 # Commutative Algebra Explained: In-Depth Examples, Equations, and Expert Insights

Recent questions in Commutative Algebra mugenziwzn6 2023-02-21

## The permutations and combinations of $abcd$ taken $3$ at a time are respectively. Ghillardi4Pi 2022-12-04

## Show that the image of ${\mathbb{P}}^{n}×{\mathbb{P}}^{m}$ under the Segre embedding $\psi$ is actually irreducible. musicbachv7 2022-08-05

## Is function composition commutative? Lillianna Andersen 2022-07-15

## If $B$ and $C$ are $A$-algebras with ring morphisms $f:A\to B$ and $g:A\to C$, and $D=B{\otimes }_{A}C$ is an $A$-algebra with morphism $a↦f\left(a\right)\otimes g\left(a\right)$, then $uf=vg$, where $u:B\to D$ is $u\left(b\right)=b\otimes 1$.The map $v:C\to D$ is not defined in the text, but my guess is it's $v\left(c\right)=1\otimes c$.I don't understand why the diagram is commutative though. That would imply $f\left(a\right)\otimes 1=1\otimes g\left(a\right)$ for all $a\in A$. Is that true, or is v something else?Added: On second thought, does this follow since $f\left(a\right)\otimes 1=a\cdot \left(1\otimes 1\right)$ and $1\otimes g\left(a\right)=a\cdot \left(1\otimes 1\right)$ where ⋅ is the $A$-module structure on $D$? $\begin{array}{ccc}A& \stackrel{f}{\to }& B\\ g↓& \mathrm{#}& ↓u& \\ C& \underset{v}{\overset{\phantom{\rule{2.75em}{0ex}}}{\to }}& D\end{array}$ slijmigrd 2022-07-14

## Let $A\subseteq B$ be rings, $B$ integral over $A$; let $\mathfrak{q},{\mathfrak{q}}^{\prime }$ be prime ideals of $B$ such that $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ and ${\mathfrak{q}}^{c}={\mathfrak{q}}^{\prime c}=\mathfrak{p}$ say. Then $\mathfrak{q}={\mathfrak{q}}^{\prime }$.Question 1. Why ${\mathfrak{n}}^{c}={\mathfrak{n}}^{\prime c}=\mathfrak{m}$?My attempt: Since $\mathfrak{p}\subseteq \mathfrak{q}$, we have $\mathfrak{m}={S}^{-1}\mathfrak{p}\subseteq {S}^{-1}\mathfrak{q}=\mathfrak{n}\subseteq {B}_{\mathfrak{p}}$. But m is maximal in ${A}_{\mathfrak{p}}$, which is not necessarily maximal in ${B}_{\mathfrak{p}}$. I can't get $\mathfrak{m}=\mathfrak{n}$ by this.Question 2. When we use that notation ${A}_{\mathfrak{p}}$, which means the localization ${S}^{-1}A$ of $A$ at the prime ideal $\mathfrak{p}$ of $A$. But in this corollary, $\mathfrak{p}$ doesn't necessarily be a prime ideal of $B$. Why can he write ${B}_{\mathfrak{p}}$? Should we write ${S}^{-1}B$ rigorously? Kyle Sutton 2022-07-13

## Proof of commutative property in Boolean algebra$a\vee b=b\vee a$$a\wedge b=b\wedge a$ orlovskihmw 2022-07-11

## The theorem is stated in the context of commutative Banach (unitary) algebras, but the proof seems to show that it is valid for any commutative algebra defined as a linear space where a commutative, associative and distributive (with respect to the addition) multiplication is defined such that $\mathrm{\forall }\alpha \in \mathbb{K}\phantom{\rule{1em}{0ex}}\alpha \left(xy\right)=\left(\alpha x\right)y=x\left(\alpha y\right)$.In any case, whether it concerns only commutative Banach unitary algebras or commutative algebras as defined above, I think we must intend contained as properly contained. Am I right? Lorena Beard 2022-07-10

## Let $H$ be a Hilbert space and $\mathcal{A}$ a commutative norm-closed unital $\ast$-subalgebra of $\mathcal{B}\left(H\right)$. Let $\mathcal{M}$ be the weak operator closure of $\mathcal{A}$.Question: For given a projection $P\in \mathcal{M}$, is the following true?$P=inf\left\{A\in \mathcal{A}:P\le A\le 1\right\}$It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $P$. Also, if the above is true, what happens if $\mathcal{A}$ is non-commutative? Kaeden Hoffman 2022-07-10

## Just a simple question. What does Eisenbud mean by $\left(x:y\right)$ where $x,y\in R$ a ring. An example on this is in the section 17 discussing the homology of the koszul complex. I assume it's something along the lines of $\left\{r\mid ax=ry\right\}$ for some $a\in R$. Brock Byrd 2022-07-10

