Is function composition commutative?

musicbachv7
2022-08-05
Answered

Is function composition commutative?

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Jake Landry

Answered 2022-08-06
Author has **22** answers

The functions g and f are said to commute with each other if g???f = f???g. In general, composition of functions will not be commutative.

Commutativity is a special property, attained only by particular functions, and often in special circumstances. For example, |x| + 3 = |x + 3| only when x ? 0.

Commutativity is a special property, attained only by particular functions, and often in special circumstances. For example, |x| + 3 = |x + 3| only when x ? 0.

asked 2022-05-29

suppose $A$ is a ${C}^{\ast}$-algebra and $x$ is a normal element in $A$. ${C}^{\ast}$-algebra generated by $x$ denote by $A[x]$. then

1) $A[x]$ is commutative.

2) $A[x]$ is the clouser of polynomials of two variable $x$ and ${x}^{\ast}$.

1) $A[x]$ is commutative.

2) $A[x]$ is the clouser of polynomials of two variable $x$ and ${x}^{\ast}$.

asked 2022-06-21

Is there an algebraic way to construct the coproduct, as for example some form of topological tensor product or similar?

Let $A$ be a commutative unital C*-Algebra and let $X=Spec(A)$ be the corresponding compact Hausdorff space of characters. By Gelfand-Naimark duality we know that

$X\times X=Spec(A\coprod A)$

or in other words

$A\coprod A=\mathcal{C}(X\times X).$

Let $A$ be a commutative unital C*-Algebra and let $X=Spec(A)$ be the corresponding compact Hausdorff space of characters. By Gelfand-Naimark duality we know that

$X\times X=Spec(A\coprod A)$

or in other words

$A\coprod A=\mathcal{C}(X\times X).$

asked 2022-06-20

Let $A$ be some $F$-Algebra, for some field $F$, with the property that $A$ is finite dimensional over $F$. Is $A$ always commutative?

asked 2022-04-10

An associative magma is said to be a semigroup. A semigroup with identity is said to be a monoid... Is there a name for an associative and commutative algebra?

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 2022-05-24

Show that $R$ can be written as a direct product of two or more (nonzero) rings iff $R$ contains a non-trivial idempotent. Show that if $e$ is an idempotent, then $R=Re\times R(1-e)$ and that $Re$ may be realized as a localization, $Re=R[{e}^{-1}]$.

asked 2022-06-24

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?