Surprise at Adjoining an Identity to an Algebra
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Surprise at Adjoining an Identity to an Algebra Ryszard R. Andruszkiewicz1 Received: 24 January 2020 / Accepted: 18 May 2020 / © The Author(s) 2020
Abstract This note contains an example of two non-isomorphic algebras A and B over an arbitrary and B obtained form A and B, respectively, by the standard field K such that the algebras A process of adjoining an identity, are isomorphic. In addition, the dimension of A may be arbitrary ≥ 2. Keywords Algebra over a field · Matrix Mathematics Subject Classification (2010) 16D25 · 13C05
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. over the field K, let Example 1 In the algebra M2 (K) ofsquare 2 × 2 matrices a b 0 x : a, b ∈ K and B0 = : x, y ∈ K . It is easily seen that A0 and A0 = 0 0 0 y a b 1 c = · B0 are subalgebras of M2 (K). Moreover, for any a, b, c ∈ K we have 0 0 0 0
Ryszard R. Andruszkiewicz
[email protected] 1
Institute of Mathematics, University of Białystok, Ciołkowskiego 1M, 15-245 Białystok, Poland
R.R. Andruszkiewicz
1 1 1 0 a b are different left identities of A0 . By Remark 1, the algeand , so 0 0 0 0 0 0 bra A0has Similarly, for any x, y, z ∈ K we obtain noran identity. aright identity neither 0 1 0 0 0 x 0 z 0 x are different right identities of the algeand , so = · 0 1 0 1 0 y 0 1 0 y bra 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, = {(a, α) : a ∈ A, α ∈ K} we define the multiplication, addition and scalar on the set A multiplication by the rules: (a1 , α1 ) · (a2 , α2 ) = (a1 · a2 + α2 a1 + α1 a2 , α1 α2 ), (a1 , α1 ) + (a2 , α2 ) = (a1 + a2 , α1 + α2 ),
(1) (2)
β · (a, α) = (βa, βα) (3) (see, [1]). A trivial verification shows that A is an algebra with identity (0, 1) and A ∼ = → K given by π((a, α)) = α is a {(a, 0) : a ∈ A} = A. Moreover, the function π : A onto K and Ker(π ) = A. Hence A is an ideal of A and homomorphism of the algebra A ∼ A/A = K. Note that the algebra A seemingly not much different from the algebra A since = A + K · (0, 1). It turns out that this is misleading! A An easy computation shows that, if f : A → B is an algebra isomorphism, then the → B given by F ((a, α)) = (f (a), α) is also an isomorphism. Therefore, function F : A and B are isomorphic the follo
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