Ultrafilter problem

Question

could you help me with this problem, please?

If $U$ is a principal ultrafilter on $I$ such that $\{a\}\in U$. Show that $Ult(\mathfrak{A}_x:x\in I)$ is isomorphic to $\mathfrak{A}_a$ and $[f]=f(a)$ for each $f$ and $j$, the embedding is the identity.

Answer 1

HINT: Recall that $Ult\models\varphi([f_1],\ldots[f_n])$ if and only if $$\{i\in I\mid\mathcal A_i\models\varphi(f_1(i),\ldots,f_n(i))\}\in U.$$

Answer 2

Hint: Recall that given $f , g \in \prod_{i \in I} A_i$, by definition we have that $f \sim_U g$ (that is, $f$ and $g$ are representatives of the same element of $\mathrm{Ult}_U ( \mathfrak{A}_i : i \in I )$) iff $\{ i \in I : f(i) = g(i) \} \in U$. In the case of a principal ultrafilter, you should be able to simplify this condition.