Nilpotent Elements and the Jordan–Chevalley Decomposition

Introduction

The eigenvalue decomposition of The Eigenvalue Decomposition of an Element writes an element as a sum of scalar multiples of orthogonal idempotents. It exists only when the minimal polynomial is split and separable. When the eigenvalues are repeated, the idempotents that separate the eigenvalues still exist, but a nilpotent part survives inside each block. The Jordan–Chevalley decomposition separates that nilpotent part from a semisimple part, and it does so in an arbitrary finite-dimensional algebra, not only in a matrix algebra.

The article assumes the Chinese remainder decomposition of The Algebra Generated by an Element and the Chinese Remainder Decomposition, the separability and splitting conditions of Separable Elements and Split Minimal Polynomials, and the generalized eigenvalue decomposition of The Eigenvalue Decomposition of an Element. The matrix Jordan form is Modules over $k[x]$ and the Jordan Form; the trace and eigenvalue material is Eigenvalues and Diagonalisation. The perfect-field hypothesis is essential in characteristic $p$, and the failure over an imperfect field is treated in §The General Field. The base is a field $k$ and $A$ is a finite-dimensional unital associative $k$-algebra.

Nilpotent Elements

Definition and index

An element $x \in A$ is nilpotent if $x^m = 0$ for some integer $m \ge 1$. The least such $m$ is the nilpotency index of $x$. The zero element is nilpotent of index one; a nonzero nilpotent has index at least two.

A nilpotent element is a zero divisor: if $x \neq 0$ and $x^m = 0$ with $m$ minimal, then $x^{m-1} \neq 0$ and $x \cdot x^{m-1} = 0$. A nilpotent element is not a unit: if $x$ is a unit and $x^m = 0$, then $1 = (x^{-1})^m x^m = 0$, a contradiction. The converse fails: a zero divisor need not be nilpotent, and the idempotents are the standard counterexamples.

Sums and products

The set of nilpotent elements is not a linear subspace. In $M_2(k)$ the matrix units

$$ E_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad E_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$

are nilpotent of index two, but their sum $E_{12} + E_{21} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ has square $I$ and is not nilpotent. The sum of two nilpotents is therefore not nilpotent in general.

Two closure properties do hold.

If $x$ and $y$ commute and are nilpotent, with $x^m = 0$ and $y^n = 0$, then $x+y$ is nilpotent: in the binomial expansion of $(x+y)^{m+n-1}$, every term has $x$-exponent at least $m$ or $y$-exponent at least $n$, so every term vanishes. The binomial coefficients are irrelevant; the vanishing comes from the exponents.
If $x$ and $y$ commute and are nilpotent, then $xy$ is nilpotent: $(xy)^m = x^m y^m = 0$ if $x^m = 0$.

Without commutativity, products of nilpotents need not be nilpotent: $E_{12}E_{21} = E_{11}$ is an idempotent, not a nilpotent.

Nilpotents and the radical

In a finite-dimensional algebra, the Jacobson radical is a nilpotent ideal, so every element of the radical is nilpotent. The converse fails: in $M_n(k)$ the radical is zero, but $E_{12}$ is a nonzero nilpotent. The nilpotent elements are therefore not the same as the radical in a noncommutative algebra.

In a commutative finite-dimensional algebra the distinction disappears: the set of nilpotent elements is an ideal, the nilradical, and it is the intersection of the prime ideals. The quotient by the nilradical is reduced, that is, it has no nonzero nilpotents. This is the commutative case of Reduced Rings and the Nilradical. For the subalgebra $k[a]$, which is commutative, the nilradical is exactly the set of nilpotent elements, and it is the radical that appears in the Jordan–Chevalley decomposition.

The nilpotent cone

In $A = M_n(k)$ the nilpotent matrices are the matrices all of whose eigenvalues are zero. They are the common zero set of the coefficients of the characteristic polynomial other than the leading one, so they form an algebraic variety, the nilpotent cone. The cone is closed under scalar multiplication but not under addition, as the example $E_{12}+E_{21}$ shows. The nilpotent cone is the fibre over zero of the characteristic polynomial map; its geometry is the subject of Modules over $k[x]$ and the Jordan Form and of the invariant theory of matrices.

