Is it true that for simple ${C}^{\ast}$-algebras, meaning that they don't have non-trivial two-sided ideals, it holds that they are necessarily non-commutative? And why?

vittorecostao1
2022-06-24
Answered

Is it true that for simple ${C}^{\ast}$-algebras, meaning that they don't have non-trivial two-sided ideals, it holds that they are necessarily non-commutative? And why?

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asked 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?

asked 2022-05-11

Let x not be quasinilpotent, so $li{m}_{n\to \mathrm{\infty}}||{x}^{n}|{|}^{1/n}=\lambda \ne 0$. Let $\pi (x)=\overline{x}$, where $\pi (x):A\to A/rad(A)$ is the canonical quotient map. Suppose that $||{\overline{x}}^{n}|{|}^{1/n}=0$. Then it's spectrum $\sigma (\overline{x})=0$, so $\overline{x}$ is not invertible in $A/rad(A)$, and thus generates a proper ideal in $A/rad(A)$. So then ${\pi}^{-1}(\overline{x})$ generates a proper ideal in $A$ containing $rad(A)$.

From here, if $rad(A)$ is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?

From here, if $rad(A)$ is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?

asked 2020-12-24

Show that quaternion multiplication is not commutative. That is, give an example to show that sometimes $\epsilon \eta$ does not equal $\eta \epsilon$ .

asked 2022-05-21

Does Aluffi's book have enough commutative algebra for algebraic geometry? I understand that traditional graduate algebra course using Hungerford's book or Lang's book provides enough background for such a course.

asked 2022-05-20

Classifying all commutative $\mathbb{R}$-algebras of matrices over $\mathbb{R}$?

I initially thought they were all isomorphic to some subring of the $n\times n$ diagonal matrices $\mathcal{D}\cong \mathbb{R}\times \cdots \times \mathbb{R}$, but this was wrong: Every commutative ring of matrices over $\mathbb{R}$ is isomorphic to the diagonals?. One counterexample is matrices of the form (using block matrix notation) $\left[\begin{array}{cc}\alpha {I}_{1}& A\\ 0& \alpha {I}_{n-1}\end{array}\right]$ for some $1\times (n-1)$ real matrix block $A$ and some $\alpha \in \mathbb{R}$, which forms a commutative ring $(\mathcal{U},+,\ast )$.

Are there other counterexamples? Can we classify all such rings up to isomorphism?

I initially thought they were all isomorphic to some subring of the $n\times n$ diagonal matrices $\mathcal{D}\cong \mathbb{R}\times \cdots \times \mathbb{R}$, but this was wrong: Every commutative ring of matrices over $\mathbb{R}$ is isomorphic to the diagonals?. One counterexample is matrices of the form (using block matrix notation) $\left[\begin{array}{cc}\alpha {I}_{1}& A\\ 0& \alpha {I}_{n-1}\end{array}\right]$ for some $1\times (n-1)$ real matrix block $A$ and some $\alpha \in \mathbb{R}$, which forms a commutative ring $(\mathcal{U},+,\ast )$.

Are there other counterexamples? Can we classify all such rings up to isomorphism?

asked 2022-01-07

Show that ${L}^{1}\left(R\right)$ a Banach algebra Commutative

asked 2022-06-20

One forms ${\mathrm{\Omega}}^{1}(M)$ is a module over ${C}^{\mathrm{\infty}}(M)$, therefore does that make ${C}^{\mathrm{\infty}}(M)$ an algebra or a commutative ring?