Why do we need ultrafilters to make sense of the cartesian product of $\mathcal{L}$-structures I'm trying understand why we need ultrafilters in model theory. Here is how I see things. Could someone tell me if this is correct ? Further explanations are always welcome.
Let $\mathcal{L}$ be a first-order language; let $I$ be a set; for all $i \in I$ let $\mathcal{M}_i := \big(M, (...)^\mathcal{M}\big)$ be a $\mathcal{L}$-structure.
Suppose I want to define turn $N:=\prod\limits_{i \in I}M_i$ into a $\mathcal{L}$-structure (say $\mathcal{N}$). I would need to be able to interpret the constant, function and relation symbols of $\mathcal{L}$. It doesn't seem to be a problem for constant and function symbols as we could define the interpretation in $\mathcal{N}$ to be the component-wise interpretation.
But for relation symbols we need some systematic way of converting things like $\langle T, F, T, T, F, F, ... \rangle$ into $T$ or $F$ (this is not formal since $I$ need not be countable nor linearly ordered but you see what I mean). Now I was wondering why couldn't we just define the relation to be true in $\mathcal{N}$ if and only is we've got $\langle T, T, T, ...\rangle$ (that is if all components are T).
But on second thought I feel it would help us much as we need the fact that the filter is ultra in the proof of Los theorem. To be more precise, it is not true that $\mathcal{N}\vDash \neg\varphi$ if and only if $\mathcal{M}_i \vDash \neg \varphi$ for all $i \in I$. Therefore the filter $\{I\}$ - which is not "ultra" - as some serious drawback that we can get rid of if we replace $\{I\}$ by an ultrafilter.
Does that make any sense ?
 A: You don't need ultrafilters to define the product of structures. You're absolutely right that we can just define the interpretation of all the symbols in the language in the obvious way.
But the question is, what are you planning to do with that? For example, if you consider the product of all finite fields, you get a ring. If you take an ultraproduct of finite fields, you get a field. And usually you get a field of characteristics $0$.
If you take a product of linear orders you will get a partial order. But if you take an ultraproduct, you get a linear order.
You can even talk about restricted product, namely use a filter rather than an ultrafilter. If you are familiar with the notion of the Adele ring, this is a wonderful example for a restricted product using the cofinite filter and all the metric completions of the rationals (namely, $\Bbb Q_p$ and $\Bbb R$).
So of course, you don't need to take a quotient to talk about products. But taking a quotient has this wonderful property that the result can be "closer" to the original structures.
