If R and S are rings then $R \times S$ is never a field

Hint: What is the inverse of $(1,0)$?

Since $R$ is nonzero, there is an $r\in R$, $r\neq 0$. Since $S$ is nonzero, there exists $s\in S$, $s\neq 0$.

Since $r\neq 0$, then $(r,0_S)\in R\times S$ is not zero; and since $s\neq 0$, then $(0_R,s)\in R\times S$ is not zero.

What is the product of $(r,0_S)$ and $(0_R,s)$?

Can that happen if nonzero elements are cancellable?

Are nonzero elements cancellable in a division ring?

(This shows a slightly stronger statement: a direct product of nontrivial rings always has zero divisors.)

Alternatively. Prove that $R\times\{0\}$ is a proper ideal of $R\times S$ (here you use that $R\neq 0$ and $S\neq 0$). Then remember that fields are simple.

Let us prove that $R\times S$ is not a field if $R$ and $S$ are any two non-zero rings. The following steps lead to a solution:

(1) Note that $(1_R,0_S)\in R\times S$ where $1_R$ is the multiplicative identity of $R$ and $0_S$ is the additive identity of $S$.

(2) If $T$ is a ring, if $0_T$ is the additive identity of $T$, and if $t\in T$, prove that $t0_T=0_T=0_Tt$. (Hint: use the distributive law and write $0_T=0_T+0_T$.)

(3) Prove that $(1_R,1_S)$ is the multiplicative identity of $R\times S$ where $1_R$ and $1_S$ are the multiplicative identities of $R$ and $S$, respectively.

(4) If $(r,s)\in R\times S$, prove that $(r,s)\cdot (1_R,0_S)=(r,0)$.

(5) Finally, prove that there does not exist $(r,s)\in R\times S$ such that $(r,s)\cdot (1_R,0_S)=(1_R,1_S)$. Deduce that the element $(1_R,0_S)\in R\times S$ has no multiplicative inverse. (Hint: note carefully where you use the fact that $R$ and $S$ are non-zero rings.)

I hope this helps!