Idempotents and Nilpotents

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Section Idempotents and Nilpotents

Worksheet Part 1: Idempotents & Nilpotents

The goal of these problems is to practice more with integral domains, zero divisors, and fields and learn about nilpotent and idempotent elements.

1.

Let \(a\) belong to a ring \(R\) with \(1\) and suppose that \(a^n=0\) for some positive integer \(n\) (such an \(a\) is called nilpotent). Show that \(1-a\) is a unit.

Hint.

Consider \((1-a)(1+a+a^2+\ldots+a^{n-1})\text{.}\)

Solution.

Using the hint, we have

\begin{equation*} (1-a)(1+a+a^2+\ldots+a^{n-1})=1-a^n=1\text{.} \end{equation*}

Thus \((1-a)\) is a unit with inverse \(1+a+a^2+\ldots+a^{n-1}\text{.}\)

2.

Argue that the set of nilpotent elements in a commutative ring form a subring.

Hint.

In a commutative ring, the Binomial Theorem holds, with the binomial coefficient denoting repeated addition of the ring elements.

Solution.

Let \(a\) and \(b\) be nilpotent elements in a commutative ring \(R\text{.}\) Then there exist positive integers \(n\) and \(m\) such that \(a^n=0\) and \(b^m=0\text{.}\) Then we have

\begin{equation*} (b-a)^{m+n}=\sum_{k=0}^{m+n} \binom{m+n}{k}(-1)^k a^k b^{m+n-k}. \end{equation*}

For each term in the sum, either \(k \geq n\) or \(k\lt n\) and so \(m+n-k \gt m\text{.}\) In either case one of \(a^k\) or \(b^{m+n-k}\) is zero, so the entire sum is zero. Thus \(b-a\) is nilpotent. Similarly, we have

\begin{equation*} (ab)^{mn}=a^{mn}b^{mn}=(a^n)^m (b^m)^n=0\text{.} \end{equation*}

Thus \(ab\) is nilpotent. Since \(0\) is nilpotent, we have that the set of nilpotent elements is closed under addition and multiplication and is non-empty, so it is a subring by the Subring Test.

3.

Show that \(0\) is the only nilpotent element in an integral domain.

Solution.

We prove the contrapositive: if a ring \(R\) has a non-zero nilpotent element, then it is not an integral domain. Let \(a\) be a non-zero nilpotent element of \(R\text{.}\) Then there exists a least positive integer \(n\geq 2\) such that \(a^n=0\text{.}\) Then we have \(a^{n-1}\neq 0\) and \(a\neq 0\text{,}\) but \(a^{n-1}a=0\text{.}\) Thus \(R\) has zero divisors and is not an integral domain.

4.

A ring element \(a\) is called an idempotent if \(a^2=a\text{.}\) Show that in an integral domain, \(1\) and \(0\) are the only idempotents.

Solution.

Suppose \(R\) is an integral domain. We have that \(0\) and \(1\) are idempotent since \(0^2=0\) and \(1^2=1\text{.}\) Now if \(a^2=a\) and \(a\neq 0\) we have by cancellation that \(a=1\text{.}\) Thus \(0\) and \(1\) are the only idempotents in an integral domain.

5.

For each \(n\) given, determine if \(\Z_n\) has any nilpotent elements: \(n=2,4,6,9,12\text{.}\) Make a conjecture about what should be true about \(n\) for \(\Z_n\) to have a nilpotent element.

Solution.

\(\Z_4\text{,}\) \(\Z_9\text{,}\) and \(\Z_{12}\) have nilpotent elements. In particular, \(2^2=0\) in \(\Z_4\text{,}\) \(3^2=0\) in \(\Z_9\text{,}\) and \(6^2=0\) in \(\Z_{12}\text{.}\) The other rings do not have nilpotent elements. A conjecture is that \(\Z_n\) has a nilpotent element if and only if \(n\) is not square-free.

6.

Find a zero-divisor and an idempotent in \(\Z_5[i]=\{a+bi\mid a,b\in \Z_5\}\text{.}\)

Solution.

To find a zero-divisor, we look for a solution to

\begin{equation*} (a+bi)(c+di)=(ac-bd)+(ad+bc)i=0\text{.} \end{equation*}

So we need \(a,b,c,d\in \Z_5\) such that not both \(a,b\) or both \(c,d\) are zero with \(ac=bd \bmod{5}\) and \(ad=-bc \bmod{5}\text{.}\) One solution is \(a=1,b=2,c=2,d=1\text{:}\)

\begin{equation*} (1+2i)(2+i)=(2+2)+(1+4)i\equiv 0\text{.} \end{equation*}

We could find all idempotents in a similar way as the above, but we can also note that \(0,1\in \Z_5[i]\) are idempotents.

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Section Idempotents and Nilpotents

Worksheet Part 1: Idempotents & Nilpotents

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Worksheet Part 1: Idempotents & Nilpotents