## 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 *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 *M*_{2}(*K*) of square 2 × 2 matrices over the field *K*, let *A*_{0} =\(\left \{\left [\begin {array}{cc}a&b\\0&0 \end {array}\right ]\colon a,b\in K\right \}\) 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 *A*_{0} and *B*_{0} are subalgebras of *M*_{2}(*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 *A*_{0}. By Remark 1, the algebra *A*_{0} 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 *B*_{0}. Applying Remark 1 again, we infer that the algebra *B*_{0} 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:

(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!

## 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* + *α**b*,*α*β)) = *a* ⋅ *b* +β *a* + *α**b* + (*α*β) ⋅ 1. But *f*((*a*,*α*)) ⋅ *f*((*b*,β)) = (*a* + *α* ⋅ 1) ⋅ (*b* +β ⋅1) = *a* ⋅ *b* +β *a* + *α**b* + (*α*β) ⋅ 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* = *M*_{2}(*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* = *A*_{0} × *C* and *B* = *B*_{0} × *C*, where the algebras *A*_{0} and *B*_{0} 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*)) = *α**f*((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*)↦*π**α* 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 *A*^{2} = {0} and *A* satisfies the assumptions of Theorem 2.

## References

- 1.
Drozd, Y.A., Kirichenko, V.V.: Finite Dimensional Algebras. Springer, Berlin (1994)

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### Cite this article

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