Proof of uniqueness and necessity

That ${\displaystyle x_{s}}$ and ${\displaystyle x_{n}}$ are polynomials in ${\displaystyle x}$ implies in particular that they commute with any operator that commutes with ${\displaystyle x}$. This observation underlies the uniqueness proof.

Let ${\displaystyle x=x_{s}+x_{n}}$ be a Jordan-Chevalley decomposition in which ${\displaystyle x_{s}}$ and (hence also) ${\displaystyle x_{n}}$ are polynomials in ${\displaystyle x}$. Let ${\displaystyle x=x_{s}'+x_{n}'}$ be any Jordan-Chevalley decomposition. Then ${\displaystyle x_{s}-x_{s}'=x_{n}'-x_{n}}$, and ${\displaystyle x_{s}',x_{n}'}$ both commute with ${\displaystyle x}$, hence with ${\displaystyle x_{s},x_{n}}$ since these are polynomials in ${\displaystyle x}$. The sum of commuting nilpotent operators is again nilpotent, and the sum of commuting potentially diagonalisable operators again potentially diagonalisable (because they are simultaneously diagonalisable over the algebraic closure of ${\displaystyle K}$). Since the only operator which is both potentially diagonalisable and nilpotent is the zero operator it follows that ${\displaystyle x_{s}-x_{s}'=0=x_{n}-x_{n}'}$.

To show that the condition that ${\displaystyle x}$ have a minimal polynomial which is a product of separable polynomials is necessary, suppose that ${\displaystyle x=x_{s}+x_{n}}$ is some Jordan-Chevalley decomposition. Letting ${\displaystyle p}$ be the minimal polynomial of ${\displaystyle x_{s}}$, one can check using the binomial theorem that ${\displaystyle p(x_{s}+x_{n})}$ can be written as ${\displaystyle x_{n}y}$ where ${\displaystyle y}$ is some polynomial in ${\displaystyle x_{s},x_{n}}$. Moreover, for some ${\displaystyle \ell \geq 1}$, ${\displaystyle x_{n}^{\ell }=0}$. Thus ${\displaystyle p(x)^{\ell }=x_{n}^{\ell }y^{\ell }=0}$ and so the minimal polynomial of ${\displaystyle x}$ must divide ${\displaystyle p^{\ell }}$. As ${\displaystyle p}$ is a product of separable polynomials (namely of ${\displaystyle p}$), so is the minimal polynomial.

Concrete example for non-existence

If the ground field is not perfect, then a Jordan–Chevalley decomposition may not exist, as it is possible that the minimal polynomial is not a product of separable polynomials. The simplest such example is the following. Let ${\displaystyle p}$ be a prime number, let ${\displaystyle k}$ be an imperfect field of characteristic ${\displaystyle p,}$ (e. g. ${\displaystyle k=\mathbb {F} _{p}(t)}$) and choose ${\displaystyle a\in k}$ that is not a ${\displaystyle p}$th power. Let ${\displaystyle V=k[X]/\left(X^{p}-a\right)^{2},}$ let ${\displaystyle x={\overline {X}}}$ be the image in the quotient and let ${\displaystyle T}$ be the ${\displaystyle k}$-linear operator given by multiplication by ${\displaystyle x}$ in ${\displaystyle V}$. Note that the minimal polynomial is precisely ${\displaystyle \left(X^{p}-a\right)^{2}}$, which is inseparable and a square. By the necessity of the condition for the Jordan-Chevalley decomposition (as shown in the last section), this operator does not have a Jordan-Chevalley decomposition. It can be instructive to see concretely why there is at least no decomposition into a square-free and a nilpotent part.

If instead of with the polynomial ${\displaystyle \left(X^{p}-a\right)^{2}}$, the same construction is performed with ${\displaystyle {X^{p}}-a}$, the resulting operator ${\displaystyle T}$ still does not admit a Jordan-Chevalley decomposition by the main theorem. However, ${\displaystyle T}$ is semi-simple. The trivial decomposition ${\displaystyle T=T+0}$ hence expresses ${\displaystyle T}$ as a sum of a semisimple and a nilpotent operator, both of which are polynomials in ${\displaystyle T}$.

Elementary proof of existence

This construction is similar to Hensel's lemma in that it uses an algebraic analogue of Taylor's theorem to find an element with a certain algebraic property via a variant of Newton's method. In this form, it is taken from (Geck 2022).