The nilpotent part of an element

Let $a \in A$ and let $k[a] \cong k[x]/(m_a)$ be its subalgebra. The nilradical $\operatorname{rad} k[a]$ is the ideal of nilpotent elements, and the quotient

$$ k[a]_{\mathrm{red}} = k[a]/\operatorname{rad} k[a] $$

is reduced, hence a product of fields. If the quotient map $k[a] \to k[a]_{\mathrm{red}}$ has a section that is a $k$-algebra homomorphism, the image $s$ of $a$ under the section is a semisimple element and $n = a - s$ is nilpotent. The Jordan–Chevalley decomposition is exactly this splitting of the quotient map. Over a perfect field the splitting always exists; over an imperfect field it can fail. This is the conceptual content of the theorem of §The Jordan–Chevalley Decomposition.

Semisimple Elements

Definition

An element $s \in A$ is semisimple if its minimal polynomial $m_s$ is squarefree, that is, if no irreducible factor of $m_s$ occurs with exponent greater than one. Equivalently, $k[s]$ is reduced, equivalently $k[s]$ is a product of fields. The element is separable if $m_s$ is separable in the sense of Separable Elements and Split Minimal Polynomials; separability implies semisimplicity, and the two notions agree over a perfect field. In characteristic $p$ they differ: the class of $x$ in $k[x]/(x^p-t)$ over $k = \mathbb{F}_p(t)$ is semisimple (its minimal polynomial is irreducible, hence squarefree) but not separable.

The distinction is important for the statement of the theorem. Over a perfect field the semisimple part is separable; over an imperfect field it may be semisimple without being separable, and the decomposition may still exist in some cases.

Examples

A scalar is semisimple: $m_s = x - \lambda$.
An idempotent is semisimple: $m_e = x(x-1)$ when $e \neq 0,1$.
A diagonalisable matrix over a perfect field is semisimple; over an imperfect field a matrix can be semisimple without being diagonalisable over the algebraic closure, as the purely inseparable example shows. This is the module-theoretic sense of semisimple; the stronger property of diagonalisability over the algebraic closure is separability, and over a perfect field the two coincide.
A nilpotent element is semisimple only when it is zero: $m_x = x^d$ is squarefree only for $d=1$.
A Jordan block of size at least two is not semisimple: its minimal polynomial is $(x-\lambda)^d$ with $d \ge 2$.

The Jordan–Chevalley Decomposition

Definition

A Jordan–Chevalley decomposition of $a \in A$ is a writing

$$ a = s + n $$

with $s, n \in A$ satisfying

$$ sn = ns, \qquad s \text{ semisimple}, \qquad n \text{ nilpotent}. $$

The element $s$ is the semisimple part and $n$ is the nilpotent part. The decomposition is additive; the multiplicative version for units is treated in §The Multiplicative Version.

Uniqueness

Theorem. If a Jordan–Chevalley decomposition exists, it is unique.

Proof over a perfect field. Suppose $a = s+n = s'+n'$ are two decompositions. Extend scalars to an algebraic closure. The eigenvalues of $a$ are the eigenvalues of $s$, because $n$ is nilpotent and commutes with $s$; the same holds for $s'$. For an eigenvalue $\lambda$ of $a$, let $V_\lambda$ be the generalized eigenspace. Since $s$ and $s'$ commute with $a$, they preserve $V_\lambda$. On $V_\lambda$ the only eigenvalue of $a$ is $\lambda$, so the only eigenvalue of $s$ on $V_\lambda$ is $\lambda$; since $s$ is semisimple, hence diagonalisable over the algebraic closure, its restriction to $V_\lambda$ is diagonalisable with the single eigenvalue $\lambda$, so it is $\lambda$ times the identity. The same holds for $s'$. Hence $s = s'$ on every $V_\lambda$, and therefore $s = s'$. Then $n = a - s = n'$.

The argument uses the perfect hypothesis to make semisimple elements diagonalisable over the algebraic closure, so that a semisimple element with a single eigenvalue is scalar. Over an imperfect field the uniqueness statement is standard but requires a little more care; it is not used in the construction below except in the perfect case.

The theorem over a perfect field

Theorem. Let $k$ be perfect and let $A$ be a finite-dimensional unital associative $k$-algebra. Every $a \in A$ has a unique Jordan–Chevalley decomposition $a = s+n$, and both $s$ and $n$ are polynomials in $a$, so $s, n \in k[a]$. The semisimple part $s$ is separable, and the minimal polynomial of $s$ is the squarefree part of $m_a$.

