Unital Algebras

Introduction

This article studies the second of the two axioms that the corpus adds to the broad sense of an algebra. An $R$-algebra in the sense fixed in Algebras is an $R$-module with a bilinear product; the previous article of this category, Associative Algebras, adds associativity; this article adds the existence of an identity $1$ with $1x = x1 = x$ for every $x$. The two axioms are independent, and most of the corpus assumes both. Associativity without unitality is the ideal $(x) \subseteq R[x]$ of §The Non-Unital Algebras of the Corpus; unitality without associativity is the symmetrised algebra of $M_2(k)$ over a field of characteristic not $2$, in Jordan Algebras, §The Symmetrisation of an Associative Algebra: with the product $x \bullet y = \tfrac12(xy + yx)$ the identity matrix is an identity, and the product is not associative.

Throughout, $R$ is a commutative ring with identity $1 \neq 0$ and $A$ is an $R$-algebra in the broad sense, unless a sentence names a stronger hypothesis. The article is algebraic throughout and uses no distance, no norm and no limit.

The aim of the article is the equivalence of §Unital Algebras as Rings over the Centre: a unital $R$-algebra is a ring with identity together with a unital ring homomorphism $R \to Z(A)$, and conversely. From it, the article treats the trivial cases and the standing hypothesis $1 \neq 0$, the unit group and its specialisations, the unitisation $A^+$ by which a unit is adjoined to any algebra, the homomorphisms that preserve the identity and those that do not, and the idempotents, whose central ones decompose a unital algebra as a product. It closes with the non-unital algebras of the corpus.

The unit theory is not repeated here. The general unit group $A^\times$, the statement that a unit is not a zero divisor and lies in no proper ideal, and the division-algebra criterion are the subject of Centre, Units, Zero Divisors and Division Algebras, §Units, §Zero Divisors and §The Regular Module and Division Algebras; the product of ideals is Ideals and Quotients of Algebras, §Products and Sums of Ideals, and the fact that a proper ideal of a unital algebra contains no unit is Ideals and Quotients of Algebras, §Ideals; the units of the matrix, group and polynomial algebras are specialised in §The Unit Group and proved in Matrix Algebras, Group Algebras and Polynomial Algebras. The convolution algebra is The Convolution Algebra $L^1(G)$, in Part III, and the operator algebras are Operator Algebras, in Part II.

The Identity and the Canonical Map

Definition. Let $A$ be an $R$-algebra. An element $1_A \in A$ is an identity of $A$ if

$$ 1_A x = x 1_A = x $$

for every $x \in A$. The algebra is unital if it has an identity, and $A$ is then a unital $R$-algebra. When no confusion arises the identity is written $1$.

Proposition (uniqueness). A unital algebra has exactly one identity.

Proof. If $1$ and $1'$ are identities then $1 = 1\,1' = 1'$, since $1'$ is an identity for the first product and $1$ for the second.

Definition. Let $A$ be a unital $R$-algebra. The canonical map is

$$ \eta : R \longrightarrow A, \qquad \eta(r) = r\,1_A , $$

the action of $r$ on the identity.

Proposition. Let $A$ be a unital $R$-algebra with canonical map $\eta$. Then:

1. $\eta$ is a unital ring homomorphism, $\eta(1_R) = 1_A$ and $\eta(r + s) = \eta(r) + \eta(s)$, $\eta(rs) = \eta(r)\eta(s)$;
2. the image $R \cdot 1_A = \eta(R)$ is contained in the centre $Z(A)$;
3. $R \cdot 1_A$ is the smallest unital $R$-subalgebra of $A$, and $\eta$ is the unique unital $R$-algebra homomorphism $R \to A$.

Proof. (1) Additivity is the module axiom $\eta(r+s) = (r+s)1_A = r1_A + s1_A$. For multiplicativity, the scalar compatibility and the bilinearity give

$$ (r1_A)(s1_A) = r\bigl(1_A(s1_A)\bigr) = r\bigl(s(1_A1_A)\bigr) = rs\,1_A = \eta(rs), $$

and $\eta(1_R) = 1_R1_A = 1_A$. (2) For $a \in A$ the first $R$-linearity axiom and the identity give $(r1_A)a = r(1_Aa) = ra$, and the second gives $a(r1_A) = r(a1_A) = ra$; hence $(r1_A)a = a(r1_A)$ and $r1_A \in Z(A)$. (3) A unital $R$-subalgebra contains $1_A$, hence contains $r1_A$ for every $r$, so it contains $R\cdot1_A$; and that set is closed under the operations because $\eta$ is a ring homomorphism and the product is $R$-bilinear. A unital homomorphism $\varphi : R \to A$ has $\varphi(r) = \varphi(r1_R) = r\varphi(1_R) = r1_A = \eta(r)$ by $R$-linearity and unitality, so it is $\eta$.

