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No. In the Gaussian integers, chooseIndeed, substituting $i=-2$ gives $\mathbb{Z}[i]/(2+i)\cong \mathbb{Z}/5\mathbb{Z}$, so $P$ is a prime ideal and its contraction is $(5)$. Put $\beta =2-i$. Modulo $P$, $\beta$ becomes $4\ne 0$ in ${\mathbb{F}}_5$, so $\beta \notin P$. ThereforeWe show that $\alpha$ is not an integral element over ${\mathbb{Z}}_{(5)}$, using a direct reduction argument. Write $B={\mathbb{Z}}_{(5)}[i]$, viewed inside $\mathbb{Q}(i)$. There is a ring homomorphismwhich reduces rational coefficients modulo $5$. It is well defined because denominators in ${\mathbb{Z}}_{(5)}$ are not divisible by $5$, and $2^2+1=0$ in ${\mathbb{F}}_5$. In particular, $\rho (\beta )=0$ and $\rho (1)=1$.

If $\alpha$ satisfied a monic equation over ${\mathbb{Z}}_{(5)}$, saymultiplication by ${\beta }^n$ would give an equality in $B$:Applying $\rho$ gives $1=0$ in ${\mathbb{F}}_5$, a contradiction. ConsequentlyThis exhibits why localization of an integral extension at a prime need not be integral. Localizing both rings by the same multiplicative subset preserves an integral extension, but $\mathbb{Z}[i{]}_P$ inverts additional elements, including $2-i$, whose image at the other prime ideal $(2-i)$ is zero. The original extension is integral because $i$ satisfies the monic equation $T^2+1=0$: the subalgebra generated by finitely many integral elements property applies to $\mathbb{Z}[i]$.