On this 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.
On
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 $C$.
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.