Let ${\displaystyle x}$ have minimal polynomial ${\displaystyle p}$ and assume this is a product of separable polynomials. This condition is equivalent to demanding that there is some separable ${\displaystyle q}$ such that ${\displaystyle q\mid p}$ and ${\displaystyle p\mid q^{m}}$ for some ${\displaystyle m\geq 1}$. By the Bézout lemma, there are polynomials ${\displaystyle u}$ and ${\displaystyle v}$ such that${\displaystyle {uq+{vq'}}=1}$. This can be used to define a recursion ${\displaystyle x_{n+1}=x_{n}-v(x_{n})q(x_{n})}$, starting with ${\displaystyle x_{0}=x}$. Letting ${\displaystyle {\mathfrak {X}}}$ be the algebra of operators which are polynomials in ${\displaystyle x}$, it can be checked by induction that for all ${\displaystyle n}$:

• ${\displaystyle x_{n}\in {\mathfrak {X}}}$ because in each step, a polynomial is applied,
• ${\displaystyle x_{n}-x\in x\cdot {\mathfrak {X}}}$ because ${\displaystyle x_{n+1}-x=(x_{n+1}-x_{n})+(x_{n}-x)}$ and both terms are in ${\displaystyle x\cdot {\mathfrak {X}}}$ by induction hypothesis,
• ${\displaystyle q(x_{n})\in x^{2^{n}}\cdot {\mathfrak {X}}}$ because ${\displaystyle q(x_{n+1})=q(x_{n})+q'(x_{n})(x_{n+1}-x_{n})+(x_{n+1}-x_{n})^{2}h}$ for some ${\displaystyle h\in {\mathfrak {X}}}$ (by the algebraic version of Taylor's theorem), where the second summand is equal to ${\displaystyle q(x_{n})(u(x_{n})q(x_{n})-1)}$ by definition of ${\displaystyle u}$ and ${\displaystyle v}$. So ${\displaystyle q(x_{n+1})=q(x_{n})^{2}(u(x_{n})+v(x_{n})^{2}h)\in x^{2\cdot 2^{n}}\cdot {\mathfrak {X}}}$.

Thus, as soon as ${\displaystyle 2^{n}\geq m}$, ${\displaystyle q(x_{n})=0}$ by the third point, so the minimal polynomial of ${\displaystyle x_{n}}$ will divide ${\displaystyle q}$ and hence be separable. Moreover, ${\displaystyle x_{n}}$ will be a polynomial in ${\displaystyle x}$ by the first point and ${\displaystyle x_{n}-x}$ will be nilpotent by the second point (in fact, ${\displaystyle (x_{n}-x)^{m}=0}$. Therefore, ${\displaystyle x=x_{n}+(x-x_{n})}$ is then the Jordan-Chevalley decomposition of ${\displaystyle x}$. Q.E.D.

This proof, besides being completely elementary, has the advantage that it is constructive: By the Cayley–Hamilton theorem, ${\displaystyle p}$ can be taken to be the characteristic polynomial of ${\displaystyle x}$, and in many contexts, ${\displaystyle q}$ can be determined from that.^[3] Then ${\displaystyle v}$ can be determined using the Euclidean algorithm. The iteration of applying the polynomial ${\displaystyle vq}$ to the matrix then can be performed until either ${\displaystyle v(x_{n})q(x_{n})=0}$ (because then all later values will be equal) or ${\displaystyle 2^{n}}$ exceeds the dimension of the vector space on which ${\displaystyle x}$ is defined (where ${\displaystyle n}$ is the number of iteration steps performed, as above).

Proof of existence via Galois theory

This proof, or variants of it, is commonly cited in context of the Jordan-Chevalley decomposition. It has the advantage that it describes quite precisely how close one can get to a Jordan-Chevalley decomposition.

Above it was observed that if ${\displaystyle x}$ has a Jordan normal form (i. e. if the minimal polynomial of ${\displaystyle x}$ splits), then it has a Jordan Chevalley decomposition. In this case, one can also see directly that ${\displaystyle x_{n}}$ (and hence also ${\displaystyle x_{s}}$) is a polynomial in ${\displaystyle x}$. Indeed, it suffices to check this for the decomposition of the Jordan matrix ${\displaystyle J=D+R}$. This is a technical argument, but does not require any tricks beyond the Chinese remainder theorem.

This fact can be used to deduce the Jordan-Chevalley decomposition in the general case. Let ${\displaystyle x}$ have minimal polynomial ${\displaystyle p}$ and assume this is a product of separable polynomials. This condition is equivalent to demanding that there is some separable ${\displaystyle q}$ such that ${\displaystyle q\mid p}$ and ${\displaystyle p\mid q^{m}}$ for some ${\displaystyle m\geq 1}$. Let ${\displaystyle L}$ be the splitting field of ${\displaystyle q}$, over which also ${\displaystyle p}$ splits. Therefore, by the argument just given, ${\displaystyle x}$ has a Jordan-Chevalley decomposition ${\displaystyle x={c(x)}+{(x-{c(x)})}}$ where ${\displaystyle c}$ is a polynomial with coefficients from ${\displaystyle L}$, <c(x)> is diagonalisable (over ${\displaystyle L}$) and ${\displaystyle x-c(x)}$ is nilpotent.