Proof. Write the primary decomposition

$$ m_a = f_1^{e_1} \cdots f_r^{e_r}, $$

with the $f_i$ distinct monic irreducibles and $e_i \ge 1$. Since $k$ is perfect, every $f_i$ is separable. The Chinese remainder theorem of The Algebra Generated by an Element and the Chinese Remainder Decomposition gives

$$ k[a] \;\cong\; \prod_{i=1}^{r} k[x]/(f_i^{e_i}), $$

with canonical idempotents $\varepsilon_i$. Work in the $i$-th factor and let $\bar a_i$ be the image of $a$. The element $f_i(\bar a_i)$ lies in the maximal ideal $(f_i)$, so it is nilpotent. Since $f_i$ is separable, $f_i'$ is coprime to $f_i$, hence $f_i'(\bar a_i)$ is a unit modulo $f_i^{e_i}$. We look for a root of $f_i$ of the form

$$ s_i = \bar a_i + f_i(\bar a_i)q_i $$

with $q_i$ a polynomial in $\bar a_i$. Taylor's formula gives

$$ f_i(\bar a_i + f_i(\bar a_i)q_i) = f_i(\bar a_i)\left(1 + f_i'(\bar a_i)q_i + f_i(\bar a_i)(\cdots)\right). $$

The bracket can be made zero modulo $f_i^{e_i}$ by choosing $q_i \equiv -f_i'(\bar a_i)^{-1} \pmod{f_i^{e_i-1}}$, because $f_i(\bar a_i)$ is nilpotent of index $e_i$. Hence $f_i(s_i) = 0$ in the factor. Since $f_i$ is irreducible, $k[s_i] \cong k[x]/(f_i)$ is a field, so $s_i$ is semisimple; and

$$ n_i = \bar a_i - s_i = -f_i(\bar a_i)q_i $$

lies in the maximal ideal $(f_i)$, so it is nilpotent. Assembling the components,

$$ \boxed{\;s = \sum_{i=1}^{r} s_i \varepsilon_i, \qquad n = \sum_{i=1}^{r} n_i \varepsilon_i\;}, $$

gives $s, n \in k[a]$, $s$ semisimple (a product of fields), $n$ nilpotent, $sn = ns$ (both lie in the commutative algebra $k[a]$), and $a = s + n$. The minimal polynomial of $s$ is $\prod_i f_i$, the squarefree part of $m_a$, and the nilpotency index of $n$ is $\max_i e_i$.

The uniqueness is the theorem of §Uniqueness.

The construction is the splitting of the quotient map $k[a] \to k[a]_{\mathrm{red}}$: the element $s$ is a lift of the image of $a$ in the reduced quotient, chosen to be a root of each $f_i$ in the local factor. The perfect hypothesis is used exactly to make each $f_i$ separable, which makes the derivative invertible modulo $f_i^{e_i}$ and the lift possible.

The general field

The proof uses the separability of each $f_i$ only when $e_i \ge 2$; when $e_i = 1$, the factor $k[x]/(f_i)$ is already a field, and one can take $s_i = \bar a_i$ without lifting. The same construction therefore proves the following.

Theorem. Let $k$ be any field and let $a \in A$. If every irreducible factor of $m_a$ that occurs with exponent at least two is separable, then $a$ has a unique Jordan–Chevalley decomposition with $s, n \in k[a]$. In particular, the decomposition exists when $m_a$ is squarefree (with $s=a$, $n=0$) and when all the repeated factors are separable, as for a nilpotent element or a power of a linear polynomial.

The condition is sufficient, not necessary for every element: an element can have a decomposition even when some simple factor of its minimal polynomial is inseparable, because the simple factor contributes a semisimple (though not separable) component. What the condition excludes is a repeated inseparable factor, and that is exactly the case in which the lift fails.

The failure over an imperfect field

Let $k = \mathbb{F}_p(t)$, let $f(x) = x^p - t$, and let

$$ A = k[x]/(f^2), \qquad a = x . $$

The polynomial $f$ is irreducible and inseparable, and $f^2$ has the repeated inseparable factor $f$. We show that $a$ has no Jordan–Chevalley decomposition in $A$.

Suppose $a = s + n$ with $s$ semisimple, $n$ nilpotent and $sn = ns$. The algebra $A$ is local with maximal ideal $(f)$, and its residue field is $k[x]/(f) = k(\alpha)$ with $\alpha^p = t$. Every element of $A$ is either a unit or nilpotent, because the non-units form the maximal ideal, which is nilpotent. If $s$ were nilpotent, then $a = s+n$ would be nilpotent, being a sum of commuting nilpotents. But $a = x$ is a unit in $A$, because $f(0) = -t \neq 0$, so $x \notin (f)$. Hence $s$ is a unit.

