Subdirect embedding of a quotient algebra If $A$ is an algebra and $\theta_i$ $\in$ $Con(A)$, i $\in$ $I$, let $\theta$ = $\cap \theta_i$. Show that $A/\theta$ can be subdirectly embedded in $\prod$$A/\theta_i$.
What intuitively I think of the homomorphism that would give me the embedding is
$h:$ $A/\theta$ $\rightarrow$ $\prod$$A/\theta_i$
$h(a/\theta)(i) = a/\theta_i$
But I don't know how to proceed and what all I need to show. 
 A: Okay, since my previous comment didn't provide enough of a hint, and since hopefully by now you have struggled with the problem for a while, I'll give you a complete outline of the proof.  However, you should fill in the details.
Define $h : A \rightarrow \prod A/\theta_i$ by $h(a) = (a/\theta_i)_i$.
By $(a/\theta_i)_i$ I mean the tuple that has $a/\theta_i$ in the $i$-th "coordinate."
Now consider the kernel of the function $h$.
Since some people have learned a definition of "kernel" which is not appropriate in this context, let me be explicit: the kernel of $h$ is the following congruence relation of $A$:
$$\operatorname{ker} h = \{(x, y) \in A\times A \mid h(x) = h(y)\}.$$
Check that, for the map $h$ defined above, $\operatorname{ker} h = \bigcap \theta_i$.
Finally, note that $h$ is a homomorphism from $A$ to $\prod A/\theta_i$, so you can apply the first isomorphism theorem: if $h$ is a homomorphism on the domain $A$, then $A/\operatorname{ker}h \cong h[A]$.
Finally, note that the image $h[A] := \{h(a) \mid a \in A\}$ is "subdirect"; that is, for each $j$, the projection $\pi_j : h[A] \to A/\theta_j$ (which maps $(a/\theta_i)_i$ to $a/\theta_j$) is surjective.
If you are still having trouble understanding this exercise, I suggest the following:

*

*A Course in Universal Algebra (Ch. II, Section 8)


*Willard's Overview of Universal Algebra (Section 3)
