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.

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 ea = a (ae = a) for each aA.

Remark 1

If e is a left identity of A and f is a right identity of A, then e = f. Indeed, ea = a and bf = b for all a,bA. Substituting a = f and b = e yields ef = 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,cK 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,zK 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 : AB 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!

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,bA and α,β∈ K, we obtain f((a,α) ⋅ (b,β)) = f((aba + αb,αβ)) = aba + αb + (αβ) ⋅ 1. But f((a,α)) ⋅ f((b,β)) = (a + α ⋅ 1) ⋅ (b +β ⋅1) = aba + 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 AB.

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 αK. For any aA and αK 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 AB. □

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.

References

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    Drozd, Y.A., Kirichenko, V.V.: Finite Dimensional Algebras. Springer, Berlin (1994)

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Correspondence to Ryszard R. Andruszkiewicz.

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Andruszkiewicz, R.R. Surprise at Adjoining an Identity to an Algebra. Vietnam J. Math. (2020). https://doi.org/10.1007/s10013-020-00437-9

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Keywords

  • Algebra over a field
  • Matrix

Mathematics Subject Classification (2010)

  • 16D25
  • 13C05