Principal Ultrafilter implies Isomorphic Ultraproduct

Question

Let $\mathfrak{F}=\{X\subseteq \mathbb {N} \mid 17\in X \}$ (Note that $\mathfrak {F}$ is principal ultrafilter) and:

Let $\mathfrak{N}$ be the standard model for arithmatic and $\mathfrak{N}^*=\mathfrak{N}^{\mathbb{N}}/ \mathfrak{F}$

Can someone explain (in details) why $\mathfrak{N} \simeq \mathfrak{N}^*$?

Answer 1

1. There is a simple lemma saying that if F is an ultrafilter on I and $G \in F$, then $F|G$ is an ultrafilte on G, where $F|G = \{ X \cap G : X \in F \}$.
2. One proves then a bit more involved statement that $(\prod_{i\in I} A_i)/F \simeq (\prod_{i\in G} A_i)/(F|G)$.
3. The claim follows then taking $G = \{j\}$ - the 17th coordinate in your case. The right-hand-side in 2. becomes then isomorphic to $A_j$.