# Suppose R is a commutative ring, M is the maximum ideal of R, and R/M is not a field. Prove: (R/M)² = 0.

It is a well known theorem that if R is a commutative ring with identity not equal to zero then R/M is a field. So R in this question doesn't have a non-zero identity. For example, if R = 2Z, which is a ring with no identity, and whose maximal ideal is 4Z, then 2Z/4Z = {4Z, 2 + 4Z}. I'm not sure what exactly does the notation (R/M)² mean or how it is defined. If (R/M)² means the set {ab∈R/M | a,b∈R/M}, then (2Z/4Z)² = {4Z} (= 0?) Can anyone help to prove?

• That's a product of ideals. See en.wikipedia.org/wiki/Ideal_(ring_theory)#Ideal_operations for a clear description of the sum and product of ideals. Nov 25 at 9:09
• In your example, viewing $\{4\mathbb{Z},2+4\mathbb{Z}\}$ as a ring without identity, then $4\mathbb{Z}$ is the $0$ element. Thus, $\{4\mathbb{Z}\}=\{0\}$, as required. Nov 25 at 9:37
• Eric and Sam, thank you very much for your clarification. May I know how to prove the question? Can you suggest any reading or relevant part of any textbook? Nov 25 at 10:20
• math.meta.stackexchange.com/questions/5020/… Nov 25 at 22:21

The question can be rephrased as:

Suppose $$R$$ is a nonzero rng with no nontrivial ideals, and $$R$$ has no multiplicative identity. Then $$R^2=\{0\}$$.

The $$R^2$$ should be interpreted as the ideal product, that is, the set $$\{\sum r_is_i\mid r_i, s_i\in R\}$$ with finite sums only, of course.

Since this is always an ideal, we know either $$R^2=\{0\}$$ or $$R^2=R$$. If $$R^2=R$$, we can show $$R$$ has an identity, and then $$R$$ must be a field.

Suppose $$R^2=R$$. Then there must exist an $$x$$ such that $$xR\neq \{0\}$$, and in fact, $$xR=R$$ again since there are no nontrivial ideals. In particular $$xr=x$$ for some $$r\in R$$.

Here's an important thing to notice: the annihilator $$ann(z):=\{r\in R\mid zr=0\}$$ is also an ideal of $$R$$. It can again be only one of two things, $$R$$ or $$\{0\}$$.

At this point, we have above that $$x\neq 0$$, and $$xr\neq 0$$, so $$ann(x)=\{0\}$$. Also, $$ann(r)=\{0\}$$ since $$xr\neq 0$$.

Because $$xr^2-xr=xr-x=0$$, we can see $$x(r^2-r)=0$$. But as $$ann(x)=\{0\}$$, it must be that $$r^2=r$$.

Then for any other $$y\in R$$, $$yr-y\neq 0$$ would imply also that $$yrr-yr\neq 0$$, but of course that isn't the case because $$y(r^2-r)=y0=0$$. This means then that $$yr-y=0$$ for all $$y$$, and that just says that $$r$$ is the identity of $$R$$.

So going back to your original statement: if $$R$$ is a nonzero ring with no nontrivial ideals, and $$R$$ is not a field, then it must be that $$R^2=\{0\}$$.