# Find idempotents in a set of mappings

Find idempotent elements of $$S$$ the set of maps $$f:X \rightarrow R$$ where $$X$$ is a given set and $$R$$ is a given ring with the operations defined by $$(f+g)(r)=f(r)+g(r)$$ and $$(fg)(r)=f(r)g(r)$$ where the sum and the product in these expressions are taken in $$R$$.

$$f\in S$$ is idempotent if $$f^2=f$$. Then for any element $$x\in X$$, we have that $$f^2(x)=(f\cdot f)(x)=f(x)f(x)=[f(x)]^2$$.

Hence the problem is when the equality $$[f(x)]^2=f(x)$$ holds. Since $$R$$ is a ring, we can add the opposite in both sides and we have that $$[f(x)]^2-f(x)=0$$ where $$0$$ is the neutral element in $$R$$.

But, the only nilpotent elements are $$0$$ and $$1$$ that are the neutral elements for the addition and multiplication on $$R$$ respectively? Or there is any more?

Is there a way to proove that if $$R$$ does not have idempotents, the ring $$S$$ does not have idempotents either? And about nilpotents?

• So at least all the $2^{|X|}$ maps $X\to\{0,1\}$ are idempotent ... Apr 13, 2020 at 13:44
• In a Boolean ring every element is an idempotent. For instance, if $R=\Bbb Z/2\Bbb Z$, then $R$ is Boolean, and every element of $S$ is idempotent. Apr 13, 2020 at 14:40
• @BrianM.Scott but if we don't know if the ring is boolean? In this case it is a ring $R$ but nothing is said about it Apr 13, 2020 at 15:01
• @Claudia: You asked whether $R$ could have idempotents besides $0_R$ and $1_R$, and I’m just answering that question: yes, it can though it certainly need not. I really don’t think that one can say much more than quasi said below. Apr 13, 2020 at 15:27

Let $$E$$ be the set of idempotent elements of $$R$$.

Then $$S=\{f:X\to R\mid f(x)\in E,\forall x\in X\}$$.

In other words, $$S$$ is the set of functions from $$X$$ to $$R$$ whose range is a subset of $$E$$.

As noted in the comments, $$R$$ always has $$0$$ as an idempotent, and if $$R$$ has a multiplicative identity $$1$$ with $$1\ne 0$$, then $$1$$ is another idempotent of $$R$$.

You mentioned "nilpotent elements" but this problem is about idempotent elements, not nilpotent elements. Note also that $$1$$ is idempotent but not nilpotent (provided $$1\ne 0$$).

I hope that clears up a few things, but feel free to ask for further clarification.

Let's look at a few examples . . .

Example $$(1)$$:$$\;$$Let $$R=\mathbb{Z}$$ and let $$X=\{a,b,c\}$$.

Then $$E=\{0,1\}$$ and $$S$$ is the set of functions from $$X$$ to $$R$$ such that $$\begin{cases} f(a)\in\{0,1\}\\[4pt] f(b)\in\{0,1\}\\[4pt] f(c)\in\{0,1\}\\ \end{cases}$$ so $$|S|=2^3=8$$.

Example $$(2)$$:$$\;$$Let $$R=\mathbb{Z_6}=\{0,1,2,3,4,5\}$$ and let $$X=\{a,b\}$$.

Then $$E=\{0,1,3,4\}$$ and $$S$$ is the set of functions from $$X$$ to $$R$$ such that $$\begin{cases} f(a)\in \{0,1,3,4\}\\[4pt] f(b)\in \{0,1,3,4\}\\[4pt] \end{cases}$$ so $$|S|=4^2=16$$.

Example $$(3)$$:$$\;$$Let $$R$$ be the ideal $$(2)=\{0,2,4\}$$ of $$\mathbb{Z_6}=\{0,1,2,3,4,5\}$$ and let $$X=\{a,b\}$$.

Then $$E=\{0,4\}$$ and $$S$$ is the set of functions from $$X$$ to $$R$$ such that $$\begin{cases} f(a)\in \{0,4\}\\[4pt] f(b)\in \{0,4\}\\[4pt] \end{cases}$$ so $$|S|=2^2=4$$. Note that for this example, $$R$$ does not have a multiplicative identity.

• Thanks for the answer. But what happens if $R$ has not a multiplicative identity 1 with 1≠0? We have f(1-f)=0. Now if f is not 0 or 1, then f and 1-f are both nonzero and hence zero divisors. But if one of them is 0, is necesarily to the other be 1? Apr 13, 2020 at 15:06
• A ring without a multiplicative identity can can still have an idempotent other than $0$. Apr 13, 2020 at 15:09
• @Claudia If there is no $\color{red}1$, then you do not have $f\cdot(\color{red}1-f)=0$. Apr 13, 2020 at 15:09