How split prime ideal $(2)$ in the particular splitting field of the polynomial

Question

Now, Let $$ f(x)=x^3+x^2−2x+8 $$ and, $K$ be the the splitting field of $f(x), O_K$ be the ring of integers of it.then $$(2)=P_1P_2P_3\iff f(x) \text{ has three roots in } \mathbb{Q}_2$$ Of course $P_i$ means different prime ideal of $O_K$. I don't understand this equivalent and where they appear.

Answer 1

For number field $K/\mathbb{Q}$, if $p$ decomposes in $\mathcal{O}_K$ as $(p)=\mathfrak{p}_1^{e_1}\cdots \mathfrak{p}_g^{e_g}$, then $$\begin{aligned}\mathcal{O}_K \otimes_\mathbb{Z} \mathbb{Z}_p &= \lim_{\longleftarrow} \mathcal{O}_K \otimes_\mathbb{Z} \frac{\mathbb{Z}}{p^n\mathbb{Z}}\\ &=\lim_{\longleftarrow} \frac{\mathcal{O}_K}{p^n\mathcal{O}_K} = \lim_{\longleftarrow} \frac{\mathcal{O}_K}{\mathfrak{p}_1^{ne_1}\mathcal{O}_K}\times\cdots\times\frac{\mathcal{O}_K}{\mathfrak{p}_g^{ne_g}\mathcal{O}_K} \\&= \mathcal{O}_{K,\mathfrak{p}_1}\times \cdots \times \mathcal{O}_{K,\mathfrak{p}_g}\end{aligned}$$ Tensoring both sides by $\mathbb{Q}$ gives $K\otimes_\mathbb{Q} \mathbb{Q}_p = K_{\mathfrak{p}_1}\times\cdots\times K_{\mathfrak{p}_g}$, where $K_{\mathfrak{p}}$ means completion at that prime. If $K$ is defined by $f\in \mathbb{Q}[x]$, then $$K\otimes_\mathbb{Q} \mathbb{Q}_p = \frac{\mathbb{Q}[x]}{(f)}\otimes_\mathbb{Q} \mathbb{Q}_p = \mathbb{Q}_p[x]/(f)$$ RHS is a product of a number of fields, which equals the number of irreducible factors of $f$ in $\mathbb{Q}_p$. By the isomorphism above, equals the number of primes lying above $p$. Your assertion follows.