The centrality in statement (2) is not a technicality: it is exactly what makes the multiplication by scalars compatible with the multiplication of $A$ in both variables.

Unital Algebras as Rings over the Centre

The canonical map converts the algebra structure into a ring structure with a distinguished central homomorphism, and the conversion is reversible. This is the spine of the article.

Theorem (associative unital algebras are rings over their centre). Let $R$ be a commutative ring with identity $1 \neq 0$. The following are equivalent:

1. an associative unital $R$-algebra structure on a set $A$, with product $A \times A \to A$;
2. a ring structure with identity on $A$, associative, together with a unital ring homomorphism $\eta : R \to Z(A)$ into its centre.

The two constructions are inverse to one another, and they establish an equivalence between the category of associative unital $R$-algebras with unital $R$-algebra homomorphisms and the category of associative rings with identity equipped with a unital ring homomorphism from $R$ into the centre, with the ring homomorphisms that make the evident triangle commute.

Proof. (1) $\Rightarrow$ (2). An $R$-algebra is a ring, associative when the algebra is; the identity of the algebra is the identity of the ring; and the canonical map $\eta : r \mapsto r1_A$ is a unital ring homomorphism with image in $Z(A)$ by the proposition of §The Identity and the Canonical Map.

(2) $\Rightarrow$ (1). Let $A$ be an associative ring with identity $1_A$ and let $\eta : R \to Z(A)$ be a unital ring homomorphism. Define the action of $R$ on $A$ by

$$ r \cdot a = \eta(r)\,a . $$

This is an $R$-module structure: distributivity in $a$ is the distributivity of the ring, distributivity in $r$ is the additivity of $\eta$, the associativity $(rs)\cdot a = \eta(rs)a = \eta(r)\bigl(\eta(s)a\bigr) = r\cdot(s\cdot a)$ is the multiplicativity of $\eta$, and $1_R \cdot a = \eta(1_R)a = 1_Aa = a$ is the unitality of $\eta$. The ring multiplication is $R$-bilinear with respect to this action: for $a, b \in A$,

$$ (r \cdot a)b = \bigl(\eta(r)a\bigr)b = \eta(r)(ab) = r \cdot (ab) $$

by the associativity of $A$, and

$$ a(r \cdot b) = a\bigl(\eta(r)b\bigr) = \bigl(a\eta(r)\bigr)b = \bigl(\eta(r)a\bigr)b = r \cdot (ab), $$

where the middle equality uses $\eta(r) \in Z(A)$. Hence $A$ is an $R$-algebra with identity $1_A$. The two passages are mutually inverse: the canonical map of the algebra constructed from $\eta$ is $r \mapsto r\cdot1_A = \eta(r)1_A = \eta(r)$, and the ring structure of an algebra is the one it started with. A unital algebra homomorphism is a ring homomorphism commuting with the algebra action, which is the condition that it intertwines the canonical maps.

Corollary (rings as algebras). Every associative ring with identity is a unital $\mathbb{Z}$-algebra in exactly one way, the canonical map being $n \mapsto n1_A$; the category of such rings is the category of unital $\mathbb{Z}$-algebras. More generally a ring $A$ with identity carries a unital $R$-algebra structure exactly when a unital ring homomorphism $R \to Z(A)$ is given, so a ring may carry several distinct $R$-algebra structures with the same product.

Example (centrality is a real condition). Let $A = \mathbb{H}$ be the quaternions, a unital $\mathbb{R}$-algebra with $Z(\mathbb{H}) = \mathbb{R}$, as computed in Centre, Units, Zero Divisors and Division Algebras, §The Centre. There is no unital ring homomorphism $\mathbb{C} \to \mathbb{H}$ with image in the centre, since $Z(\mathbb{H}) = \mathbb{R}$ contains no copy of $\mathbb{C}$; hence $\mathbb{H}$ is not a $\mathbb{C}$-algebra, although it is a left $\mathbb{C}$-module through any embedding of $\mathbb{C}$ in $\mathbb{H}$. The module structure is not enough: the two sides of the product must commute with the scalars, which is the centrality.

Remark. The theorem explains the convention, stated in Conventions in Mathematics, that a module is written with its scalars on the left and that the side begins to matter when the ring is non-commutative. For a unital algebra over a commutative $R$ the scalars lie in the centre, so the left and right actions agree, and no side is preferred; the centrality of $\eta$ is what makes the notation unambiguous.

Trivial Cases and the Hypothesis $1 \neq 0$

Theorem. In a unital algebra $A$, if $1 = 0$ then $A = \{0\}$.

Proof. For every $a \in A$, $a = a1 = a0 = 0$.

Corollary. A nonzero unital algebra has $1 \neq 0$. The zero algebra is the only algebra in which the identity equals the zero element, and in it $0$ is a unit, $0 \cdot 0 = 0 = 1$, with unit group $A^\times = \{0\}$ the trivial group.

