Let $R$ be a commutative ring with unity such that for some $n\ge 2$, $\prod_{1\le i <j\le n} (x_i-x_j)=0$ for every $x_1,...,x_n \in R$. Then how to show that $|R/M| \le n$ for every maximal ideal $M$ of $R$ and $\mathrm{Spec}(R)$ is totally disconnected in Zariski topology ?
UPDATE: From the answer of Don Antonio and the comments, it follows that $|R/P| \le n$ for every prime ideal $P$ of $R$. In particular, every prime ideal of $R$ is maximal, hence $\mathrm{Spec}(R)$ is totally disconnected.
Suppose you have $\;n\;$ elements in $\;R/M\;$ different from zero (and thus at least $\;n+1\;$ elements, with zero), say $\;a_1+M,..,a_n+M\in R/M\;,\;a_i+M\neq M$ , but then
$$\prod_{1\le i<k\le1}^n\left((a_k+M) -(a_i+M)\right)=\prod_{k=1}^n(a_k-a_i)+M=0+M=M$$
Since $\;R/M\;$ is an integral domain (if $\;M\;$ is prime) the above means that at least there's one pair with $\;a_k-a_i\in M\iff a_k+M=a_i+M\;$ , which contradicts the assumption that the above are different elements in the quotient...