In a ring that is the direct product of two fields, why isn't it a field? Suppose that we have the direct product of two fields, defined as: $\mathbb{F}^2 = \mathbb{F} \ \times\ \mathbb{F}$. Then, this is obviously not a field because we can take $(1,0)$, and this doesn't have a multiplicative inverse. However, I saw a mention to this today in a book, and the reasoning was:

"$\mathbb{F} \ \times\ \mathbb{F}$ has the zero-divisor $(1,0) \neq
 (0,0)$, hence it is not a field".

(Lectures on Rings and Modules by Joachim Lambek, pg. 17).
I am not sure what they mean by zero-divisor here nor what $(1,0) \neq
 (0,0)$ has anything to do with the direct product of fields not being a field. I always assumed there wasn't a multiplicative inverse, and that was it. Am I missing something deeper?
 A: $(1,0)$ has a zero-divisor because:
$$(1,0) * (0,1)=(0,0)$$
A: A zero divisor is a non-zero element for which another non-zero element can be multiplied and give zero.
That is $a,b\neq 0$ and we have $a\cdot b = 0$. The importans lies in not equal to zero because by the definition of rings and the axioms, $0\cdot a = 0$ for all $a$ and hence it must be excluded as it otherwise doesn't supply any relevant information.
A: A zero divisor is a non-zero element $a$ of the ring for which there is some other non-zero element $b$ with $ab = 0$. For instance, in the ring of integers modulo $6$, the element $2$ is a zero divisor since $2 \cdot 3 \equiv 0 \pmod 6$.
In your example, the element $(1,0)$ is a zero divisor as $(1,0) \cdot (0,1) = (0,0)$, which is the zero element.
Fields have no zero divisors, as zero divisors are not invertible (you can prove this yourself, and you should if you are learning the material).

Am I missing something deeper?

I wouldn't say so --- this is just another requirement, and is more or less the same thing in different words.