Remark (the standing hypothesis). The corpus writes $1 \neq 0$ throughout, and it is a hypothesis on the coefficient structure rather than on the algebra: it is part of the definition of a field in Conventions in Mathematics; it opens the standing hypotheses of the algebra articles, among them Ideals and Quotients of Algebras, Centre, Units, Zero Divisors and Division Algebras, Tensor Powers and the Free Algebra and Matrix Algebras, each of which fixes a commutative ring $R$ with identity $1 \neq 0$ before anything else; and Non-Associative Algebras and the Property Ladder builds it into the word unital, an identity being written $1 \neq 0$. The corollary is the reason: $1 \neq 0$ excludes exactly the zero algebra, the one algebra in which the identity carries no information, and it is the zero algebra that the definition of a simple algebra in Ideals and Quotients of Algebras, §Simplicity, Maximality and the Radical, excludes when it speaks of a nonzero algebra. The hypothesis is therefore not a matter of taste but the exclusion of the one degenerate case.

Remark (subalgebras). If $B$ is a subalgebra of a unital algebra $A$ and $1_A \in B$, then $B$ is unital with $1_B = 1_A$, by the uniqueness of the identity. If $1_A \notin B$, then $B$ may still be unital with an identity of its own, and then that identity is an idempotent of $A$ different from $1_A$; a subalgebra of this kind is not a unital subalgebra of $A$. The algebra $R \times 0$ inside $R \times R$ is the simplest instance.

The Unit Group

Definition. An element $u$ of a unital algebra $A$ is a unit if there is $u^{-1} \in A$ with $uu^{-1} = u^{-1}u = 1$; the unit group $A^\times$ is the set of units.

That $A^\times$ is a group under the multiplication of $A$, that the inverse is unique and two-sided, that a unit is not a zero divisor, and that a unit lies in no proper ideal are the subject of Centre, Units, Zero Divisors and Division Algebras, §Units and §Zero Divisors, and are not repeated. What follows is the specialisation to the algebras in which the corpus computes it.

Theorem (specialisations).

1. In the matrix algebra $M_n(R)$ the units are the matrices of invertible determinant, $M_n(R)^\times = \mathrm{GL}_n(R) = \{A : \det A \in R^\times\}$; over a field this is the general linear group (Matrix Algebras).
2. In the group algebra $k[G]$ over a field $k$ the group $G$ and the scalars $k^\times$ lie in $k[G]^\times$, so $k^\times G \subseteq k[G]^\times$, and the inclusion is strict in general (Group Algebras, §Units and Zero Divisors).
3. In the polynomial algebra $k[x]$ the units are the nonzero constants, $k[x]^\times = k^\times$; the proof is the degree function, and it is in Polynomial Algebras.
4. In a division algebra $A \neq 0$, $A^\times = A \setminus \{0\}$; this characterises the division algebras among the unital algebras, and it is the criterion of Centre, Units, Zero Divisors and Division Algebras, §The Regular Module and Division Algebras. Over $\mathbb{R}$ the examples are $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$, with the last computed from the norm in Centre, Units, Zero Divisors and Division Algebras. Remark. The unit group is a functor on unital algebras, by the theorem of §Homomorphisms and the Identity, and it is the group through which the multiplicative structure of an algebra is visible: for a group algebra $k[G]$ it contains $G$, and the representation theory of Group Algebras is the representation theory of the algebra through it. The division algebras are exactly the unital algebras for which $A^\times$ is as large as it can be, $A^\times = A \setminus \{0\}$.

The Unitisation

A non-unital algebra is embedded in a unital one in a universal way, and the construction is functorial.

Definition. Let $A$ be an $R$-algebra. The unitisation of $A$ is the $R$-module

$$ A^+ = A \oplus R $$

with the product

$$ (a, r)(b, s) = (ab + rb + sa,\, rs), \qquad a, b \in A, \quad r, s \in R , $$

where $rb$ and $sa$ denote the $R$-module action. The symbol $A^+$ for the unitisation is the one used by K-Theory of Operator Algebras; Jordan Algebras uses the same symbol for a different construction, the symmetrised algebra of an associative algebra, and the two are not to be confused.

Theorem (the unitisation is unital, with $A$ an ideal). In the algebra $A^+$:

1. the element $(0, 1)$ is an identity;
2. the inclusion $a \mapsto (a, 0)$ is an injective algebra homomorphism whose image $A \oplus 0$ is a two-sided ideal;
3. the quotient map $A^+ \to R$, $(a, r) \mapsto r$, is a unital algebra homomorphism with kernel $A \oplus 0$, so that $A^+/A \cong R$;
4. $A^+$ is associative if and only if $A$ is.

