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.

  • 2
    $\begingroup$ For the first part, recall that $R/M$ is a field. $\endgroup$
    – Wojowu
    May 21, 2018 at 20:20
  • $\begingroup$ Continuing Wojowu nice hint: the same is true if $\;M\;$ is just prime instead of maximal... $\endgroup$
    – DonAntonio
    May 21, 2018 at 20:21
  • $\begingroup$ @Wojowu: Yes ... so any polynomial of degree $k$ can have atmost $k$ many roots in $R/M$ (does this help ?) ... I don't know what to do ... $\endgroup$
    – user
    May 21, 2018 at 20:25
  • $\begingroup$ The equation holds in any quotient of $R$. If $P$ is a prime ideal, then $R/P$ is an integral domain and the equation says that if you take a list of elements $x_1, \ldots x_n$, then there are $i \neq j$ with $x_i - x_j = 0$. I.e., $R/P$ has less than $n$ elements. $\endgroup$
    – Rob Arthan
    May 21, 2018 at 20:30
  • $\begingroup$ @RobArthan: Ah yes, thanks ... it was very silly of me ... do you have any idea about the total disconnected ness ? $\endgroup$
    – user
    May 21, 2018 at 20:33

1 Answer 1


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...

  • $\begingroup$ thanks ... do you have any ideas about $Spec (R)$ being totally disconnected ? $\endgroup$
    – user
    May 21, 2018 at 20:44
  • $\begingroup$ where do we really need that $a_i+M \ne M$ for every $i$ ? $\endgroup$
    – user
    May 21, 2018 at 20:49
  • $\begingroup$ Why must you present something like this as a proof by contradiction? Please have a look at the constructive way of presenting the same proof in my comment. (And, by the way, $R/M$ must have less than $n$ elements.) $\endgroup$
    – Rob Arthan
    May 21, 2018 at 20:53

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