Is K^I with an arbitrary index set and pointwise product always an unital algebra?

Question

I don't see any reason why it might not be correct, but the "arbitrary index set" bothers me. Is it wrong for infinite dimensions?

Answer 1

I think you might be confusing $K^I$, which also gets written as $\prod_I K$, with the finite support version $\bigoplus_I K$.

The first is the family of all functions $I \to K$, where we multiply and add functions in the obvious way (pointwise). This is, in general, an enormous algebra. In particular, for a big enough set $I$, the characteristic functions

$$\chi_i(x) = \begin{cases} 1 & x=i \\ 0 & x \neq i \end{cases}$$

do not form a basis.

Contrast this with the vector space $\bigoplus_I K$, which is defined to be the vector space that does have the $\chi_i$ as a basis. In particular, since every element in this space can be written as a finite linear combination of the $\chi_i$, any function $f \in \bigoplus_I K$ takes nonzero values on only a finite set. We say that $f$ has finite support.

Now we see the difference: You are correct in recognizing that $\bigoplus_I K$ is not a unital algebra for $I$ infinite. This is because the unit must be $f_1(x) = 1$ the constant $1$ function. But when $I$ is infinite, $f_1$ does not have finite support.

But in $K^I$, we are allowing all functions. We pay the price of losing an obvious basis - in fact, we can only find bases by using the Axiom of Choice in general. But we gain access to all functions, in particular $f_1$ from before. So this algebra is, in fact, unital.

As a brief aside, there are other subtle differences between these algebras. $K^I$ is a contravariant thing to do, while $\bigoplus_I K$ is a covariant thing to do. This leads to some different categorical properties that don't exist when $I$ is finite (because in the finite case the two constructions coincide).

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I hope this helps ^_^