Proof. (1) $(0,1)(b,s) = (0\cdot b + 1\cdot b + s\cdot 0,\, 1\cdot s) = (b,s)$ and $(a,r)(0,1) = (a\cdot 0 + r\cdot 0 + 1\cdot a,\, r\cdot 1) = (a,r)$. (2) The map is $R$-linear and $(a,0)(b,0) = (ab, 0)$, so it is a homomorphism, injective because $a \mapsto (a,0)$ is. For the ideal property, $(b,s)(a,0) = (ba + sa, 0)$ and $(a,0)(b,s) = (ab + ra, 0)$ lie in $A \oplus 0$. (3) The map is $R$-linear, multiplicative because the second coordinate of a product is the product of the second coordinates, and unital because $(0,1) \mapsto 1$; its kernel is the set of pairs with second coordinate zero, which is $A \oplus 0$. (4) The computation is

$$ \bigl((a,r)(b,s)\bigr)(c,t) = (ab + rb + sa,\, rs)(c,t) = \bigl((ab + rb + sa)c + rs\,c + t(ab + rb + sa),\, rst\bigr), $$

$$ (a,r)\bigl((b,s)(c,t)\bigr) = (a,r)(bc + sc + tb,\, st) = \bigl(a(bc + sc + tb) + r(bc + sc + tb) + st\,a,\, rst\bigr), $$

and the two differ by the difference of $a(bc)$ and $(ab)c$, which is zero exactly when $A$ is associative, all other terms agreeing.

Theorem (universal property). Let $A$ be an $R$-algebra and let $B$ be a unital associative $R$-algebra. Then every $R$-algebra homomorphism $\varphi : A \to B$ extends uniquely to a unital $R$-algebra homomorphism $\varphi^+ : A^+ \to B$, given by

$$ \varphi^+(a, r) = \varphi(a) + r\,1_B . $$

Equivalently, restriction along the inclusion $A \hookrightarrow A^+$ is a bijection $\operatorname{Hom}^{1}_{R\text{-alg}}(A^+, B) \to \operatorname{Hom}_{R\text{-alg}}(A, B)$, natural in $A$ and in $B$: the unitisation is left adjoint to the forgetful functor from unital associative $R$-algebras to $R$-algebras.

Proof. The map $\varphi^+$ is $R$-linear, and it is multiplicative:

$$ \varphi^+\bigl((a,r)(b,s)\bigr) = \varphi(ab + rb + sa) + rs\,1_B = \varphi(a)\varphi(b) + r\varphi(b) + s\varphi(a) + rs\,1_B, $$

while

$$ \varphi^+(a,r)\varphi^+(b,s) = \bigl(\varphi(a) + r1_B\bigr)\bigl(\varphi(b) + s1_B\bigr) = \varphi(a)\varphi(b) + s\varphi(a) + r\varphi(b) + rs\,1_B , $$

the two being equal, the middle two terms differing only in their order. It is unital because $\varphi^+(0,1) = 1_B$. If $\psi : A^+ \to B$ is a unital homomorphism that restricts to $\varphi$ on $A$, then $\psi(0,1) = 1_B$ and hence $\psi(a,r) = \psi(a,0) + \psi(0,r) = \varphi(a) + r\,1_B$ by linearity, which is $\varphi^+$; so the extension is unique. Naturality is the identity $\varphi^+(f^+(a,r)) = (\varphi f)^+(a,r)$ for a homomorphism $f : A \to A'$, immediate from the definitions.

Corollary (functoriality). For an $R$-algebra homomorphism $f : A \to A'$, the assignment $f^+(a,r) = (f(a), r)$ is a unital $R$-algebra homomorphism $A^+ \to A'^+$, and the unitisation is a functor. The unitisation of a unital algebra gains nothing and loses nothing: for unital $A$ the map

$$ (a, r) \longmapsto (a + r\,1_A,\, r) $$

is an isomorphism of $R$-algebras $A^+ \to A \times R$ onto the direct product, with inverse $(c,r) \mapsto (c - r\,1_A,\, r)$; under it $A$ corresponds to $A \times 0$, and the identity $(0,1)$ of $A^+$ corresponds to $(1_A, 1)$, while the original identity $(1_A, 0)$ of $A$ is not the identity of $A^+$.

Proof. Multiplicativity of $f^+$ is the multiplicativity of $f$, and $f^+(0,1) = (0,1)$. For the isomorphism, the displayed map $\theta(a,r) = (a + r1_A, r)$ is $R$-linear and bijective with the stated inverse, and it is multiplicative because

$$ \theta\bigl((a,r)(b,s)\bigr) = \bigl(ab + rb + sa + rs\,1_A,\, rs\bigr) = \bigl((a + r1_A)(b + s1_A),\, rs\bigr) = \theta(a,r)\,\theta(b,s), $$

the middle equality being the expansion of the product in the unital algebra $A$.

