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Rudin PMA p.161

It says, A family $\mathscr{A}$ of complex functions defined on a set $E$ is said to be an algebra if (i) $f+g\in\mathscr{A}$, (ii)$fg\in\mathscr{A}$, and (iii) $cf\in\mathscr{A}$ for all $f\in\mathscr{A}, g\in\mathscr{A}, c\in\mathbb{C}$.

I guess it is definitely not a definition of 'algebra' generally, but it is an example of 'algebra'. And how do i call it? $\mathscr{A}$ is an algebra on $E$?

I'm asking this because it seems Rudin doesn't make definitions very clearly.

(He even defined 'equicontinuous' as the definition of 'uniformly equicontinuous' on wikipedia, which are totally different)

I tried to understand definition on wikipedia myself, but then there are many terms i'm not familiar with, so i tried to find meanings of those terms, then again there are terms i don't know.

So far, i saw that algebra has different meanings depend on situation, and don't even know what 'algebra' here in my book means.

Please someone explain what kind of algebra it is, and what is the definition of that algebra in relatively easy terms.

Thank you in advance

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The appropriate wikipedia link here is en.wikipedia.org/wiki/Algebra_over_a_field –  Learner Dec 17 '12 at 12:20

1 Answer 1

up vote 3 down vote accepted

What you're mentioning there is an example of an algebra over the complex $\,\Bbb C\,$: it is a ring which is also a complex vector space. As simple and important as this. You can sum and also multiply vectors among themselves, and you can multiply then by scalars in some field.

Polynomials rings (which would then be called algebras), square matrices and functions (with some possible restrictions here and there) are some classical examples of algebras

Now, there is other, more formal, definition of an algebra (usually over a field or a division ring, but not necessarily) $\,A\,$ over $\,B\,$, which talks of an injection of rings $\,i:B\to A\,$ s.t. some conditions are fulfilled (e.g., compatible with scalar multiplication or, to make things simple, into $\,A'$s center, etc.), but to begin with I don't think you need to mess up with this.

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This explains very nice! Thank you –  Katlus Dec 17 '12 at 12:30
    
In the general definition, $A\to B$ needs not be injective. –  user18119 Dec 17 '12 at 12:42
    
Yes, that's true. Thanks –  DonAntonio Dec 17 '12 at 14:10

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