Constructing a field from a direct product of fields Let $F_1$ and $F_2$ be fields. It is well-known that the direct product $F_1\times F_2$ is never a field (the element $(1,0)$ is not invertible). I'm curious, however, if there is a natural way to construct a field out of this direct product.
My first thought was to somehow quotient out by all elements for which one component is zero - however, this set does not form an ideal in $F_1\times F_2$, since it isn't a subgroup. If you instead quotient out one by one the collection of elements for which a particular component is zero, you just get $0$ (I think). So maybe there is a more subtle way to construct a field out of $F_1\times F_2$ (if there is indeed such a construction).
 A: Whilst the product of two fields is never a field, an ultraproduct always is.  If you try taking an ultraproduct of two fields, or any finite number of fields, then you will just get one of the fields back.
However, things get exciting when you take an ultraproduct of infinitely many fields.  For example if you take an ultraproduct of all fields of prime order: $\mathbb{F}_2, \mathbb{F}_3,\mathbb{F}_5,\cdots$, you will get a field of characteristic $0$.  This goes against the intuition mentioned in comments that fields of different characteristics do not talk to each other.
In simple terms, if we have an infinite sequence of fields $k_1,k_2,k_3,\cdots$, we can first take their product, and then say that two elements of this product are the same if and only if they agree on a large set of components.  The result is a field.
What do we mean by large though?  Given a subset of $\mathbb{N}$ we need to define what it means to be large.  Then given two elements in the product, we can look at the set of indices where they agree, and we will know if they are the same element in our field based on whether or not this set is large.
Unfortunately, we cannot actually specify what we mean by large as we need the axiom of choice to prove that such a concept exists.  What we can do is specify what properties we want largeness to satisfy, and then prove using the axiom of choice that such a notion exists (in fact many such notions).
Properties:


*Being large cannot be equivalent to containing some number $i$.  If it was then we would just get $k_i$ as the result of our ultraproduct.


*If $A\subseteq B$ and $A$ is large, then $B$ is large.


*If $A,B$ both large, then $A\cap B$ large.


*If $A$ not large, then its complement $A^c$ is large.


*The empty set $\emptyset$ is not large.
Note that if we partition a large set $A$ into two pieces $B,C$, one of them must be large, as otherwise by (2) and (3) we would have $A\cap B^c\cap C^c=\emptyset$ large, contradicting (4).
If a single element set was large, then any set containing that element would be large by (1), and any set not containing the element would not be large (by (2) and (4)), so we would contradict (0).
If a finite set were large, we could keep removing one element at a time, till we reached the empty set.  At each stage we have a large set, as either we have a large set, or the singleton set we removed is large, which cannot happen. This contradicts (4).
Proof that we get a field:
Let $K$ denote the product of the $k_i$ and let $I$ denote the set of elements which are $0$ on a large set of components.  We know $I$ is closed under addition by (2).  We know it is closed under multiplication by elements of $K$ by (1).  Thus $I$ is an ideal, and our ultraproduct is just the ring $K/I$.  Note this ring is non-zero by (4).
To see that this ring is a field, take any element $x\in K$.  Let $y\in K$ be the inverse of $x$ on every component where $x$ is non-zero, and $0$ elsewhere.  Then by (3), either $x$ is $0$ on a large set, or $xy$ agrees with the identity $(1,1,1\cdots)$ on a large set.  That is every non-zero element of $K/I$ has an inverse.
Finally lets return to our example: an ultraproduct of all fields of prime order, $\mathbb{F}_2, \mathbb{F}_3,\mathbb{F}_5,\cdots$.  To see that it has characteristic $0$, just note that $(m,m,m,\cdots)$ agrees with $0$ only on the finite set of components $\mathbb{F}_p$ where $p|m$.  Finite sets cannot be large, so we have $(m,m,m,\cdots)\neq 0$ in $K/I$.