Remark. The unitisation is the smallest unital algebra through which every homomorphism from $A$ to a unital algebra factors: it is initial among the unital algebras receiving an $R$-algebra homomorphism from $A$, and every such algebra receives a unique unital homomorphism from $A^+$ extending it. Applied to an algebra that already has an identity it adjoins a new one rather than preserving the old, by the corollary, so $A^+$ is not $A$ when $A$ is unital; what the corollary says is that $A^+$ is then the direct product of $A$ with the base ring. The construction is used without comment in Part II, where the proof of the Gelfand–Naimark theorem applies the GNS construction of Operator Algebras, §The Gelfand–Naimark Theorems, to $A^+$ when $A$ has no identity, and in Part III, where K-Theory of Operator Algebras records that $A \to A^+$ is an isomorphism on $K_1$ and that $K_0(A) = \ker\bigl(K_0(A^+) \to K_0(\mathbb{C})\bigr)$; the universal property is stated here because it is a Part I statement.

Homomorphisms and the Identity

Definition. A homomorphism $\varphi : A \to B$ of unital $R$-algebras is unital if $\varphi(1_A) = 1_B$.

This is the definition used in Ideals and Quotients of Algebras, and it is not automatic.

Theorem. Let $\varphi : A \to B$ be an $R$-algebra homomorphism of unital algebras.

1. $\varphi(1_A)$ is an idempotent of $B$, and $\varphi$ is unital if and only if that idempotent is $1_B$.
2. If $\varphi$ is unital then $\varphi(A^\times) \subseteq B^\times$, and $\varphi$ restricts to a group homomorphism $A^\times \to B^\times$, $u \mapsto \varphi(u)$, with $\varphi(u^{-1}) = \varphi(u)^{-1}$.
3. Conversely, if $\varphi$ maps some unit of $A$ to a unit of $B$, then $\varphi$ is unital.

Proof. (1) $\varphi(1_A)^2 = \varphi(1_A1_A) = \varphi(1_A)$, so the image of the identity is idempotent; it equals $1_B$ exactly when $\varphi$ is unital. (2) If $u \in A^\times$ then $\varphi(u)\varphi(u^{-1}) = \varphi(uu^{-1}) = \varphi(1_A) = 1_B$ and likewise on the other side, so $\varphi(u) \in B^\times$ with inverse $\varphi(u^{-1})$. (3) Suppose $u \in A^\times$ and $\varphi(u) \in B^\times$, and let $e = \varphi(1_A)$. Since $\varphi(u1_A) = \varphi(u)e$ and $\varphi(u1_A) = \varphi(u)$, the idempotent $e$ satisfies $\varphi(u)e = \varphi(u)$; multiplying this on the left by the inverse of $\varphi(u)$ gives $e = 1_B$, so $\varphi$ is unital.

Example (homomorphisms that do not preserve the identity). The zero homomorphism $A \to B$ between nonzero unital algebras is not unital; it sends $1_A$ to the idempotent $0$. For an associative unital algebra $A$ and a central idempotent $e \in Z(A)$ the map $\varphi(a) = eae$ is an algebra homomorphism, since

$$ \varphi(a)\varphi(b) = eaebe = e^2abe = eab e = \varphi(ab) $$

using $e$ central and idempotent, with $\varphi(1_A) = e$; it is unital exactly when $e = 1_A$. Its image is the ideal $Ae$, and for $A = R \times R$ and $e = (1,0)$ it is the projection onto the first factor.

Remark (why the corpus insists). The distinction between the two notions of homomorphism is not a convention of convenience. A unital homomorphism is what makes the passage to units a functor, as statement (2) shows; it is what the universal properties of the unitisation and of the enveloping algebra of Universal Enveloping Algebras quantify over; it is what makes a quotient map $A \to A/I$ unital and therefore makes the quotient of a unital algebra unital; and it is what makes the augmentation $k[G] \to k$ of Group Algebras, $g \mapsto 1$, a unital homomorphism whose kernel is the augmentation ideal. A module over a unital algebra is required to be unital, $1 \cdot m = m$, for the same reason: the identity of the algebra must act as the identity of the module, and the representation theory of Group Algebras and Matrix Algebras is the theory of unital modules.

Idempotents and the Peirce Decomposition

Definition. An element $e$ of an associative algebra $A$ is an idempotent if $e^2 = e$. Two idempotents $e, f$ are orthogonal if $ef = fe = 0$, a family $e_1, \dots, e_n$ is complete if $e_1 + \cdots + e_n = 1_A$, and an idempotent is central if it lies in the centre, $e \in Z(A)$.