Let ${\displaystyle \sigma }$ be a field automorphism of ${\displaystyle L}$ which fixes ${\displaystyle K}$. Then

$${\displaystyle c(x)+(x-{c(x)})=x={\sigma (x)}={\sigma ({c(x)})}+{\sigma (x-{c(x)})}}$$

Here ${\displaystyle \sigma (c(x))=\sigma (c)(x)}$ is a polynomial in ${\displaystyle x}$, so is ${\displaystyle x-c(x)}$. Thus, ${\displaystyle \sigma (c(x))}$ and ${\displaystyle \sigma (x-c(x))}$ commute. Also, ${\displaystyle \sigma (c(x))}$ is potentially diagonalisable and ${\displaystyle \sigma ({x-c(x)})}$ is nilpotent. Thus, by the uniqueness of the Jordan-Chevalley decomposition (over ${\displaystyle L}$), ${\displaystyle \sigma (c(x))=c(x)}$ and ${\displaystyle \sigma (c(x))=c(x)}$. Because ${\displaystyle q}$ is separable, the extension ${\displaystyle L/K}$ is Galois, meaning an element of ${\displaystyle L}$ fixed by all automorphisms of ${\displaystyle L}$ which fix ${\displaystyle K}$ must lie in ${\displaystyle K}$. That is, ${\displaystyle x_{s},x_{n}}$ are endomorphisms (represented by matrices) over ${\displaystyle K}$. Finally, since ${\displaystyle \left\{1,x,x^{2},\dots \right\}}$ contains an ${\displaystyle L}$-basis that spans the space containing ${\displaystyle x_{s},x_{n}}$, by the same argument, we also see that ${\displaystyle c}$ has coefficients in ${\displaystyle K}$. This completes the proof. Q.E.D.

This argument describes precisely why the condition on ${\displaystyle x}$ is required in the theorem and how close to a Jordan-Chevalley decomposition one can get in general: If ${\displaystyle p}$ is the minimal polynomial of ${\displaystyle x}$, and ${\displaystyle L}$ is the splitting field of ${\displaystyle p}$, then ${\displaystyle x}$ admits a Jordan-Chevalley decomposition over the subfield of ${\displaystyle L}$ of elements fixed by the entire Galois group ${\displaystyle \operatorname {Gal} \left(L/K\right)}$, but not over smaller subfields. (Note that obtaining this latter formulation of the result does not require Galois theory. The Galois theory is only needed to establish that the field is ${\displaystyle K}$ when the minimal polynomial of ${\displaystyle x}$ is a product of separable polynomials.)

Separable algebras

The Jordan-Chevalley decomposition is very closely related to the Wedderburn principal theorem in the following formulation:^[4]

Usually, the term „separable“ in this theorem refers to the general concept of a separable algebra and the theorem might then be established as a corollary of a more general high-powered result.^[5] However, if it is instead interpreted in the more basic sense that every element have a separable minimal polynomial, then this statement is essentially equivalent to the Jordan-Chevalley decomposition as described above. This gives a different way to view the decomposition, and for instance (Jacobson 1979) takes this route for establishing it.

Over perfect fields, this result simplifies. Indeed, ${\displaystyle A/J}$ is then always separable in the sense of minimal polynomials: If ${\displaystyle a\in A}$, then the minimal polynomial ${\displaystyle p}$ is a product of separable polynomials, so there is a separable polynomial ${\displaystyle q}$ such that ${\displaystyle q\mid p}$ and ${\displaystyle p\mid q^{m}}$ for some ${\displaystyle m\geq 1}$. Thus ${\displaystyle q(a)\in J}$. So in ${\displaystyle A/J}$, the minimal polynomial of ${\displaystyle a+J}$ divides ${\displaystyle q}$ and is hence separable. The crucial point in the theorem is then not that ${\displaystyle A/J}$ is separable (because that condition is vacuous), but that it is semisimple, meaning its radical is trivial.

The same statement is true for Lie algebras, but only in characteristic zero. This is the content of Levi’s theorem. (Note that the notions of semisimple in both results do indeed correspond, because in both cases this is equivalent to being the sum of simple subalgebras or having trivial radical, at least in the finite-dimensional case.)

