$Center of algebra of functions$

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$Center of algebra of functions$

Let $A$ be an algebra over a field (not necessarily commutative) with multiplication $m$ and unit $1_A$, and $X$ be a finite set. The set of functions from $X$ to $A$, $Fun(X,A)$, can be given an algebra structure with the obvious pointwise definitions: $\forall x \in X, (f+g)(x):=f(x)+g(x), (\lambda f)(x):=\lambda f(x), (fg)(x):=f(x)g(x)$. Clearly the center $Z=\{f \in Fun(X,A): fg=gf, \forall g \in Fun(X,A)\}$ of $Fun(X,A)$ is the whole $Fun(X,A)$ if $A$ is commutative (since $Fun(X,A)$ is also commutative), but what is the center if $A$ is not commutative?

Algebra

Jbuck

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Let $Z_A$ denote the center of the algebra $A$, that is, $Z_A=\{z\in A | za = az, \forall a \in A\}$.

We claim that $Z$, the center of $Fun(X,A)$, is the set of functions from $X$ to the center of $A$: $Fun(X,Z_A)$, i.e, $Z =Fun(X,Z_A)$.

We prove this by showing double inclusion:
$\subset$ :
Let $f\in Z$ and consider any arbitrary element $w \in A$. Define the constant map $g_w \in Fun(X,A)$ by $g_w(x)= w, \forall x \in X$. Since $f\in Z$, it commutes with $g_w$: $fg_w = g_wf$. This means that for any fixed $x \in X$, $f(x)g_w(x) = g_w(x) f(x)$, that is, $f(x)w=wf(x)$. Since $w$ is arbitrary, this implies that $f(x) \in Z_A$, and since $x$ was arbitrary, $f\in Fun(X,Z_A)$.

$\supset$:
Let $f\in Fun(X,Z_A)$ and consider any $g\in Fun(X,A)$. For any $x\in X,$ $(fg)(x) = f(x)g(x) = g(x)f(x) = (gf)(x)$, since $f(x) \in Z_A$, so it commutes with $g(x) \in A$. Since $g$ was arbitrary, $f\in Z$.

Comment:
We have shown an equivalence between $Z$ and $Fun(X,Z_A)$, with the latter depending fundamentally on the algebra $A$, regardless of the cardinality of $X$.

Mathe

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Jbuck

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Hey there, I hope you're alright! I was wondering if you could help me with some questions this Friday, 10a.m. GMT. I will have around a 2-2.5 hour window and I'm willing to pay $60 per question if you're online and able to answer within the time limit, would you be interested?
Mathe

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Yes, I am interested. Did you mean GMT +0?
Jbuck

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Sorry, it is actually GMT +1. Perfect then, I will be posting the required questions around Friday, 10 a.m. GMT+1, thank you!
Mathe

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I just realized I won't be available at that time. I'm sorry.
Jbuck

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I see, that is alright? Do you know if there is a way to send a private message to a user in this site instead of commenting in old posts?
Mathe

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There isn't as far as I know. But if you can try the next day that would be great.
Jbuck

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Are you available Monday, May 23?
Mathe

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Yes!
Jbuck

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Great, do you have a particular preference regarding time?
Mathe

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11 am GMT +1!
Jbuck

0

Alright great, I will be posting the questions around that time, thanks!
Jbuck

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Hello, I'm posting this to confirm that you will be available Monday, May 23, 11 a.m. GMT +1, which is in around 1 hour and 40 minutes from now, thank you
Mathe

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Hi, I'm here.

$Center of algebra of functions$

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