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Krull's principal ideal theorem states that

$A$ is a Noetherian ring and (x) is a principal, proper ideal of $A$, then each minimal prime ideal over (x) has height at most one.

According to proof which is available on Wikipedia, it starts in the following way-

Let $A$ be Noetherian ring, x be an element of it and $\mathfrak p$ be minimal prime over x. Replacing $A$ by localization $A_{\mathfrak p}$.We can assume that $A$ is local with maximal ideal $\mathfrak p$.....

Now, I know that $A_{\mathfrak p}$ is local with maximal as well as prime ideal $\mathfrak pA_{\mathfrak p}$. But I'm not able to understand the part in the proof where it says we can assume that $A$ is local ring with maximal ideal $\mathfrak p$.

Can someone please tell me how we can assume that $A$ is local ring with maximal ideal $\mathfrak p$?

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In general, if $R$ is a ring and $U$ is a multiplicative set in $R$, then the prime ideals of the localization $U^{-1}R$ are in one-to-one correspondence with the prime ideals of $R$ which do not intersect $U$.

To be more precise, if $$T = \{\mathfrak{q} \in \text{Spec}(R): \mathfrak{q} \cap U = \emptyset \}$$ Then $$\text{Spec}(U^{-1}R) = \{U^{-1}\mathfrak{q}:\mathfrak{q}\in T \}$$ In particular, if $\mathfrak{p}$ is a prime ideal of $R$, then the prime ideals of $R_{\mathfrak{p}}$ are in one-to-one correspondence with the prime ideals of $R$ which are contained in $\mathfrak{p}$.

Therefore, the height of $\mathfrak{p}$ in $R$ is equal to the height of $\mathfrak{p}R_{\mathfrak{p}}$ in the localized ring $R_{\mathfrak{p}}$. So, showing that $\text{ht}(\mathfrak{p})\le 1$ is equivalent to showing that $\text{ht}(\mathfrak{p}R_{\mathfrak{p}})\le 1$.

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