Preservation under representations

The crucial point in the proof for the Wedderburn principal theorem above is that an element ${\displaystyle x\in A}$ corresponds to a linear operator ${\displaystyle T_{x}:A\to A}$ with the same properties. In the theory of Lie algebras, this corresponds to the adjoint representation of a Lie algebra ${\displaystyle {\mathfrak {g}}}$. This decomposed operator has a Jordan-Chevalley decomposition ${\displaystyle \operatorname {ad} (x)=\operatorname {ad} (x)_{s}+\operatorname {ad} (x)_{n}}$. Just as in the associative case, this corresponds to a decomposition of ${\displaystyle x}$, but polynomials are not available as a tool. One context in which this does makes sense is the restricted case where ${\displaystyle {\mathfrak {g}}}$ is contained in the Lie algebra ${\displaystyle {\mathfrak {gl}}(V)}$ of the endomorphisms of a finite-dimensional vector space ${\displaystyle V}$ over the perfect field ${\displaystyle K}$. Indeed, any Lie algebra can be realised in this way.

If ${\displaystyle x=x_{s}+x_{n}}$ is the Jordan decomposition, then ${\displaystyle \operatorname {ad} (x)=\operatorname {ad} (x_{s})+\operatorname {ad} (x_{n})}$ is the Jordan decomposition of the adjoint endomorphism ${\displaystyle \operatorname {ad} (x)}$ on the vector space ${\displaystyle {\mathfrak {g}}}$. Indeed, first, ${\displaystyle \operatorname {ad} (x_{s})}$ and ${\displaystyle \operatorname {ad} (x_{n})}$ commute since ${\displaystyle [\operatorname {ad} (x_{s}),\operatorname {ad} (x_{n})]=\operatorname {ad} ([x_{s},x_{n}])=0}$. Second, in general, for each endomorphism ${\displaystyle y\in {\mathfrak {g}}}$, we have:

1. If ${\displaystyle y^{m}=0}$, then ${\displaystyle \operatorname {ad} (y)^{2m-1}=0}$, since ${\displaystyle \operatorname {ad} (y)}$ is the difference of the left and right multiplications by y.
2. If ${\displaystyle y}$ is semisimple, then ${\displaystyle \operatorname {ad} (y)}$ is semisimple, since semisimple is equivalent to potentially diagonalisable over a perfect field (if ${\displaystyle y}$ is diagonal over the basis ${\displaystyle \{b_{1},\dots ,b_{n}\}}$, then ${\displaystyle \operatorname {ad} (y)}$ is diagonal over the basis consisting of the maps ${\displaystyle M_{ij}}$ with ${\displaystyle b_{i}\mapsto b_{j}}$ and ${\displaystyle b_{k}\mapsto 0}$ for ${\displaystyle k\neq 0}$).^[6]

Hence, by uniqueness, ${\displaystyle \operatorname {ad} (x)_{s}=\operatorname {ad} (x_{s})}$ and ${\displaystyle \operatorname {ad} (x)_{n}=\operatorname {ad} (x_{n})}$.

The adjoint representation is a very natural and general representation of any Lie algebra. The argument above illustrates (and indeed proves) a general principle which generalises this: If ${\displaystyle \pi$ :{\mathfrak {g}}\to {\mathfrak {gl}}(V)} is any finite-dimensional representation of a semisimple finite-dimensional Lie algebra over a perfect field, then ${\displaystyle \pi }$ preserves the Jordan decomposition in the following sense: if ${\displaystyle x=x_{s}+x_{n}}$, then ${\displaystyle \pi (x_{s})=\pi (x)_{s}}$ and ${\displaystyle \pi (x_{n})=\pi (x)_{n}}$.^[7][8]

Nilpotency criterion

The Jordan decomposition can be used to characterize nilpotency of an endomorphism. Let k be an algebraically closed field of characteristic zero, ${\displaystyle E=\operatorname {End} _{\mathbb {Q} }(k)}$ the endomorphism ring of k over rational numbers and V a finite-dimensional vector space over k. Given an endomorphism ${\displaystyle x:V\to V}$, let ${\displaystyle x=s+n}$ be the Jordan decomposition. Then ${\displaystyle s}$ is diagonalizable; i.e., ${\textstyle V=\bigoplus V_{i}}$ where each ${\displaystyle V_{i}}$ is the eigenspace for eigenvalue ${\displaystyle \lambda _{i}}$ with multiplicity ${\displaystyle m_{i}}$. Then for any ${\displaystyle \varphi \in E}$ let ${\displaystyle \varphi (s):V\to V}$ be the endomorphism such that ${\displaystyle \varphi (s):V_{i}\to V_{i}}$ is the multiplication by ${\displaystyle \varphi (\lambda _{i})}$. Chevalley calls ${\displaystyle \varphi (s)}$ the replica of ${\displaystyle s}$ given by ${\displaystyle \varphi }$. (For example, if ${\displaystyle k=\mathbb {C} }$, then the complex conjugate of an endomorphism is an example of a replica.) Now,

