Surprise at Adjoining an Identity to an Algebra

Abstract

This note contains an example of two non-isomorphic algebras A and B over an arbitrary field K such that the algebras $\widetilde {A}$ and $\widetilde {B}$ obtained form A and B, respectively, by the standard process of adjoining an identity, are isomorphic. In addition, the dimension of A may be arbitrary ≥ 2.

1 Introduction

Throughout this paper, all considered algebras are algebras over an arbitrary field K. An element e of an algebra A is called a left (right) identity of A if e ⋅ a = a (a ⋅ e = a) for each a ∈ A.

Remark 1

If e is a left identity of A and f is a right identity of A, then e = f. Indeed, e ⋅ a = a and b ⋅ f = b for all a,b ∈ A. Substituting a = f and b = e yields e ⋅ f = f = e and e = f.

Therefore, if e and f are different left identities (right identities) of an algebra A, then A has no right identity (left identity). In particular, A has no identity.

A straightforward computation shows that any isomorphism of algebras preserves both the left and right identities.

Example 1

In the algebra M₂(K) of square 2 × 2 matrices over the field K, let A₀ =$\left \{\left [\begin {array}{cc}a&b\\0&0 \end {array}\right ]\colon a,b\in K\right \}$ and $B_{0}=\left \{\left [\begin {array}{cc}0&x\\0&y \end {array}\right ]\colon x,y\in K\right \}$. It is easily seen that A₀ and B₀ are subalgebras of M₂(K). Moreover, for any a,b,c ∈ K we have $\left [\begin {array}{cc}1&c\\0&0 \end {array}\right ]\cdot \left [\begin {array}{cc}a&b\\0&0 \end {array}\right ]=\left [\begin {array}{cc}a&b\\0&0 \end {array}\right ]$, so $\left [\begin {array}{cc}1&0\\0&0 \end {array}\right ]$ and $\left [\begin {array}{cc}1&1\\0&0 \end {array}\right ]$ are different left identities of A₀. By Remark 1, the algebra A₀ has neither a right identity nor an identity. Similarly, for any x,y,z ∈ K we obtain $\left [\begin {array}{cc}0&x\\0&y \end {array}\right ]\cdot \left [\begin {array}{cc}0&z\\0&1 \end {array}\right ]=\left [\begin {array}{cc}0&x\\0&y \end {array}\right ]$, so $\left [\begin {array}{cc}0&0\\0&1 \end {array}\right ]$ and $\left [\begin {array}{cc}0&1\\0&1 \end {array}\right ]$ are different right identities of the algebra B₀. Applying Remark 1 again, we infer that the algebra B₀ has neither a left identity nor an identity.

In the theory of algebras, there is a well-known method of adjoining an identity. Namely, on the set $\widetilde {A}=\{(a,\alpha )\colon a\in A,\alpha \in K\}$ we define the multiplication, addition and scalar multiplication by the rules:

$$ \begin{array}{@{}rcl@{}} (a_{1},\alpha_{1})\cdot (a_{2},\alpha_{2})&=&(a_{1}\cdot a_{2}+\alpha_{2}a_{1}+\alpha_{1}a_{2},\alpha_{1}\alpha_{2}), \end{array} $$

(1)

$$ \begin{array}{@{}rcl@{}} (a_{1},\alpha_{1})+(a_{2},\alpha_{2}) &=& (a_{1}+a_{2},\alpha_{1}+\alpha_{2}), \end{array} $$

(2)

$$ \begin{array}{@{}rcl@{}} \upbeta\cdot (a,\alpha)&=&(\upbeta a,\upbeta\alpha) \end{array} $$

(3)

(see, [1]). A trivial verification shows that $\widetilde {A}$ is an algebra with identity (0,1) and $A\cong \{(a,0)\colon a\in A\}=\overline {A}$. Moreover, the function $\pi \colon \widetilde {A}\to K$ given by π((a,α)) = α is a homomorphism of the algebra $\widetilde {A}$ onto K and $\text {Ker} (\pi )=\overline {A}$. Hence $\overline {A}$ is an ideal of $\widetilde {A}$ and $\widetilde {A}/\overline {A}\cong K$. Note that the algebra $\widetilde {A}$ seemingly not much different from the algebra A since $\widetilde {A}=\overline {A}+K\cdot (0,1)$. It turns out that this is misleading!

An easy computation shows that, if f : A → B is an algebra isomorphism, then the function $F\colon \widetilde {A}\to \widetilde {B}$ given by F((a,α)) = (f(a),α) is also an isomorphism. Therefore, the following natural question arises: does the fact that algebras $\widetilde {A}$ and $\widetilde {B}$ are isomorphic imply that algebras A and B are isomorphic? A positive answer to this question is suggested by [1, p. 12, Exercise 12]. We will show that this is not true!