## Let $R$ be a commutative $k$-algebra, where $k$ is a field of characteristic zero.Could one please give an example of such $R$ which is also:(i) Not affine (= infinitely generated as a $k$-algebra).and(ii) Not an integral domain (= has zero divisors).My first thought was $k\left[{x}_{1},{x}_{2},\dots \right]$, the polynomial ring over k in infinitely many variables, but unfortunately, it satisfies condition (i) only. It is not difficult to see that it is an integral domain: If $fg=0$ for some $f,g\in k\left[{x}_{1},{x}_{2},\dots \right]$, then there exists $M\in \mathbb{N}$ such that $f,g\in k\left[{x}_{1},\dots ,{x}_{M}\right]$, so if we think of $fg=0$ in $k\left[{x}_{1},\dots ,{x}_{M}\right]$, we get that $f=0$ or $g=0$, and we are done. rmd1228887e 2022-07-10

## Let $A$ be a commutative Banach algebra. Let ${\chi }_{1}$ and ${\chi }_{2}$ be characters of $A$.I am having some difficulty seeing why the following statement is true:If $\mathrm{ker}{\chi }_{1}=\mathrm{ker}{\chi }_{2}$, then since ${\chi }_{1}\left(\mathbf{\text{1}}\right)={\chi }_{2}\left(\mathbf{\text{1}}\right)=\mathbf{\text{1}}$, we have that ${\chi }_{1}={\chi }_{2}$. Wisniewool 2022-07-09

## Need to find a functor $T:$ Set $\to$ Set such that Alg(T) is concretely isomorphic to the category of commutative binary algebras.The first idea is that the functor is likely to map object $X\in Ob\left($ to the $X×X$ because then we have to get a binary algebra, i.e., the operation $X×X\to X$, which have to be commutative.So the question (if these thoughts are right) is: how to map $X$ to $X×X$ to get later a commutative binary algebra? therightwomanwf 2022-07-08

## I would like to know, under what condition on the group $G$ (abelian, compact or localement compact ...), the algebra ${L}^{1}\left(G\right)$ is commutative? Maliyah Robles 2022-07-07

## Let $\mathcal{A}$ be a ${C}^{\ast }$-algebra.(i) Let $\phi$ be a state on a $u\in \mathcal{A}$-algebra $\mathcal{A}$. Suppose that $|\phi \left(u\right)|=1$ for all unitary elements u∈A. Show that φ is a pure state. [Hint: $\mathrm{span}\mathcal{U}\left(\mathcal{A}\right)=\mathcal{A}$ ](ii) Let $\phi$ be a multiplicative functional on a ${C}^{\ast }$-algebra $\mathcal{A}$. Show that $\phi$ is a pure state on $\mathcal{A}$.(iii) Show that the pure and multiplicative states coincide for commutative $\mathcal{A}$.I managed to work out the first two problems but I have no idea about the last one. How to see from being an extreme element in the state space of a commutative $\mathcal{A}$ that the extreme element is multiplicative? Yesenia Obrien 2022-07-07

## I am at a very initial stage of commutative algebra. I want to know whether the power of a prime ideal in a commutative ring is prime ideal or not? Callum Dudley 2022-07-03

## $A\subset B$ is a ring extension. Let $y,z\in B$ elements which satisfy quadratic integral dependance ${y}^{2}+ay+b=0$ and ${z}^{2}+cz+d=0$ over $A$. Find explicit integral dependance relations for $y+z$ and $yz$. Sam Hardin 2022-07-02

## $\text{Content}\left(fg\right)\subset \text{Content}\left(f\right)\text{Content}\left(g\right)\subset \text{rad}\left(\text{Content}\left(fg\right)\right)$to deduce that if $\text{Content}\left(f\right)$ contains a nonzerodivisor of $R$, then $f$ is nonzerodivisor of $S=R\left[{x}_{1},...,{x}_{r}\right]$. ntaraxq 2022-07-02

## Let $\mathfrak{g}$ be a complex linear Lie algebra. Assume that the center $\mathfrak{z}$ of $\mathfrak{g}$ is trivialLet $\mathfrak{r}$ be the radical of $\mathfrak{g}$. If $\mathfrak{r}$ is abelian, then $\mathfrak{g}$ is semisimple?What if $\mathfrak{g}$ is the Lie algebra of an algebraic complex linear group? Aganippe76 2022-07-02
## Is it true that every commutative Hopf algebra is related to a Group in such a way that the co-multiplication is originated from the multiplication of the group, the antipode from the inverse?Making it more explicitly, can all commutative and co-comutative Hopf algebra $H$ be written in this form: $H=\mathbb{C}\left[G\right]$, with the usual group algebra structure$\eta :1\to e,\phantom{\rule{1em}{0ex}}m:g\otimes h↦gh$the coalgebra structure$\epsilon :g↦1,\phantom{\rule{1em}{0ex}}\mathrm{\Delta }:g↦g\otimes g$where all of the maps above are defined on the basis of group elements? lilmoore11p8 2022-07-01