Recent questions in Commutative Algebra

Commutative AlgebraAnswered question

mugenziwzn6 2023-02-21

The permutations and combinations of $abcd$ taken $3$ at a time are respectively.

Commutative AlgebraAnswered question

Ghillardi4Pi 2022-12-04

Show that the image of ${\mathbb{P}}^{n}\times {\mathbb{P}}^{m}$ under the Segre embedding $\psi $ is actually irreducible.

Commutative AlgebraAnswered question

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 $a\mapsto f(a)\otimes g(a)$, then $uf=vg$, where $u:B\to D$ is $u(b)=b\otimes 1$.The map $v:C\to D$ is not defined in the text, but my guess is it's $v(c)=1\otimes c$.I don't understand why the diagram is commutative though. That would imply $f(a)\otimes 1=1\otimes g(a)$ for all $a\in A$. Is that true, or is v something else?Added: On second thought, does this follow since $f(a)\otimes 1=a\cdot (1\otimes 1)$ and $1\otimes g(a)=a\cdot (1\otimes 1)$ where ⋅ is the $A$-module structure on $D$?

$\begin{array}{ccc}A& \stackrel{f}{\to}& B\\ {\scriptstyle g}\downarrow & \mathrm{\#}& \downarrow {\scriptstyle u}& \\ C& \underset{v}{\overset{\phantom{\rule{2.75em}{0ex}}}{\to}}& D\end{array}$

$\begin{array}{ccc}A& \stackrel{f}{\to}& B\\ {\scriptstyle g}\downarrow & \mathrm{\#}& \downarrow {\scriptstyle u}& \\ C& \underset{v}{\overset{\phantom{\rule{2.75em}{0ex}}}{\to}}& D\end{array}$

Commutative AlgebraAnswered question

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?

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?

Commutative AlgebraAnswered question

Kyle Sutton 2022-07-13

Proof of commutative property in Boolean algebra

$a\vee b=b\vee a$

$a\wedge b=b\wedge a$

$a\vee b=b\vee a$

$a\wedge b=b\wedge a$

Commutative AlgebraAnswered question

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 $\mathrm{\forall}\alpha \in \mathbb{K}\phantom{\rule{1em}{0ex}}\alpha (xy)=(\alpha x)y=x(\alpha y)$.

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?

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?

Commutative AlgebraAnswered question

Lorena Beard 2022-07-10

Let $H$ be a Hilbert space and $\mathcal{A}$ a commutative norm-closed unital $\ast $-subalgebra of $\mathcal{B}(H)$. Let $\mathcal{M}$ be the weak operator closure of $\mathcal{A}$.

Question: For given a projection $P\in \mathcal{M}$, is the following true?

$P=inf\{A\in \mathcal{A}:P\le A\le 1\}$

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?

Question: For given a projection $P\in \mathcal{M}$, is the following true?

$P=inf\{A\in \mathcal{A}:P\le A\le 1\}$

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?

Commutative AlgebraAnswered question

Kaeden Hoffman 2022-07-10

Just a simple question. What does Eisenbud mean by $(x:y)$ 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 $\{r\mid ax=ry\}$ for some $a\in R$.

Commutative AlgebraAnswered question

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 $k[{x}_{1},{x}_{2},\dots ]$, 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 $f,g\in k[{x}_{1},{x}_{2},\dots ]$, then there exists $M\in \mathbb{N}$ such that $f,g\in k[{x}_{1},\dots ,{x}_{M}]$, so if we think of $fg=0$ in $k[{x}_{1},\dots ,{x}_{M}]$, we get that $f=0$ or $g=0$, and we are done.

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 $k[{x}_{1},{x}_{2},\dots ]$, 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 $f,g\in k[{x}_{1},{x}_{2},\dots ]$, then there exists $M\in \mathbb{N}$ such that $f,g\in k[{x}_{1},\dots ,{x}_{M}]$, so if we think of $fg=0$ in $k[{x}_{1},\dots ,{x}_{M}]$, we get that $f=0$ or $g=0$, and we are done.

Commutative AlgebraAnswered question

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 ${\chi}_{1}(\mathbf{\text{1}})={\chi}_{2}(\mathbf{\text{1}})=\mathbf{\text{1}}$, we have that ${\chi}_{1}={\chi}_{2}$.