The identity $1_A$ and the zero element are idempotents, and they are the only idempotents of a division algebra: if $e^2 = e$ with $e \neq 0$ then $e = 1_A$ after multiplying by $e^{-1}$. An idempotent different from $0$ and $1_A$ is neither a unit nor nilpotent. The idempotents of a ring are treated in Rings, §Idempotents, and collected ring by ring in List of Rings by Their Idempotents; the statement that a one-sided ideal is generated by an idempotent exactly when it is a direct summand of the regular module is Modules, §Examples and Direct Sums, Free Modules and Rank, §Direct Summands and Idempotents.

Theorem (Peirce decomposition, one idempotent). Let $A$ be an associative unital $R$-algebra, let $e \in A$ be an idempotent and put $f = 1_A - e$. Then $f$ is an idempotent orthogonal to $e$, and every $a \in A$ expands as

$$ a = eae + eaf + fae + faf , $$

a direct sum of $R$-submodules

$$ A = eAe \oplus eAf \oplus fAe \oplus fAf . $$

The summands are the Peirce spaces of $e$. Writing $e_1 = e$ and $e_2 = f$, the products obey

$$ (e_i A e_j)(e_k A e_l) \subseteq \delta_{jk}\, e_i A e_l . $$

The corners $eAe$ and $fAf$ are subalgebras with identities $e$ and $f$ respectively; the off-diagonal spaces $eAf$ and $fAe$ are not subalgebras, but each is a bimodule over the two corners.

Proof. Since $1_A = e + f$ acts as an identity, $a = (e+f)a(e+f) = eae + eaf + fae + faf$. The element $f$ is idempotent because $f^2 = (1-e)^2 = 1 - 2e + e^2 = 1 - e = f$, and $ef = e(1-e) = 0 = (1-e)e = fe$. For the directness, suppose $x_{11} + x_{12} + x_{21} + x_{22} = 0$ with $x_{ij} \in e_i A e_j$. Multiplying the relation on the left by $e_1 = e$ and on the right by $e_1$ annihilates the three terms carrying a factor $e_2 = f$, leaving $x_{11} = 0$; the same computation with $e_1$ on the left and $e_2$ on the right leaves $x_{12} = 0$, and symmetrically $x_{21} = x_{22} = 0$. The product rule follows from $e_je_k = \delta_{jk}e_j$, which gives $e_iAe_j \cdot e_kAe_l = e_iA(e_je_k)Ae_l \subseteq \delta_{jk}e_iAe_l$. In particular $eAe \cdot eAe \subseteq e^2Ae^2 = eAe$, and $e \cdot eae \cdot e = eae$, so $eAe$ is a subalgebra with identity $e$; the same argument applies to $f$.

Corollary (a central idempotent splits the algebra). If $e$ is central then $eAf = Aef = 0$ and $fAe = Afe = 0$, the Peirce decomposition reduces to

$$ A = Ae \oplus Af = Ae \oplus A(1_A - e), $$

both summands are two-sided ideals, they annihilate one another, and the map $a \mapsto (ae, af)$ is an isomorphism of unital $R$-algebras

$$ A \cong Ae \times A(1_A - e) $$

onto the product of $Ae$ and $Af$ with the identity $e$ and $1_A - e$ respectively.

Proof. If $e \in Z(A)$ then $eAf = (eA)f = (Ae)f = A(ef) = 0$, and similarly $fAe = 0$, so the four-corner sum collapses to the two diagonal corners; also $a \mapsto ae$ is $R$-linear, and $a(xe) = (ax)e$ and $(xe)a = (xa)e$ show that $Ae$ is an ideal, likewise $Af$. The map $\varphi(a) = (ae, af)$ is additive and multiplicative, since $(ae)(be) = aebe = ab e^2 = abe$ using the centrality of $e$, and the same for $f$; the cross terms never occur because $\varphi(ab) = (abe, abf)$ already has the two components separate. It is unital, $\varphi(1_A) = (e, f)$, which is the identity of the product. It is injective because $ae = 0$ and $af = 0$ give $a = a(e+f) = 0$, and surjective because an element of $Ae \times Af$ is $(xe, yf)$ and is the image of $xe + yf$, with $\varphi(xe+yf) = (xe, yf)$ since $fe = 0$ and $ef = 0$.

Remark (why centrality is needed). Without it the Peirce pieces are not ideals and no product decomposition follows, so the Peirce decomposition is strictly finer than the decomposition by central idempotents. For $A = M_2(k)$ the idempotent $E_{11}$ does not commute with $E_{12}$ and so is not central, and the decomposition is $M_2(k) = kE_{11} \oplus kE_{12} \oplus kE_{21} \oplus kE_{22}$, four one-dimensional Peirce spaces none of which is two-sided; $M_2(k)$ is simple, so its only two-sided ideals are $0$ and $M_2(k)$, and no product decomposition exists (Matrix Algebras, §Ideals and Simplicity). The Peirce spaces of $E_{11}$ are the matrix units, the observations $E_{11}M_2E_{22} = kE_{12}$ and $E_{22}M_2E_{11} = kE_{21}$ being the two off-diagonal corners of Matrix Algebras, §Matrix Units. The endomorphism ring of a module gives the same corners in the general form $M = eM \oplus (1-e)M$ (Direct Sums, Free Modules and Rank, §Direct Summands and Idempotents).

