Nile potent element

Nilpotent and Nilpotent are redirects to this article. Further meanings are listed under Nilpotenz (disambiguation) .

A nilpotent element is a term from ring theory , a branch of mathematics . An element of a ring is called nilpotent if it is multiplied enough often by itself to produce the zero element .

definition

An element of a ring is called nilpotent if there is a positive natural number such that . An ideal is said to be nilpotent if there is a positive natural number such that holds. ${\displaystyle x}$ ${\displaystyle R}$ ${\displaystyle n}$ ${\displaystyle x^{n}=0}$ ${\displaystyle I\subseteq R}$ ${\displaystyle n}$ ${\displaystyle I^{n}=(0)}$

Examples

For example, the matrix is

${\displaystyle A={\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}}$

nilpotent, because it applies

${\displaystyle A^{3}=0}$.

(For special properties of nilpotent matrices see the article nilpotent matrix .)

• In the remainder class ring , the remainder classes of 0, 2, 4 and 6 are nilpotent, since their third power is congruent to 0 modulo 8. In this ring, each element is either nilpotent or a unity . ${\displaystyle \mathbb {Z} /8\mathbb {Z} }$

• In the residual class ring , the nilpotent elements are exactly the residual classes of 0 and 6. ${\displaystyle \mathbb {Z} /12\mathbb {Z} }$

• The zero element of a ring is always nilpotent, there is. ${\displaystyle 0^{1}=0}$

properties

The set of all nilpotent elements of a commutative ring forms an ideal, the so-called Nile radical .

The intersection of all prime ideals in a commutative ring with 1 is exactly the Nile radical.

In the following be a ring, a nilpotent element of and the smallest natural number with . ${\displaystyle R}$ ${\displaystyle a}$ ${\displaystyle R}$ ${\displaystyle n}$ ${\displaystyle a^{n}=0}$

• Is , then is and is zero divisor , because and . ${\displaystyle a\neq 0}$ ${\displaystyle n>1}$ ${\displaystyle a}$ ${\displaystyle aa^{n-1}=0}$ ${\displaystyle a^{n-1}\neq 0}$

If there is also a ring with 1 and not the zero ring , then: ${\displaystyle R}$

• ${\displaystyle a}$is not invertible (with regard to the multiplication), because the contradiction follows for a ring element ( was chosen minimally!). ${\displaystyle ab=1}$ ${\displaystyle b}$ ${\displaystyle 0=a^{n}b=a^{n-1}}$ ${\displaystyle n}$
• ${\displaystyle 1-a}$is invertible because it applies . ${\displaystyle (1-a)\left(1+a+a^{2}+\dotsb +a^{n-1}\right)=1-a^{n}=1=\left(1+a+a^{2}+\dotsb +a^{n-1}\right)(1-a)}$
• If there is a unit of that also commutes, then what one sees as by looking at the representation can also be inverted . ${\displaystyle b}$ ${\displaystyle R}$ ${\displaystyle a}$ ${\displaystyle b+a}$ ${\displaystyle b+a=b\cdot \left(1-\left(-b^{-1}a\right)\right)}$

Let be a residue class ring and the product of all prime divisors of , ie all prime numbers that appear in the prime factorization of . Eg for is . Then the nilpotent elements of are exactly the remainder classes of integers that are multiples of . The idea of ​​the proof is as follows: If the largest exponent that occurs in the prime factorization of , then is a multiple of ; every number for which a power is a multiple of must itself have every prime divisor of . ${\displaystyle R}$ ${\displaystyle \mathbb {Z} /m\mathbb {Z} }$ ${\displaystyle p}$ ${\displaystyle m}$ ${\displaystyle m}$ ${\displaystyle m=12=2^{2}\cdot 3}$ ${\displaystyle p=6=2\cdot 3}$ ${\displaystyle R}$ ${\displaystyle p}$ ${\displaystyle k}$ ${\displaystyle m}$ ${\displaystyle p^{k}}$ ${\displaystyle m}$ ${\displaystyle m}$ ${\displaystyle m}$

A ring that contains no nilpotent elements other than zero is called reduced .

Individual evidence

1. ^ Serge Lang : Algebra , 3rd edition, Graduate Texts in Mathematics, Springer Verlag 2005, ISBN 978-0387953854 , p. 417.