On definition of fibers in ring theory: why one does not need to consider taking radical?

Let $\varphi: X\to Y$ be a dominant morphism of affine varieties over algebraically closed field $k$, $\varphi*: k[Y]\to k[X]$ be the induced $k$-algebra monomorphism. Let $\mathfrak{m}_y$ be the ideal of a point $y\in Y$. I believe the ideal $I(\varphi^{-1}(y))$ of the fiber $\varphi^{-1}(y)$ over $y$ is exactly the radical $\sqrt{k[X]\varphi^*(\mathfrak{m}_y)}$ of the extension of the maximal ideal $\mathfrak{m}_y$.

Based on the observation above, I would like to define the fiber over the maximal ideal $\mathfrak{m}\subseteq A$ to be $B/(\sqrt{B\mathfrak{m}})$ for the algebra extension $A\subseteq B$. However, Eisenbud's commutative algebra gives the correct definition as $B/(B\mathfrak{m})$, without taking into consideration the case where $B\mathfrak{m}$ does not equal to its radical (which can be shown to be possible given examples). Could anyone explain to me why we need not consider that? Thanks in advance.