A semisimple unit $s$ in a local algebra has $k[s]$ a product of fields; since $k[s]$ is a subalgebra of the local algebra $A$, it has no nontrivial idempotents, so $k[s]$ is a field. The residue of $s$ in $A/(f) = k(\alpha)$ is $\alpha$, because $s = a - n$ and $n$ is nilpotent, hence lies in $(f)$. The minimal polynomial of $s$ is therefore an irreducible polynomial vanishing at $\alpha$, so it is $f$; in particular $s$ is a root of $f$ in $A$.

But $f$ has no root in $A$. A root $r$ of $f$ in $A$ must be congruent to $x$ modulo $(f)$: reducing modulo $(f)$ gives $f(r) = 0$ in $k(\alpha)$, and the only root of $f$ in $k(\alpha)$ is $\alpha$, because $f = (x-\alpha)^p$ there. Hence $r = x + fq$ for some polynomial $q$, and in characteristic $p$,

$$ f(x + fq) = (x+fq)^p - t = x^p + f^p q^p - t = f + f^p q^p = f\left(1 + f^{p-1}q^p\right). $$

The factor $1 + f^{p-1}q^p$ is congruent to $1$ modulo $f$, hence is a unit, so the product is a nonzero multiple of $f$ and is not zero modulo $f^2$. Hence no root exists, a contradiction. Therefore $a$ has no Jordan–Chevalley decomposition.

The same failure occurs for the companion matrix of $f^2$ in $M_{2p}(k)$. Over the algebraic closure, $f^2 = (x-\alpha)^{2p}$ and the companion matrix is a single Jordan block of size $2p$ at $\alpha$; its semisimple part would be $\alpha I$, which is not a matrix over $k$ because $\alpha \notin k$. So the matrix has no Jordan–Chevalley decomposition over $k$ either. The example shows that the perfect hypothesis, or the separability hypothesis of §The general field, cannot be dropped.

The Multiplicative Version

Units and unipotents

Let $a \in A$ be a unit, with Jordan–Chevalley decomposition $a = s+n$. Then $s$ is a unit: $s = a - n = a(1 - a^{-1}n)$, and $1 - a^{-1}n$ is a unit because $a^{-1}n$ is nilpotent, with inverse the finite geometric series

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

Conversely, if $s$ is a unit, then $a = s(1 + s^{-1}n)$ is a unit. Hence $a$ is a unit if and only if $s$ is a unit.

When $a$ is a unit, put

$$ u = 1 + s^{-1}n . $$

Then $u - 1 = s^{-1}n$ is nilpotent, so $u$ is unipotent, and

$$ \boxed{\;a = s u = u s\;} . $$

This is the multiplicative Jordan–Chevalley decomposition: every unit is the product of a commuting semisimple unit and a unipotent element. It is unique, because the additive decomposition is unique and $u = s^{-1}a$. Conversely, a multiplicative decomposition $a = su$ with $s$ semisimple, $u$ unipotent and $su = us$ gives the additive decomposition $a = s + s(u-1)$, with $s(u-1)$ nilpotent. The two versions carry the same information for units.

Eigenvalues and the unipotent part

Over an algebraic closure, the semisimple part $s$ has the same eigenvalues as $a$, and the unipotent part $u$ has all eigenvalues equal to $1$: $u = 1 + (\text{nilpotent})$, so its only eigenvalue is $1$. The multiplicative decomposition therefore separates the eigenvalues (in $s$) from the unipotent thickening (in $u$). For a single Jordan block $J_n(\lambda)$ with $\lambda \neq 0$,

$$ J_n(\lambda) = \lambda I \cdot \left(I + \lambda^{-1}N\right), $$

with semisimple part $\lambda I$ and unipotent part $I + \lambda^{-1}N$.