Proof: First, since ${\displaystyle n\varphi (s)}$ is nilpotent,

${\displaystyle 0=\operatorname {tr} (x\varphi (s))=\sum _{i}\operatorname {tr} \left(s\varphi (s)|V_{i}\right)=\sum _{i}m_{i}\lambda _{i}\varphi (\lambda _{i})}$.

If ${\displaystyle \varphi }$ is the complex conjugation, this implies ${\displaystyle \lambda _{i}=0}$ for every i. Otherwise, take ${\displaystyle \varphi }$ to be a ${\displaystyle \mathbb {Q} }$-linear functional ${\displaystyle \varphi :k\to \mathbb {Q} }$ followed by ${\displaystyle \mathbb {Q} \hookrightarrow k}$. Applying that to the above equation, one gets:

${\displaystyle \sum _{i}m_{i}\varphi (\lambda _{i})^{2}=0}$

and, since ${\displaystyle \varphi (\lambda _{i})}$ are all real numbers, ${\displaystyle \varphi (\lambda _{i})=0}$ for every i. Varying the linear functionals then implies ${\displaystyle \lambda _{i}=0}$ for every i. ${\displaystyle \square }$

A typical application of the above criterion is the proof of Cartan's criterion for solvability of a Lie algebra. It says: if ${\displaystyle {\mathfrak {g}}\subset {\mathfrak {gl}}(V)}$ is a Lie subalgebra over a field k of characteristic zero such that ${\displaystyle \operatorname {tr} (xy)=0}$ for each ${\displaystyle x\in {\mathfrak {g}},y\in D{\mathfrak {g}}=[{\mathfrak {g}},{\mathfrak {g}}]}$, then ${\displaystyle {\mathfrak {g}}}$ is solvable.

Proof:^[10] Without loss of generality, assume k is algebraically closed. By Lie's theorem and Engel's theorem, it suffices to show for each ${\displaystyle x\in D{\mathfrak {g}}}$, ${\displaystyle x}$ is a nilpotent endomorphism of V. Write ${\textstyle x=\sum _{i}[x_{i},y_{i}]}$. Then we need to show:

${\displaystyle \operatorname {tr} (x\varphi (s))=\sum _{i}\operatorname {tr} ([x_{i},y_{i}]\varphi (s))=\sum _{i}\operatorname {tr} (x_{i}[y_{i},\varphi (s)])}$

is zero. Let ${\displaystyle {\mathfrak {g}}'={\mathfrak {gl}}(V)}$. Note we have: ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(x):{\mathfrak {g}}\to D{\mathfrak {g}}}$ and, since ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(s)}$ is the semisimple part of the Jordan decomposition of ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(x)}$, it follows that ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(s)}$ is a polynomial without constant term in ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(x)}$; hence, ${\displaystyle \operatorname {ad} _{{\mathfrak {g}}'}(s):{\mathfrak {g}}\to D{\mathfrak {g}}}$ and the same is true with ${\displaystyle \varphi (s)}$ in place of ${\displaystyle s}$. That is, ${\displaystyle [\varphi (s),{\mathfrak {g}}]\subset D{\mathfrak {g}}}$, which implies the claim given the assumption. ${\displaystyle \square }$

Linear algebraic groups

Let ${\displaystyle G}$ be a linear algebraic group over a perfect field. Then, essentially by definition, there is a closed embedding ${\displaystyle G\hookrightarrow \mathbf {GL} _{n}}$. Now, to each element ${\displaystyle g\in G}$, by the multiplicative Jordan decomposition, there are a pair of a semisimple element ${\displaystyle g_{s}}$ and a unipotent element ${\displaystyle g_{u}}$ a priori in ${\displaystyle \mathbf {GL} _{n}}$ such that ${\displaystyle g=g_{s}g_{u}=g_{u}g_{s}}$. But, as it turns out,^[11] the elements ${\displaystyle g_{s},g_{u}}$ can be shown to be in ${\displaystyle G}$ (i.e., they satisfy the defining equations of G) and that they are independent of the embedding into ${\displaystyle \mathbf {GL} _{n}}$; i.e., the decomposition is intrinsic.

When G is abelian, ${\displaystyle G}$ is then the direct product of the closed subgroup of the semisimple elements in G and that of unipotent elements.^[12]