Suppose $R$ is a commutative ring with unity, then which of the following is a subring of $R$:

A. $\{r\in R: r^2 = 0\}$

B. $\{r\in R: r^n = 0, n\geq 1\}$

C. $\{r^2: r\in R\}$

I think it suffices to show that the set is closed under minus and product.

As I continued, I found that none of them are closed under subtraction so none of them are right, which confused me.

Thanks for any suggestions.

  • $\begingroup$ Can you state the definition of "subring" (in this context)? (This is for your benefit in solving the problem, not the benefit of the readers.) $\endgroup$ – Will R Apr 4 '17 at 13:50
  • $\begingroup$ What you say is correct. It suffices to show that the sets are closed under subtraction and multiplication i.e. $a-b$ and $ab$ are in the set whenever $a,b$ are in the same set. $\endgroup$ – Error 404 Apr 4 '17 at 13:50
  • $\begingroup$ @WillR the subset of the ring which forms a ring with respect to the addition and multiplication in the ring $\endgroup$ – Jason Apr 4 '17 at 13:51
  • $\begingroup$ Okay, now the next step is to take each set in turn, go through the conditions for being a subring (which you know) and see which of them satisfies all the conditions. $\endgroup$ – Will R Apr 4 '17 at 13:52
  • 1
    $\begingroup$ @TobiasKildetoft oh I got it, is it by binomial theorem $\endgroup$ – Jason Apr 4 '17 at 14:22

Since you specify that $R$ is a commutative ring with unity, I'm going to assume your definition of "ring" does not require the existence of a multiplicative identity, and therefore that a subring $S\subseteq R$ need not satisfy $1\in S.$

A. This is not a subring. Take $R=\mathbb{R}[X,Y]/(X^{2},Y^{2}).$ Then $X^{2}=0$ and $Y^{2}=0$ but $(X+Y)^{2}=2XY\neq0,$ so $R$ is not closed under addition.

B. This is a subring. The key is that $n$ is not fixed (that is, it is allowed to depend on $r$), and, since $R$ is commutative, we can use the binomial theorem.

Let $S$ be the set in question, and let $a,b\in S.$ Then there exist $m,n\in\mathbb{N}$ such that $a^{m}=b^{n}=0.$ Without loss of generality, assume $n\leq m.$ Then, since $R$ is commutative, $(ab)^{n} = a^{n}b^{n} = 0$ also. Meanwhile, $$(a-b)^{m+n} = \sum_{k=0}^{m+n}{m+n\choose k}a^{k}b^{n+(m-k)} = \sum_{k=0}^{m}a^{k}b^{n+(m-k)} + \sum_{k=m+1}^{m+n}a^{k}b^{n+(m-k)}.$$ On the far right, we have two summations. The first summation is zero because each term contains a $b^{s}$ with $s\geq n.$ The second summation is zero because each term contains an $a^{t}$ with $t\geq m.$ Therefore $(a-b)^{m+n}=0.$

The rest of the ring axioms either can be deduced from this much or are inherited from $R.$

C. This is not a subring in general. Take $R=\mathbb{Z};$ then the set in question contains $1=1^{2}$ and $4=2^{2},$ but does not contain $5 = 1^{2}+2^{2}.$

  • $\begingroup$ Note: If you do require subrings to have multiplicative identities, then A. and B. are obviously "no, unless $R$ is the zero ring" and C. remains the same. $\endgroup$ – Will R Apr 4 '17 at 14:57

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