an integral domain, then $(R \backslash \{0\}, \times)$ is a well-defined monoid. Then you ask that $log: (R \backslash \{0\}, \times) \to (R,+)$

JS PL (he/him) (Oct 24 2023 at 01:21):

Or you define $log$ as a partial function. (side note: this is why for the differential category story, we have to work in a restriction category to define $log$ via differentiaiton)

Jean-Baptiste Vienney (Oct 24 2023 at 01:22):

JS PL (he/him) said:

Jean-Baptiste Vienney said:

I think it is better to define an exponential and a relative exponential by writing the domain as $(R \backslash \{0\}, +)$

$(R \backslash \{0\}, +)$ does not make sense. What do you with $x + (-x)$?

Ok, I see everything was just a typo, I wrote a $+$ instead of a $\times$

Jean-Baptiste Vienney (Oct 24 2023 at 01:23):

JS PL (he/him) said:

Or you define $log$ as a partial function. (side note: this is why for the differential category story, we have to work in a restriction category to define $log$ via differentiaiton)

Ok!

JS PL (he/him) (Oct 24 2023 at 01:24):

Oh I see, well I think you should keep the definition of an exponential ring as $exp: (R,+) \to (R,\times)$.
Then show that $im(exp) \subset R^\times$, so every $exp(x)$ is a unit.
Then define you can define a log ring as being on the units only. Or something like that. (but you probably still need partiality of some sort)

Jean-Baptiste Vienney (Oct 24 2023 at 01:27):

Hmm, so you can define directly an exponential ring as $(R,+) \rightarrow (R^{\times},\times)$ also

JS PL (he/him) (Oct 24 2023 at 01:47):

Jean-Baptiste Vienney said:

Hmm, so you can define directly an exponential ring as $(R,+) \rightarrow (R^{\times},\times)$ also

I think its better to have this as a nice lemma. Or explain that it is somehow equivalent.

Jean-Baptiste Vienney (Oct 24 2023 at 01:54):

Yes, yes I would say that it is equivalent

Jean-Baptiste Vienney (Oct 24 2023 at 15:24):

I think there is probably some stuff to write about relative exp / log, for instance: Do Hurwitz ring admit a logarithm for the exponential we were talking about? Maybe we could define a logarithm using the Taylor expansion in the same way than the exponential and by some computation with series, find that it is a logarithm, exactly like in calculus. And this is not so obvious by looking at the two Taylor series, that $exp(ln(x))=x$, I mean I have never done this computation... So, what I want to say is that if you guys want to try to write more seriously stuff about these questions, I would be glad to. Nobody as written anything in the litterature about relative exponential rings or combining exp and log ring, as far as Google tells me, so I think it would be valuable to study such an algebraic setting. But, not right now, I say this for one day in the future, I must focus on my courses until the end of May :smiling_face_with_tear:

Jean-Baptiste Vienney (Oct 24 2023 at 15:26):

I would be glad to obtain results about kinda classifying these structures as much as we can

Jean-Baptiste Vienney (Oct 24 2023 at 15:27):

I'd like to understand if there are super weird examples or not for instance

Jean-Baptiste Vienney (Oct 24 2023 at 15:27):

By the way, we can work with rigs and not only rings

Jean-Baptiste Vienney (Oct 24 2023 at 15:27):

I believe there are nice examples from tropical analysis kind of stuff

Jean-Baptiste Vienney (Oct 24 2023 at 15:28):

I'm very attracted by understanding phenoma which appear at the same time in finite fields or algebra in general, calculus and tropical anaylsis stuff

Jean-Baptiste Vienney (Oct 24 2023 at 15:35):

By the way, JS as probably did stuff about combining logarithm and exponential in a differential category but I'm not against working simply with rigs

Jean-Baptiste Vienney (Oct 24 2023 at 15:36):

I mean if there is an agreement that it's possible to do valuable work

Jean-Baptiste Vienney (Oct 24 2023 at 15:38):

I'm wondering if we could define free exp rings, free log rings for instance also

Jean-Baptiste Vienney (Oct 24 2023 at 15:40):

I'm not even sure that someone as written anything about the category of exponential rings...