There is an exercise in Burris and Sankappanavar's "A Course in Universal Algebra":

Problem: Find two algebras $\mathbf{A}_1$, $\mathbf{A}_2$ such that neither can be embedded in $\mathbf{A}_1 \times \mathbf{A}_2$.

My solution: Consider $\mathbb{Z}_2$ and $\mathbb{Z}_3$ as a rings with unity (in $\langle +, \cdot, -, 0, 1 \rangle$ signature).

Assume $\varphi \colon \mathbb{Z}_2 \to \mathbb{Z}_2 \times \mathbb{Z}_3$ is such embedding.

Hence $(0, 0) = \varphi(0) = \varphi(1 + 1) = \varphi(1) + \varphi(1) = (1, 1) + (1, 1) = (0, 2)$. So we get a contradiction. Same with $\mathbb{Z}_3$, we should consider $\varphi(0) = \varphi(1 + 1 + 1)$.

Question 0: Is my example correct?

Main question: Is there a special name for this embedding property (or more general: for any family of algebras $(\mathbf{A}_i, i \in I)$ every $\mathbf{A}_i$ can be embedded in $\Pi_{i \in I} \mathbf{A}_i$) and results or works concerning classes of algebras satisfying it? For example, this property implies the Weak Amalgamation Property and so on. I'm interested in deeper results. Thank you in advance!

  • $\begingroup$ Your example is correct, given that ring maps send $1$ to $1$ $\endgroup$ – zcn Jul 27 '14 at 20:58
  • $\begingroup$ Thank you, @zcn! I've added the signature to be more precise. $\endgroup$ – Random Jack Jul 27 '14 at 21:18

The following is hardly deep, but it's another characterization of the property you're interested in.

Claim: Let $(A_i)_{i\in I}$ be a family of algebras. For each $i$, the following are equivalent:

  1. $A_i$ embeds in $\prod_{j\in I}A_j$.
  2. There is a homomorphism (not necessarily an embedding) $A_i\rightarrow \prod_{j\in I} A_j$.
  3. For all $j\in I$, there is a homomorphism $f_j: A_i\rightarrow A_j$.

Proof: (1 $\rightarrow$ 2) An embedding is a homomorphism.

(2 $\rightarrow$ 3) Let $g:A_i\rightarrow \prod_{j\in I}A_j$ be a homomorphism. For each $j$, let $\pi_j:\prod_{j\in I} A_j$ be the projection on the $j^\text{th}$ coordinate. Then $\pi_j\circ g$ is a homomorphism $A_i\rightarrow A_j$.

(3 $\rightarrow$ 1) Consider the map $g:A_i\rightarrow \prod_{j\in I}A_j$ given by $f_j$ on the $j^\text{th}$ coordinate for all $j\neq i$ and $\text{id}_{A_i}$ on the $i^\text{th}$ coordinate. $g$ is an embedding, since if $a\neq b$ in $A_i$, $g(a)$ and $g(b)$ differ on the $i^\text{th}$ coordinate.

Edit: After a bit more thought, I realized that something much stronger than what I had originally written below the break is true. I'm sure this is well-known by the universal algebraists, but it hadn't occurred to me before.

Let $V$ be a variety of algebras in a signature $L$. We denote the initial algebra by $0$ and the terminal algebra by $1$.

Claim: The following are equivalent:

  1. For every family $(A_j)_{j\in I}$ and every $i\in I$, there is an embedding $A_i \hookrightarrow \prod_{j\in I} A_j$.
  2. For every $A,B\in V$, there is a homomorphism $A\rightarrow B$.
  3. There is a homomorphism $1\rightarrow 0$.
  4. $1\cong 0$.
  5. There is exactly one constant, $c\in L$ up to provable equality (i.e. for all constants $d\in L$, the equation $d = c$ holds in $V$), and for all operations $f(x_1,\dots,x_n)\in L$, the equation $f(c,\dots,c) = c$ holds in $V$.

Note that if you're willing to view constants as $0$-ary operations, we can view the condition $d = c$ for all constants $d$ as a special case of the equation $f(c,\dots,c) = c$ for all operations $f$.

Proof: (1 $\leftrightarrow$ 2) Assuming (1), Given $A$ and $B$, $A$ embeds in the product $A\times B$, so by the previous part there is a homomorphism $A\rightarrow B$. Conversely, given $A_i$ and $\prod_{j\in I}A_j$, by (2) there is a homomorphism $A_i\rightarrow A_j$ for all $j\in I$, so by the previous part $A_i$ embeds in the product.

(2 $\leftrightarrow$ 3) Note that (3) is a special case of (2). Conversely, given $A$ and $B$, we compose the homomorphism $1\rightarrow 0$ with the unique maps into $1$ and out of $0$: $A\rightarrow 1 \rightarrow 0 \rightarrow B$.

(3 $\rightarrow$ 5) If there are no constants in $L$, then $0$ is empty, so there is no function $1\rightarrow 0$. Pick a constant $c\in L$. Then for every constant $d\in L$ and every function $f\in L$, the equations $c = d$ and $f(c,\dots,c) = c$ hold in $1$, so they must hold in $0$ (since $1$ maps to $0$, sending $c$ to $c$), and hence they hold everywhere in $V$ (since $0$ maps to everything, sending $c$ to $c$).

(5 $\rightarrow$ 4) The initial algebra $0$ can be constructed as the set of all $L$-terms without any variables, modulo provable equivalence. By the equations assumed to hold in $V$, every term is provably equal to $c$, and hence $0$ has exactly one element, and the obvious map $1\rightarrow 0$ is an isomorphism.

(4 $\rightarrow$ 3) This is clear.

When these conditions hold, we call the objects which are both initial and terminal "zero objects". Then, strengthening 2, for any $A,B\in V$, there is a canonical zero homomorphism $A\rightarrow B$, the unique map which factors as $A\rightarrow 0\rightarrow B$. And strengthening 1, there is a canonical embedding $A_i\rightarrow \prod_{j\in I} A_j$, which is the identity map on the $i^\text{th}$ coordinate and the zero map on every other coordinate.

This is what happens for example, in the case of groups, monoids, semilattices, rngs (rings without unit), etc.

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  • $\begingroup$ Thank you, @Alex, for such a detailed answer! Interesting characterization. I've thought about property 5, but only as of sufficient condition. I also like the point about initial and terminal objects. After such a characterization I think this property is not of a big interest. $\endgroup$ – Random Jack Jul 29 '14 at 7:07

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