T.A springer lemma on dominant morphisms between irreducible varieties I am trying to understand the following lemma in T.A springer's linear algebraic groups.

My problem is with understanding the last sentence shown above.
Firstly why should the fact that the ideal in the above argument goes to zero imply that the algebra of regular functions on the fiber is K[T]?
Is it because the kernel of the map in question is the ideal of regular functions vanishing on the fiber? If so how do we know the function b does not vanish on the fiber as well?
secondly why should the isomorphism of said algebra to K[T] imply the fiber is isomorphic to the affine line? shouldn't this line of reasoning only apply in the case of an affine variety?
 A: OK, first of all there must be some missing ambient assumptions. Are $X$ and $Y$ assumed to be reduced? For example, the map $\mathrm{Spec}(k)\to\mathrm{Spec}(k[\varepsilon]/(\varepsilon^2))$ is dominant (it's a homeomorphism) map of irreducible varieties but certainly $k[\varepsilon]/(\varepsilon^2)$ does not embed into $k$ as $k$-algebras. So, let's assume that everything is reduced. Then, it is true that the induced map
$$A:=k[Y]\to k[X]=:B$$
is injective.
If we now assume that $B=A[b]$ then note that for any prime ideal $\mathfrak{p}$ of $A$ one has that
$$\begin{aligned}\phi^{-1}(\mathfrak{p}) &=\mathrm{Spec}(B\otimes_A k(\mathfrak{p}))\\ &= \mathrm{Spec}(A[b]\otimes_A k(\mathfrak{p}))\\ &= \mathrm{Spec}(k(\mathfrak{p})[T]/\overline{I_b})\end{aligned}$$
where $k(\mathfrak{p}):=A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is the residue field of $A$ at $\mathfrak{p}$, $I_b$ is the kernel of the $A$-algebra map $A[t]\to B$ given by $t\mapsto b$, and $\overline{I_b}$ is the image of $I_b$ under the quotient map $A[t]\to k(\mathfrak{p})[T]$.
From this we see that there is a closed embedding
$$\phi^{-1}(\mathfrak{p})\hookrightarrow \mathbb{A}^1_{k(\mathfrak{p})}$$
which clearly implies, thinking about the closed subschemes of $\mathbb{A}^1_{k(\mathfrak{p})}$, that $\phi^{-1}(\mathfrak{p})$ is either finite or isomorphic to $\mathbb{A}^1_{k(\mathfrak{p})}$.