The Generalized Eigenvalue Decomposition

Splitting field

Suppose $k$ is a splitting field for $m_a$, so that

$$ m_a = \prod_{i=1}^{r}(x-\lambda_i)^{e_i} $$

with the $\lambda_i$ distinct. The generalized eigenvalue decomposition of The Eigenvalue Decomposition of an Element gives idempotents $\varepsilon_i$ and nilpotent elements $n_i = (a-\lambda_i)\varepsilon_i$ with

$$ \boxed{\;a = \sum_{i=1}^{r}\left(\lambda_i \varepsilon_i + n_i\right), \qquad n_i^{e_i} = 0\;} . $$

The semisimple part is

$$ s = \sum_{i=1}^{r} \lambda_i \varepsilon_i, $$

and the nilpotent part is $n = \sum_i n_i$. This is the Jordan–Chevalley decomposition in the split case. The $\varepsilon_i$ are the projections onto the generalized eigenspaces, the $n_i$ are the nilpotent parts inside the blocks, and the decomposition is the algebraic form of the Jordan form: over a splitting field, the semisimple part is diagonal and the nilpotent part carries the off-diagonal Jordan structure.

Relation to the eigenvalue decomposition

The eigenvalue decomposition is the special case $n = 0$, equivalently $e_i = 1$ for every $i$, equivalently $m_a$ squarefree. The generalized decomposition allows $n \neq 0$ and absorbs the repeated roots into the nilpotent parts. The two decompositions agree exactly when $a$ is semisimple. This is the same dichotomy as in The Eigenvalue Decomposition of an Element: distinct roots give idempotents without nilpotents, repeated roots give idempotents with nilpotent parts.

Functoriality

An algebra homomorphism $\varphi : A \to B$ sends the Jordan–Chevalley decomposition of $a$ to that of $\varphi(a)$: $\varphi(s)$ is semisimple because its minimal polynomial divides $m_s$, $\varphi(n)$ is nilpotent because $\varphi(n)^m = \varphi(n^m) = 0$, and they commute. When the decomposition is unique, it is the decomposition of $\varphi(a)$. In particular, a representation of $A$ carries the decomposition of $a$ to the matrix decomposition of the image.

The Matrix Case

The semisimple and nilpotent parts

Let $A = M_n(k)$ with $k$ perfect, and let $a \in A$. The Jordan–Chevalley decomposition $a = S + N$ has the following properties.

$S$ is diagonalisable over the algebraic closure $\bar k$; $N$ is nilpotent.
$SN = NS$, and both $S$ and $N$ are polynomials in $a$, hence commute with every matrix commuting with $a$.
The eigenvalues of $S$ are the eigenvalues of $a$, with the same algebraic multiplicities; the only eigenvalue of $N$ is $0$.
The minimal polynomial of $S$ is the squarefree part of $m_a$; the nilpotency index of $N$ is the largest exponent in $m_a$.
Over $\bar k$, in a basis in which $S$ is diagonal, $N$ is block diagonal, with a nilpotent block on each eigenspace; the Jordan blocks of $a$ are the Jordan blocks of these nilpotent blocks, shifted by the eigenvalues. The decomposition is the invariant form of the Jordan form: it does not depend on the choice of basis, and it exists over $k$ even when the Jordan form itself requires an extension.

The matrix $S$ is the semisimple part and $N$ the nilpotent part of the matrix. The decomposition is the matrix case of the theorem of §The theorem over a perfect field; the explicit construction via the Chinese remainder theorem is the one given there.

A Jordan block

For $a = J_n(\lambda) = \lambda I + N$ with $N$ the nilpotent Jordan block,

$$ s = \lambda I, \qquad n = N . $$

The semisimple part is scalar, the nilpotent part is the off-diagonal part, and the decomposition is the familiar separation of the diagonal from the superdiagonal. The minimal polynomial is $(x-\lambda)^n$, the squarefree part is $x-\lambda$, and the nilpotency index of $n$ is $n$.

A block with two eigenvalues

For

$$ a = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}, $$

the minimal polynomial is $(x-1)^2(x-2)$. The semisimple part is

$$ s = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}, $$

and the nilpotent part is $n = E_{12}$. Then $s+n = a$, $sn = ns$, $s$ is semisimple with minimal polynomial $(x-1)(x-2)$, and $n^2 = 0$. Both $s$ and $n$ are polynomials in $a$: the Lagrange idempotents of the eigenvalues $1$ and $2$ are $\varepsilon_1 = (a-2I)/(1-2) = 2I - a$ and $\varepsilon_2 = a - I$; the generalized eigenprojector of the repeated eigenvalue $1$ is $\varepsilon_1$, and

$$ n = (a-I)\varepsilon_1 = (a-I)(2I-a) = E_{12}. $$

Then $s = a - n$. Both are polynomials in $a$, as the construction guarantees.