Example (the split algebras). Let $A = \mathbb{R} \times \mathbb{R}$ and $e = (1,0)$. Then $e$ is a central idempotent, $Ae = \mathbb{R} \times 0$, $A(1-e) = 0 \times \mathbb{R}$, and the corollary is the tautology $A \cong Ae \times Af$. The same pattern with a nontrivial computation is the split-complex algebra $\mathbb{D} = \mathbb{R}[j]/(j^2-1)$, where

$$ e_+ = \tfrac{1+j}{2}, \qquad e_- = \tfrac{1-j}{2} $$

are central idempotents with $e_+ + e_- = 1$ and $e_+e_- = 0$; the corollary gives $\mathbb{D} = \mathbb{D}e_+ \oplus \mathbb{D}e_-$ with $\mathbb{D}e_\pm \cong \mathbb{R}$, that is $\mathbb{D} \cong \mathbb{R} \times \mathbb{R}$ by $a + bj \mapsto (a+b, a-b)$ (Split Complex Algebra, §The Idempotent Basis, §Idempotent Decomposition). Over $\mathbb{R}$ the field $\mathbb{C}$ has no idempotent other than $0$ and $1$, so it admits no such decomposition; this is the algebraic content of the fact that $\mathbb{D}$ splits and $\mathbb{C}$ does not, although the two are equal in dimension. The only ideals of $\mathbb{D}$ are $0$, $\mathbb{D}e_+$, $\mathbb{D}e_-$ and $\mathbb{D}$, so its ideal lattice and its Peirce decomposition are read from the same two idempotents.

Remark (in the corpus). The four-corner decomposition for an algebra that is not commutative, together with the matrix-unit computation of its corners, is carried out for the biquaternions $\mathbb{B} \cong M_2(\mathbb{C})$ in Biquaternion Ideals and Peirce Decomposition, §The Peirce Decomposition, and catalogued in List of Rings by Their Idempotents, §The Peirce Decomposition. The analogue for a Jordan algebra, where the middle space $J_{1/2}(e)$ appears in place of the two off-diagonal corners and the theorem is proved from the cubic identity $2L_e^3 - 3L_e^2 + L_e = 0$, is Jordan Algebras, §The Peirce Decomposition; the two theories agree on a symmetrised associative algebra only in the diagonal part.

The Non-Unital Algebras of the Corpus

Each of the following is an algebra of the corpus that has no identity, with the article that establishes it.

• The polynomials of positive degree. The ideal $(x) \subseteq R[x]$, consisting of the polynomials with zero constant term, is an $R$-algebra under the inherited product; it has no identity, because an identity $e$ would satisfy $xe = x$, and writing $e = xh$ with $h \in R[x]$ gives $x(xh) = x$, that is $x(xh - 1) = 0$. The coefficient of $x$ in $x(xh-1)$ is the constant term of $xh - 1$, which is $-1$; comparing with the zero polynomial gives $-1 = 0$, against the standing hypothesis $1 \neq 0$. The polynomial algebra and its ideals are Polynomial Algebras, §Ideals and Quotients, and the general statement that a proper ideal of a unital algebra contains no unit is Ideals and Quotients of Algebras, §Ideals.

• A square-zero ideal. If $I^2 = 0$ and $I$ is nonzero then $I$ has no identity: an identity $e$ would satisfy $e = e^2 \in I^2 = 0$. The maximal ideal $(\varepsilon)$ of the dual numbers $\mathbb{D}' = \mathbb{R}[\varepsilon]/(\varepsilon^2)$ is the standard instance, and $(\varepsilon)$ is exactly the set of zero divisors of $\mathbb{D}'$, the zero divisors of Centre, Units, Zero Divisors and Division Algebras, §Zero Divisors.

• A proper ideal of a unital algebra. A proper two-sided ideal $I$ of a unital algebra $A$ contains no unit of $A$, so $1_A \notin I$ (Ideals and Quotients of Algebras, §Ideals); an ideal of this kind is non-unital as a subalgebra of $A$, but it may carry an identity of its own, an idempotent of $A$ different from $1_A$, and then it is a unital algebra. The two cases occur: $(x) \subset R[x]$ is non-unital, while the augmentation ideal $I(G) \subset k[G]$ of a group algebra is spanned in the case $G = \mathbb{Z}/2$ over a field of characteristic not $2$ by $1 - g$ with $(1-g)^2 = 2(1-g)$, so it carries the identity $(1-g)/2$ and is unital (Group Algebras, §The Augmentation Ideal).

