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Example text

2. Step 2. Now suppose that the lemma holds for some pair of words α1 = at Dar , β1 = ak bs . We will show that in this case the lemma also holds for the words α2 = at Dak , β2 = ar bs . Indeed, α∗2 ◦ β2∗ = (at Dak )∗ ◦ (ar ◦ bs ) = 2[(at D)∗ ◦ ak ] ◦ (ar ◦ bs ) − (at+k D)∗ ◦ (ar ◦ bs ) = −2J1 {(at D)∗ , ak , ar , bs } + 2[(at D)∗ ◦ ak+r ] ◦ bs + 2[(at D)∗ ◦ (ak ◦ as )] ◦ ar + 2[(at D)∗ ◦ (ar ◦ bs )] ◦ ak − 2[(at D)∗ ◦ bs ] ◦ ak+r − 2[(at D)∗ ◦ ar ] ◦ (ak ◦ bs ) − (at+k D)∗ ◦ (ar ◦ bs ) = − 2J1 {at D, ak , ar , bs } + 2(at D ◦ ak+r ) ◦ bs + 2[at D ◦ (ak ◦ bs )] ◦ ar + 2[at D ◦ (ar ◦ bs )] ◦ ak − 2(at D ◦ bs ) ◦ ak+r − (at+r D) ◦ (ak ◦ bs ) − (at+k D) ◦ (ar ◦ bs ) ∗ − (at Dar )∗ ◦ (ak ◦ bs ) = −α∗1 ◦ β1∗ + 2(at D ◦ ar ) ◦ (ak ◦ bs ) + 2(at D ◦ ak ) ◦ (ar ◦ bs ) − (at+r D) ◦ (ak ◦ bs ) − (at+k D) ◦ (ar ◦ bs ) ∗ = −α∗1 ◦ β1∗ + at Dar ◦ (ak ◦ bs ) + at Dak ◦ (ar ◦ bs ) = −α∗1 ◦ β1∗ + (α1 ◦ β1 )∗ + (α2 ◦ β2 )∗ = (α2 ◦ β2 )∗ .

Lemma 6. Any associative ring Σ with characteristic diﬀerent from 2 can be embedded in a ring Σ that admits unique division by 2. Proof. Consider the set Σ of pairs (σ, 2k ) where σ ∈ Σ, and k ≥ 0 is an integer. We will consider the pairs (σ1 , 2k1 ) and (σ2 , 2k2 ) to be equivalent if 2k2 σ1 = 2k1 σ2 . We deﬁne addition and multiplication of the pairs in the familiar way: (σ1 , 2k1 ) + (σ2 , 2k2 ) = 2k2 σ1 + 2k1 σ2 , 2k1 +k2 , (σ1 , 2k1 )(σ2 , 2k2 ) = σ1 σ2 , 2k1 +k2 . Obviously, the ring Σ satisﬁes the requirements of Lemma 6.

In the case of algebras over a ﬁeld, Cohn [2] proved that a homomorphic image of a special Jordan algebra is not necessarily special. It follows that the class of special Jordan algebras cannot be deﬁned by identical relations. In the present paper, it is proved that a Jordan algebra over Σ that has a ﬁnite or countably inﬁnite set of generators is special if and only if it can be embedded into a Jordan algebra over Σ with two generators. In the last section of this paper, we remove the requirement that for every element a there exists an element b such that 2b = a.