The companion matrix over an imperfect field

The companion matrix of $f^2$ with $f = x^p - t$ over $k = \mathbb{F}_p(t)$ has no Jordan–Chevalley decomposition over $k$, as shown in §The failure over an imperfect field. Over the algebraic closure it has the decomposition $\alpha I + J_{2p}(0)$, but $\alpha \notin k$ and the semisimple part is not defined over $k$. The example is the standard warning that the Jordan–Chevalley decomposition is not a formal consequence of the existence of a Jordan form over the algebraic closure.

The biquaternion case

Under the matrix model $\mathbb{B} \cong M_2(\mathbb{C})$ of Introduction to the 2×2 Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions, the element

$$ a = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = I + E_{12} $$

is unipotent. Its Jordan–Chevalley decomposition is $s = I$ and $n = E_{12}$, and its multiplicative form is $a = I \cdot a$ with unipotent part $a$. The example is the smallest non-semisimple biquaternion: the semisimple part is scalar, the nilpotent part is the off-diagonal matrix unit, and the two commute. A biquaternion whose matrix is diagonalisable has $n = 0$; a biquaternion whose matrix has a repeated eigenvalue and a nontrivial Jordan block has $n \neq 0$. The biquaternion spectral theory of The Biquaternion Eigenvalue Decomposition reads the same decomposition through the conjugations.

Summary

A nilpotent element is an element some power of which is zero; its least exponent is the nilpotency index. Nilpotents are zero divisors and non-units. The set of nilpotents is not a subspace, but sums and products of commuting nilpotents are nilpotent. In a finite-dimensional algebra the radical is a nilpotent ideal contained in the set of nilpotents; in a commutative algebra the nilpotents form the nilradical, and the quotient by it is reduced.

A Jordan–Chevalley decomposition of $a$ is a writing $a = s+n$ with $s$ semisimple, $n$ nilpotent and $sn=ns$. It is unique when it exists. Over a perfect field it always exists, with $s, n \in k[a]$; the semisimple part is the lift of $a$ to the reduced quotient $k[a]_{\mathrm{red}}$, its minimal polynomial is the squarefree part of $m_a$, and the nilpotency index of $n$ is the largest exponent in $m_a$. Over a general field the same construction works whenever every repeated irreducible factor of $m_a$ is separable; it fails for the class of $x$ in $k[x]/((x^p-t)^2)$ over $\mathbb{F}_p(t)$, and for the companion matrix of the same polynomial.

For a unit, the multiplicative form is $a = su$ with $s$ semisimple, $u$ unipotent and $su=us$; it is equivalent to the additive form via $u = 1 + s^{-1}n$. Over a splitting field, the semisimple part is $\sum_i \lambda_i \varepsilon_i$ and the nilpotent part is $\sum_i n_i$, with $\varepsilon_i$ the generalized eigenprojectors and $n_i = (a-\lambda_i)\varepsilon_i$. In a matrix algebra over a perfect field, $S$ is diagonalisable over the algebraic closure, $N$ is nilpotent, both are polynomials in $a$, and the decomposition is the invariant form of the Jordan form.

Summary of Notation

Symbol	Meaning
$x$, $n$	Nilpotent elements; $x^m=0$
nilpotency index	The least $m \ge 1$ with $x^m=0$
$\operatorname{rad} k[a]$	The nilradical of $k[a]$, the ideal of nilpotent elements
$k[a]_{\mathrm{red}}$	The reduced quotient $k[a]/\operatorname{rad} k[a]$
$s$	The semisimple part of $a$; $m_s$ squarefree
$n$	The nilpotent part of $a$
$a = s+n$	The additive Jordan–Chevalley decomposition
$a = su$	The multiplicative Jordan–Chevalley decomposition for a unit
$u = 1 + s^{-1}n$	The unipotent part of a unit
$m_a = \prod_i f_i^{e_i}$	The primary decomposition of the minimal polynomial
$\varepsilon_i$	The generalized eigenprojector of the $i$-th primary factor
$n_i = (a-\lambda_i)\varepsilon_i$	The nilpotent part of the $i$-th block, in the split case
$S$, $N$	The semisimple and nilpotent parts of a matrix
$J_n(\lambda)$	The Jordan block of size $n$ at $\lambda$

Nilpotent Elements and the Jordan–Chevalley Decomposition

Introduction