Show every algebra can be embedded into each of it's ultrapowers Show every algebra can be embedded into each of it's ultrapowers.
Let $I$ be any set and and consider $\mathcal{P}(I)$. Let $U$ be an ultrafilter over $I$. Then, we define the ultrapower for some algebra $A$ to be $\prod_{i \in I} A_i/U$ where for all $i \in I, A_i \simeq A$. 
And, $a/U = b/U \iff  [[a=b]]$ where $[[a=b]] = \{ i \in I : a_i = b_i \}$.
So if I understand correctly then for any $U$ : ultrafilter over any $I$, the above is an ultrapower of $A$. So need to show every possible scenario gives an embedding of $A$ into $\prod_{i \in I} A_i/U$.  
Seems like all we need is a map that such that $a \in A \mapsto (a, a,a, ...) \in \prod_{i \in I }A_i$. Then modding out by $U$ gives $a=b$ iff $(a,a,...) = (b,b,...)$ which is to say $[[a=b]] = I \,\text{or} \,\emptyset$.
Is this the correct intuition?
 A: Yes, the map $$h: a\mapsto [(a,a,a,...)]$$ is an embedding of $A$ into $\prod A/\mathcal{U}$. However, it does take a slight bit of work to show this. What we need to do is show that $h$ is 


*

*injective,


and


*

*a homomorphism.


Injectivity is easy: $(a,a ,a,...)\equiv(b, b, b,...)\iff \{i\in I: a=b\}\in\mathcal{U}$. Now the set $\{i\in I: a=b\}$ is either $\emptyset$ (if $a\not=b$) or $I$ (if $a=b$), and we know $\emptyset\not\in\mathcal{U}$; so if $a\not=b$ then $h(a)\not=h(b)$.
To show that $h$ is a homomorphism is equally easy: by definition, $$f(h(a_1), . . ., h(a_n))=f([(a_1, a_1, . . .)], ..., [(a_n, a_n, . . .)])=[(f(a_1, . . . , a_n), f(a_1, . . . , a_n), ...)]=h(f(a_1, . . ., a_n))$$ for each function $f$ in the signature of the algebra. (Note that the second equality holds because of how we interpret $\prod A/\mathcal{U}$ as an algebra.)

Interestingly, the map $h$ is much better than just an embedding of algebras. First, having relation symbols kicking around doesn't wreck things - we can take the ultrapower of an arbitrary first-order structure, and $h$ still defines an embedding. Second, this embedding is in fact an elementary embedding: suppose $\varphi(a_1, . . ., a_n)$ is a sentence in the relevant language with parameters $a_i\in A$. Then $\varphi(a_1, . . ., a_n)$ is true about $A$ iff $\varphi(h(a_1), . . . , h(a_n))$ is true about $\prod A/\mathcal{U}$. This is essentially the content of Los' theorem.
