De Branges's theorem

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In mathematics, an element x of a ring R is called nilpotent if there exists some positive integer n such that xⁿ = 0.

The term was introduced by Benjamin Peirce in the context of elements of an algebra that vanish when raised to a power.

Examples

• This definition can be applied in particular to square matrices. The matrix

$A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)$

is nilpotent because A³ = 0. See nilpotent matrix for more.

• In the factor ring Z/9Z, the equivalence class of 3 is nilpotent because 3² is congruent to 0 modulo 9.

• Assume that two elements a, b in a (non-commutative) ring R satisfy ab = 0. Then the element c = ba is nilpotent (if non-zero) as c² = (ba)² = b(ab)a = 0. An example with matrices (for a, b):

$A=\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right),B=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right).$

Here AB = 0, BA = B.

• The ring of coquaternions contains a cone of nilpotents.

Properties

No nilpotent element can be a unit (except in the trivial ring {0} which has only a single element 0 = 1). All non-zero nilpotent elements are zero divisors.

An n-by-n matrix A with entries from a field is nilpotent if and only if its characteristic polynomial is tⁿ.

If x is nilpotent, then 1 − x is a unit, because xⁿ = 0 entails

$(1-x)(1+x+x^2+\cdots +x^{n-1})=1-x^n=1.$

More generally, the sum of a unit element and a nilpotent element is a unit when they commute.

Commutative rings

The nilpotent elements from a commutative ring $R$ form an ideal $\mathfrak{N}$; this is a consequence of the binomial theorem. This ideal is the nilradical of the ring. Every nilpotent element $x$ in a commutative ring is contained in every prime ideal $\mathfrak{p}$ of that ring, since $x^n=0\in \mathfrak{p}$. So $\mathfrak{N}$ is contained in the intersection of all prime ideals.

If $x$ is not nilpotent, we are able to localize with respect to the powers of $x$: $S=\{1,x,x^2,...\}$ to get a non-zero ring $S^{-1}R$. The prime ideals of the localized ring correspond exactly to those primes $\mathfrak{p}$ with $\mathfrak{p}\cap S=\mathrm{\varnothing }$. As every non-zero commutative ring has a maximal ideal, which is prime, every non-nilpotent $x$ is not contained in some prime ideal. Thus $\mathfrak{N}$ is exactly the intersection of all prime ideals.

Nilpotent elements in Lie algebra

Let $\mathfrak{g}$ be a Lie algebra. Then an element of $\mathfrak{g}$ is called nilpotent if it is in $[\mathfrak{g},\mathfrak{g}]$ and $\mathrm{ad}x$ is a nilpotent transformation. See also: Jordan decomposition in a Lie algebra.

Nilpotency in physics

An operand Q that satisfies Q² = 0 is nilpotent. Grassmann numbers which allow a path integral representation for Fermionic fields are nilpotents since their squares vanish. The BRST charge is an important example in physics. As linear operators form an associative algebra and thus a ring, this is a special case of the initial definition. More generally, in view of the above definitions, an operator Q is nilpotent if there is n∈N such that Qⁿ = 0 (the zero function). Thus, a linear map is nilpotent iff it has a nilpotent matrix in some basis. Another example for this is the exterior derivative (again with n = 2). Both are linked, also through supersymmetry and Morse theory, as shown by Edward Witten in a celebrated article.

The electromagnetic field of a plane wave without sources is nilpotent when it is expressed in terms of the algebra of physical space.

Algebraic nilpotents

The two-dimensional dual numbers contain a nilpotent space. Other algebras and numbers that contain nilpotent spaces include split-quaternions (coquaternions), split-octonions, biquaternions $\mathbb{ℂ}\otimes \mathbb{ℍ}$, and complex octonions $\mathbb{ℂ}\otimes \mathbb{𝕆}$.

See also

• Idempotent element
• Unipotent
• Reduced ring
• Nil ideal

References

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