I am having some difficulty seeing why the following statement is true:

If $\mathrm{ker}{\chi}_{1}=\mathrm{ker}{\chi}_{2}$, then since ${\chi}_{1}(\mathbf{\text{1}})={\chi}_{2}(\mathbf{\text{1}})=\mathbf{\text{1}}$, we have that ${\chi}_{1}={\chi}_{2}$.

Commutative AlgebraAnswered question

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 $X\in Ob($ to the $X\times X$ because then we have to get a binary algebra, i.e., the operation $X\times X\to X$, which have to be commutative.

So the question (if these thoughts are right) is: how to map $X$ to $X\times X$ to get later a commutative binary algebra?

The first idea is that the functor is likely to map object $X\in Ob($ to the $X\times X$ because then we have to get a binary algebra, i.e., the operation $X\times X\to X$, which have to be commutative.

So the question (if these thoughts are right) is: how to map $X$ to $X\times X$ to get later a commutative binary algebra?

Commutative AlgebraAnswered question

therightwomanwf 2022-07-08

I would like to know, under what condition on the group $G$ (abelian, compact or localement compact ...), the algebra ${L}^{1}(G)$ is commutative?

Commutative AlgebraAnswered question

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 $|\phi (u)|=1$ for all unitary elements u∈A. Show that φ is a pure state. [Hint: $\mathrm{span}\mathcal{U}(\mathcal{A})=\mathcal{A}$ ]

(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?

(i) Let $\phi $ be a state on a $u\in \mathcal{A}$-algebra $\mathcal{A}$. Suppose that $|\phi (u)|=1$ for all unitary elements u∈A. Show that φ is a pure state. [Hint: $\mathrm{span}\mathcal{U}(\mathcal{A})=\mathcal{A}$ ]

(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?

Commutative AlgebraAnswered question

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?

Commutative AlgebraAnswered question

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 $yz$.

Commutative AlgebraAnswered question

Sam Hardin 2022-07-02

$\text{Content}(fg)\subset \text{Content}(f)\text{Content}(g)\subset \text{rad}(\text{Content}(fg))$

to deduce that if $\text{Content}(f)$ contains a nonzerodivisor of $R$, then $f$ is nonzerodivisor of $S=R[{x}_{1},...,{x}_{r}]$.

to deduce that if $\text{Content}(f)$ contains a nonzerodivisor of $R$, then $f$ is nonzerodivisor of $S=R[{x}_{1},...,{x}_{r}]$.

Commutative AlgebraAnswered question

ntaraxq 2022-07-02

Let $\mathfrak{g}$ be a complex linear Lie algebra. Assume that the center $\mathfrak{z}$ of $\mathfrak{g}$ is trivial

Let $\mathfrak{r}$ 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?

Let $\mathfrak{r}$ 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?

Commutative AlgebraAnswered question

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: $H=\mathbb{C}[G]$, with the usual group algebra structure

$\eta :1\to e,\phantom{\rule{1em}{0ex}}m:g\otimes h\mapsto gh$

the coalgebra structure

$\epsilon :g\mapsto 1,\phantom{\rule{1em}{0ex}}\mathrm{\Delta}:g\mapsto g\otimes g$

where all of the maps above are defined on the basis of group elements?

Making it more explicitly, can all commutative and co-comutative Hopf algebra $H$ be written in this form: $H=\mathbb{C}[G]$, with the usual group algebra structure

$\eta :1\to e,\phantom{\rule{1em}{0ex}}m:g\otimes h\mapsto gh$

the coalgebra structure

$\epsilon :g\mapsto 1,\phantom{\rule{1em}{0ex}}\mathrm{\Delta}:g\mapsto g\otimes g$

where all of the maps above are defined on the basis of group elements?

Commutative AlgebraAnswered question

lilmoore11p8 2022-07-01

Hamiltons quaternion rule states that ij=k and ji=-k. How can the commutative rule just be broken to make this true?

In basic terms, commutative algebra represents a complex study of the rings that take place in algebraic number theory. It also relates to solving problems and questions based on algebraic geometry. You will see that there are solutions related to Dedekind rings and various commutative patterns. Commutative algebra solutions will be quite short in most cases, yet one should start with examples that are represented and link them with various questions. Take your time to post your commutative algebra example problem as well and always compare your question to similar questions as it will help you provide the required data.
When you are dealing with inequality and graph solution tasks as you are seeking commutative equ