Given $F$ a field, $F \times F$ need not be a field I'm trying to prove that the Cartesian product of a field $F$ is not necessarily a field. It's clear that $\mathbb{R}^2$ is not a field, but in order to even talk about $\mathbb{R}^2$ lacking multiplicative inverses for nonzero elements, I need to actually define a multiplication. But that doesn't prove, in general, that $\mathbb{R}^2$ is not a field, but only, in the case where multiplication is defined component-wise, $\mathbb{R}^2$ is not a field.
Is there a way to prove this without having to define an (incorrect) form of multiplication? This is the most natural multiplication, but there isn't a guarantee that there are no other forms.
 A: There seems to be some confusion regarding the multiplication operation for your product. Given rings $A$ and $B$, we have the direct product $A \times B$ which is a ring under coordinate-wise operations. That is to say, $$(a, b) + (a', b') := (a + a', b + b')$$
and similarly for multiplication.
Now, given a field $F$, it is automatically a ring. So when you're asked to prove that $F \times F$ is not a field, you are meant to show that the ring $F \times F$ is not a field. And as already mentioned, the ring structure is already forced to be the coordinate wise one.
In fact, there is something stronger which is true: It's not just that "$F \times F$ need not be a field", we actually have that $F \times F$ is never a field. (I assume that $1 \neq 0$ for fields.)
Indeed, note that $(1, 0) \cdot (0, 1) = (0, 0)$. (Both the elements on the left are nonzero but their product is the zero element of the ring.)
The above actually shows that the product of any two fields is never a field. (Even more generally, the product of integral domains is never an integral domain.)

To emphasise a bit more: To ask about $F \times F$ being a field or not, we should be given some operations to check with. The first paragraph shows how the operations make it a ring and then it makes sense to ask whether a given ring is a field.
If we are not given any operations, then the question would become something like: "Can we define some operations on $F \times F$ which makes it a field?"
This becomes a bit less interesting because then we are just treating $F$ as a set and forgetting any structure that it had to begin with.