2 Main Results

Lemma 1

Let A be an algebra without identity. If A is a subalgebra of an algebra S with the identity 1, then $\widetilde {A}\cong A+K\cdot 1$.

Proof

Since 1∉A, we get A ∩ (K ⋅ 1) = {0}. Therefore, the function $f\colon \widetilde {A}\to S$ given by f((a,α)) = a + α ⋅ 1 is injective. From (2) and (3) we conclude that f is K-linear. Moreover, (1) implies that for arbitrary a,b ∈ A and α,β∈ K, we obtain f((a,α) ⋅ (b,β)) = f((a ⋅ b +β a + αb,αβ)) = a ⋅ b +β a + αb + (αβ) ⋅ 1. But f((a,α)) ⋅ f((b,β)) = (a + α ⋅ 1) ⋅ (b +β ⋅1) = a ⋅ b +β a + αb + (αβ) ⋅ 1, so f((a,α) ⋅ (b,β)) = f((a,α)) ⋅ f((b,β)). Thus f is an embedding of algebras. Moreover, $f(\widetilde {A})=A+K\cdot 1$, so $\widetilde {A}\cong A+K\cdot 1$. □

Theorem 1

There exist two non-isomorphic algebras A and B of arbitrary dimension ≥ 2 and without identities for which $\widetilde {A}\cong \widetilde {B}$.

Proof

Let C be an arbitrary algebra with identity 1. Then S = M₂(K) × C is an algebra with the identity (I,1), where $I=\left [\begin {array}{cc}1&0\\0&1 \end {array}\right ]$. Let A = A₀ × C and B = B₀ × C, where the algebras A₀ and B₀ are as in Example 1. Then A and B are subalgebras of the algebra S. Moreover, by Example 1, $\left (\left [\begin {array}{cc}1&0\\0&0 \end {array}\right ],1\right )$ and $\left (\left [\begin {array}{cc}1&1\\0&0 \end {array}\right ],1\right )$ are different left identities of the algebra A. It follows from Remark 1 that the algebra A has neither a right identity nor an identity. Similarly, the algebra B has at least two right identities and has neither a left identity nor an identity. Hence, by Remark 1, the algebras A and B are not isomorphic.

By Lemma 1, $\widetilde {A}\cong A+K\cdot (I,1)$ and $\widetilde {B}\cong B+K\cdot (I,1)$. But one can easily check that $A+K\cdot (I,1)=B+K\cdot (I,1)=\left \{\left [\begin {array}{cc}p&q\\0&r \end {array}\right ]\colon p,q,r\in K\right \}\times C$, so $\widetilde {A}\cong \widetilde {B}$.

Since the algebra C is arbitrary and $\dim (A_{0})=2$ we have that $\dim (A)$ can take arbitrary values ≥ 2. □

Theorem 2

Let A be an algebra without identity, which cannot be homomorphically mapped onto the field K. If B is an algebra without identity such that $\widetilde {A}\cong \widetilde {B}$, then A≅B.

Proof

Let $f\colon \widetilde {A}\to \widetilde {B}$ be any isomorphism of algebras. Then f((0,1)) = (0,1) and, consequently, f((0,α)) = αf((0,1)) = α(0,1) = (0,α) for each α ∈ K. For any a ∈ A and α ∈ K we have (a,α) = (a,0) + (0,α), so f((a,α)) = f((a,0)) + (0,α). Hence $\widetilde {B}=f(\overline {A})+K\cdot (0,1)$.

Suppose, contrary to our claim, that $f(\overline {A})\not \subseteq \overline {B}$. Since $\overline {B}$ is the kernel of the natural epimorphism (b,α)↦πα of algebra $\widetilde {B}$ onto the field K, $\pi (f(\overline {A}))$ is a non-zero ideal of K. Thus $\pi (f(\overline {A}))=K$. But $A\cong \overline {A}$, so K is a homomorphic image of the algebra A. This contradicts our assumption.

Hence $f(\overline {A})\subseteq \overline {B}$. But $\widetilde {B}=f(\overline {A})+K\cdot (0,1)$, so by the modularity of the lattice of K-subspaces of the algebra $\widetilde {B}$, we get $\overline {B}=f(\overline {A})+[\overline {B}\cap K\cdot (0,1)]=f(\overline {A})$. It follows that $f(\overline {A})=\overline {B}$ and thus $\overline {A}\cong \overline {B}$. But $A\cong \overline {A}$ and $B\cong \overline {B}$, so A≅B. □

Note that the assumptions of Theorem 2 are satisfied by a large class of algebras; for instance, every nil-algebra satisfies these assumptions. Moreover, if an algebra A of dimension 1 has no identity, then A² = {0} and A satisfies the assumptions of Theorem 2.