Intuitively I think that since $R/M$ will be a field and can't have zero divisors, the set of zero-divisors must lie inside $M$ that they vanish in $R/M$.

I tried to prove this, but I got stuck, so I'm afraid that my intuition is wrong.

Is this a correct statement? If not, does it hold when $R$ is a finite unital commutative ring?

  • $\begingroup$ Is the implication $"M\subseteq R$ maximal $\Rightarrow$ $R/M$ field" even true if $R$ has zero divisors? $\endgroup$ – mathmax May 14 '14 at 16:12
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    $\begingroup$ @mathmax: Yes. Ideal correspondence theorem still holds and the proof of the implication remains just the same I think. $\endgroup$ – math.n00b May 14 '14 at 16:17
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    $\begingroup$ @mathmax All we need for that is $R$ commutative with $1$. $\endgroup$ – mez May 14 '14 at 18:31
  • $\begingroup$ I thought we needed to have an integral domain somewhere in the proof, but now I've checked that we don't. $\endgroup$ – mathmax May 14 '14 at 18:44

For the field of two elements $F_2$, the ring $F_2\times F_2$ contains the maximal ideal $F_2\times \{0\}$. Does this ideal contain all zero divisors?

  • $\begingroup$ I sure do get a lot of mileage out of this example... $\endgroup$ – rschwieb May 14 '14 at 18:33
  • $\begingroup$ Can we say that in a finite ring any maximal ideal is a subset of zero divisors? Because any element is either a unit or a zero divisor. I'm talking about a unital commutative ring. $\endgroup$ – math.n00b May 15 '14 at 18:22
  • $\begingroup$ @math.n00b You can do better than that: in any commutative Artinian ring, the elements of maximal ideals are zero divisors. $\endgroup$ – rschwieb May 15 '14 at 20:05
  • $\begingroup$ Interesting. Why is it so? Is it true that in any commutative Artinian ring all elements are either units or zero-diviors? If yes, why? $\endgroup$ – math.n00b May 16 '14 at 16:03
  • $\begingroup$ @math.n00b Let $x$ be any element of a commutative Artinian ring $R$. Then the chain $xR\supseteq x^2R\supseteq\ldots$ stabilizes. This means there is an $n$ and an $r$ such that $x^n=x^{n+1}r$. Rearranging you get $x^n(1-rx)=0$. Suppose $x$ isn't a zero divisor. Then you can cancel the $x^n$ from the left to conclude $1-rx=0$, whence $x$ is a unit. Thus the elements that aren't zero divisors are units. $\endgroup$ – rschwieb May 16 '14 at 17:27

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