• The zero algebra. The algebra $\{0\}$ has $1 = 0$ and is the degenerate case of §Trivial Cases and the Hypothesis $1 \neq 0$; it is the unique unital algebra whose identity is also its zero element.

• The commutator algebra $A^-$. It has no identity for nonzero $A$: an identity $e$ of any algebra satisfies $e \cdot e = e$, while the product of $A^-$ is the commutator, so $e \cdot e = [e,e] = 0$ and hence $e = 0$, which forces $A = 0$. The construction is that of Associative Algebras, §The Universal Enveloping Algebra, and it shows that a non-associative algebra of the corpus may be non-unital as well, for a reason independent of the failure of associativity.

These are the non-unital algebras of the corpus, and they are not deficient for it: each is an algebra in the sense fixed in Algebras, and a construction that needs an identity is applied to its unitisation, §The Unitisation.

Summary

A unital $R$-algebra is an $R$-algebra with an element $1$ satisfying $1x = x1 = x$; the identity is unique. The canonical map $\eta(r) = r1_A$ is a unital ring homomorphism whose image lies in the centre, and the resulting equivalence is the article's spine: a unital $R$-algebra is the same thing as an associative ring $A$ with identity together with a unital ring homomorphism $R \to Z(A)$, the two constructions being inverse; in particular the rings with identity are exactly the unital $\mathbb{Z}$-algebras. If $1 = 0$ then $A = \{0\}$, so a nonzero unital algebra has $1 \neq 0$, the standing hypothesis under which the corpus states its ring- and field-based results and the reason the definition of a simple algebra in Ideals and Quotients of Algebras speaks of a nonzero algebra.

The unit group $A^\times$ is a group; it is specialised to $\mathrm{GL}_n(R)$ for $M_n(R)$, to the subgroup containing $k^\times G$ for $k[G]$, to $k^\times$ for $k[x]$, to $A \setminus \{0\}$ for a division algebra, the general theory being that of Centre, Units, Zero Divisors and Division Algebras. The unitisation $A^+ = A \oplus R$ with $(a,r)(b,s) = (ab + rb + sa, rs)$ is unital with identity $(0,1)$, contains $A$ as a two-sided ideal with $A^+/A \cong R$, is associative exactly when $A$ is, and is characterised by the universal property that every homomorphism $A \to B$ into a unital associative algebra extends uniquely to a unital homomorphism $A^+ \to B$; it is left adjoint to the forgetful functor, and for unital $A$ it is $A \times R$. A homomorphism of unital algebras sends $1_A$ to an idempotent and is unital exactly when that idempotent is $1_B$, equivalently when it carries some unit to a unit; a unital homomorphism restricts to a group homomorphism of the unit groups.

An idempotent $e$, one with $e^2 = e$, gives the Peirce decomposition $A = eAe \oplus eAf \oplus fAe \oplus fAf$ with $f = 1_A - e$, a direct sum of $R$-submodules whose diagonal corners $eAe$ and $fAf$ are subalgebras with identities $e$ and $f$ and whose products obey $(e_iAe_j)(e_kAe_l) \subseteq \delta_{jk}e_iAe_l$. The decomposition is generally finer than the decomposition of $A$ by its ideals: when $e$ is central the two off-diagonal corners vanish and the sum becomes a product $A \cong Ae \times A(1_A - e)$ of unital algebras with identities $e$ and $1_A - e$, and when $e$ is not central, as for $E_{11} \in M_2(k)$, the Peirce pieces are not ideals and no product is obtained. The trivial idempotents $0$ and $1_A$ give nothing, the division algebras and the field $\mathbb{C}$ have no others, and the split-complex algebra with its two central idempotents $(1 \pm j)/2$ is the smallest case in which the product is nontrivial.

The non-unital algebras of the corpus are the polynomials of positive degree, the square-zero ideals, the proper ideals that carry no identity of their own, the commutator algebra $A^-$, and the zero algebra; each is pointed to its own article, and a proper ideal that does carry an identity, such as the augmentation ideal of $k[\mathbb{Z}/2]$ over a field of characteristic not $2$, is a unital algebra and not on the list.

Summary of Notation

Further Reading

• T. Y. Lam, A First Course in Noncommutative Rings (Springer, 2nd ed. 2001), for the unit group, the idempotents and the rings with identity as algebras over their centre.
• Richard S. Pierce, Associative Algebras (Springer, 1982), for the unitisation, the central idempotents and the decomposition of a unital algebra by them.
• Frank W. Anderson and Kent R. Fuller, Rings and Categories of Modules (Springer, 2nd ed. 1992), for unital modules, the unitisation as a left adjoint and the categorical form of the equivalence.