Is $\mathbb{Z}_p \times \mathbb{Z}_p$ a field when $p$ is prime?

Question

Initially I thought that it is but in a solution to a problem, it was said that $\mathbb{Z}_7 \times \mathbb{Z}_7$ is not a field, because it is not an integral domain due to the fact that: $(7,7) \cdot (1,1)=(0,0)$, which seems wrong to me because $(7,7)=(0,0)$.

Answer 1

It is not an integral domain, since $(1,0).(0,1)=(0,0)$.

Answer 2

The direct product of a field $K$, $L=K\times K = \{(a,b)\mid a,b\in K\}$ with component-wise operations (addition and multiplication), is generally not a field, since $(a,0)\cdot (0,b) = (0,0)$ and so the ring $L$ contains zero-divisors.

Answer 3

Whenever you have two integral domains, their product is not an integral domain anymore by the argument of José (For me the zero ring is not an integral domain). That is also why many classes of rings are not closed under products as we often want them to be integral domains.

Answer 4

So, the cross-product of fields isn't a field again, because of the zero divisors $(a,0)$ and $(0,b)$, for $a,b\neq0$.

But the reason you gave was incorrect, as (as you noted) $(7,7)=(0,0)\in\Bbb Z_7×\Bbb Z_7$. You have to take two nonzero elements whose product is zero. $(1,0)$ and $(0,1)$, for instance.

A field is an integral domain, since $ab=0$ implies one of $a$ and $b$ isn't a unit (hence is zero).

In other words, the category of fields, $\bf{Field}$, does not have products.