nLab nilradical

Not to be confused with nilpotent ideal, which is a nilpotent element in the lattice of ideals of a ring.

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Definition

For $R$ a commutative ring, its nilradical $I \subset R$ is the ideal of nilpotent elements: the collection of those elements $a \in R$ such that there is $n \in \mathbb{N}$ with $a^n = 0$.

The quotient $R/I$ is also called the reduced part of $R$.

(If $R$ is not commutative there are different generalization of the notion of nilradical. See Wikipedia, for the moment.)

Examples

Every local Artinian ring is a commutative ring whose set of non-invertible elements is a nilradical.

With rings regarded as formal duals of affine schemes, the canonical inclusion

$$Spec R/I \to Spec R$$

is to be thought of as exhibiting the inclusion of $Spec R/I$ into an infinitesimal thickening of itself.

For $X : CRing \to Set$ a presheaf on the category of commutative rings, the presheaf

$$X_{dR} : Spec R \mapsto X(Spec R/I)$$

is called the de Rham space of $X$.

Properties

Relation to prime ideals

Theorem

Assuming the axiom of choice, if $R$ is a commutative ring, then the nilradical equals the Jacobson radical, the intersection of all prime ideals of $R$.

Proof

In one direction, it it elementary that if $x$ is nilpotent, then $x$ belongs to any prime $\mathfrak{p}$.

For the other direction, we assume the axiom of choice, or slightly more sharply, the ultrafilter principle, which implies that any non-trivial ring $R$ has a proper prime ideal $\mathfrak{p}$: see prime ideal theorem. We must show that if $x$ is not nilpotent, then there is some prime ideal $\mathfrak{p}$ that does not contain $x$. But if $x$ is not nilpotent, then the set $1, x, x^2, \ldots$ is a multiplicative set that does not contain $0$. It follows that the localization $R[x, \frac1{x}]$ with respect to this multiplicative set is a non-trivial ring, and hence has a prime ideal $\mathfrak{q}$. The image of $x$ under the canonical map $\phi: R \to R[x, \frac1{x}]$ is invertible, hence is not contained in $\mathfrak{q}$. Then the inverse image $\mathfrak{p} = \phi^{-1}(\mathfrak{q})$ is a prime ideal that does not contain $x$.

Related concepts

• infinitesimal extension

• reduced ring

• Artinian local ring

References

See also:

• Wikipedia, Nilradical