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?

gutinyalk

Beginner2022-07-08Added 11 answers

An ideal $I\subset R$ is prime iff $R/I$ is an integral domain (that is, it contains no nontrivial zero divisors).

If $I$ is a prime ideal such that ${I}^{2}\ne I$, then in the quotient ring $R/{I}^{2}$ the elements of the form $x+{I}^{2}$, where $x\in I$, will be nilpotents. Thus, $R/{I}^{2}$ is not a domain and ${I}^{2}$ is not prime.

For a concrete and illustrative example, consider $I=(p)\subset \mathbb{Z}$ - the ideal of the ring of integers generated by a prime $p$. Then, ${I}^{2}=({p}^{2})$, and is not prime (since it is generated by a non-prime integer).

If $I$ is a prime ideal such that ${I}^{2}\ne I$, then in the quotient ring $R/{I}^{2}$ the elements of the form $x+{I}^{2}$, where $x\in I$, will be nilpotents. Thus, $R/{I}^{2}$ is not a domain and ${I}^{2}$ is not prime.

For a concrete and illustrative example, consider $I=(p)\subset \mathbb{Z}$ - the ideal of the ring of integers generated by a prime $p$. Then, ${I}^{2}=({p}^{2})$, and is not prime (since it is generated by a non-prime integer).

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

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

Is function composition commutative?

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 $D$?

$\begin{array}{ccc}A& \stackrel{f}{\to}& B\\ {\scriptstyle g}\downarrow & \mathrm{\#}& \downarrow {\scriptstyle u}& \\ C& \underset{v}{\overset{\phantom{\rule{2.75em}{0ex}}}{\to}}& D\end{array}$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?Proof of commutative property in Boolean algebra

$a\vee b=b\vee a$

$a\wedge b=b\wedge a$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?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?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 $a\in R$.

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.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 ${\chi}_{1}={\chi}_{2}$.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?I would like to know, under what condition on the group $G$ (abelian, compact or localement compact ...), the algebra ${L}^{1}(G)$ is commutative?

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?$A\subset B$ 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 $yz$.