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On the page I have read, that we may embed $\boldsymbol{A}$ into $\boldsymbol{A}\times \boldsymbol{B}$ if there exists homomorphism $\sigma:\boldsymbol{A}\to \boldsymbol{B}$. Can anyone give an example of two algebras that are not homomorphic and consequently not embeddable into their direct product?

Thanks in advance.

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Simple example: $A=\mathbb Q$ and $B=\mathbb Z$ – Thomas Andrews Aug 28 '12 at 12:34
If there exists an isomorphism from an algebra A to an algebra B, then we say that A and B are isomorphic. However we don't say that they are "homomorphic" if there exists a homomorphism $A\to B$. One reason is that this relation isn't symmetric: for example, there exists a homomorphism $\mathbb{Z}\to\mathbb{Z}/2$, but there does not exist a homomorphism $\mathbb{Z}/2\to\mathbb{Z}$. So it doesn't make sense to say whether these algebras "are homomorphic". – Brad Aug 28 '12 at 12:53
(That's assuming you require homomorphisms to preserve the multiplicative identity, but then again if you don't then there is always the zero homomorphism, and everything is "homomorphic to" everything else.) – Brad Aug 28 '12 at 12:54
up vote 3 down vote accepted

If you are using ordinary algebras, and you are not assuming that your homomorphisms have to preserve the identity of the ring, then there is always the map $a\mapsto (a,0)$ that injects $A$ into $A\times B$. However, I'm guessing you want your maps to preserve identity, since otherwise zero homomorphisms would be everywhere.

If you want your map to preserve identity, then finding an example of $A$ that does not embed into $A\times B$ becomes really simple. Pick $A$ and $B$ to have two different coprime finite characteristics, say $7\cdot 1_A=0_A$ and $10\cdot 1_B=0_B$. Then the characteristic of the product ring is 70. But then if $\phi$ was an additive map preserving identity, $\phi(0)=\phi(7\cdot 1_A)=7\phi(1_A)=7\cdot 1_{A\times B}\neq 0$, a contradiction.

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You are right, sorry. So if I dont want the existence of such embedding I need to search e.g. for nilpotent algebras, is that right? On the same page there is stated that homomorphism $\boldsymbol{A}\to \boldsymbol{A}\times\boldsymbol{B}$ doesnt exist unless every element in $\boldsymbol{B}$ is idempotent for every operation (and algebras without idempotent elements are nilpotent, if I am correct). – mrImposs Aug 28 '12 at 12:24
@mrImposs Certainly 0 is (additively and multiplicatively) idempotent in any algebra. So, I'm not really sure what you mean by "algebras without idempotent elements". – rschwieb Aug 28 '12 at 12:28
@mrImposs A ring without identity may have no nonzero idempotents and yet not be nilpotent, so that statement doesn't make much sense to me either. Take, for example, any nontrivial ideal of an integral domain. – rschwieb Aug 28 '12 at 12:29
tnx for your answer! – mrImposs Aug 28 '12 at 12:35
@mrImposs Anytime! – rschwieb Aug 28 '12 at 12:37

Let $A$ and $B$ be algebras for the signature comprising just one unary operator $u$. Then a homomorphism from $A$ to $B$ is a function $f : A \rightarrow B$ that commutes with $u$, i.e., for all $a \in A$, $f(u(a)) = u(f(a))$. Now take $A = \{1, 2\}$ with $u$ the cyclic permutation $(1\;2)$ and take $B = \{1, 2, 3\}$ with $u$ the cyclic permutation $(1\;2\;3)$. Then $A \times B$ is isomorphic to $C = \{1, 2, 3, 4, 5, 6\}$ with $u$ the cyclic permutation $(1\;2\;3\;4\;5\;6)$ and there are no homomorphisms from $A$ to $C$.

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