So this is a follow up question to an earlier question about finite-type, quasi-finite, generically finite morphisms of integral schemes. In fact, it was based on exercise 3.7 of chapter II of Hartshorne.
The setup for the question is as follows: I am given a finite-type, generically finite, dominant morphism $f: X \longrightarrow Y$ of integral schemes. I am asked to show that there is a dense open subset $U \subseteq Y$ such that the induced morphism $f^{-1}(U) \longrightarrow U$ is finite.
I am given a hint that I should first show that the function field of X is a finite extension of the function field of Y which was the subject of the question I referenced above. I am fairly confident that I have sorted that out and understand it now. I have made a solid attempt at the rest of the question and I was hoping someone could again go through this part and pick apart any flaws in my reasoning. I feel like I have a habit of overlooking subtleties. Any feedback would be greatly appreciated.
My attempt is as follows:
Let $\text{Spec }B \subseteq Y $ and let $\text{Spec }A \subset X$ such that $$ \text{Spec }A \subseteq f^{-1}({\text{Spec }B}) $$ Let $\kappa(\varepsilon)$ and $\kappa(\eta)$ be the field of fractions of $X$ and $Y$ respectively. We have that $A$ is finitely generated as a $B$-algebra. Let $\{ a_{1}, a_{2}, \ldots , a_{m} \}$ be a collection of generators for $A$ over $B$. Consider these generators as elements in $\kappa(\varepsilon)$ under the localization map: $$\left\lbrace \frac{a_{1}}{1}, \frac{a_{2}}{1} , \ldots , \frac{a_{m}}{1} \right\rbrace .$$ We will abuse notation and write these without the denominator when it is clear from context that we are considering them as elements of $\kappa(\varepsilon)$. Since any finite field extension is algebraic, we have that $\kappa(\varepsilon)$ is algebraic over $\kappa(\eta)$. Let $P_{i}$ denote the minimal polynomial for $a_{i}$ over $\kappa(\eta)$ so that $$ P_{i}\left( \frac{a_{i}}{1} \right) = 0 \qquad 1 \leq i \leq m $$ These are polynomials with coefficients in $\kappa(\eta)$, so they take the form $$ P_{i}\left( \frac{a_{i}}{1} \right) = a_{i}^{n_{i}} + \frac{b'_{n-1, i}}{b_{n-1, i}}a_{i}^{n_{i}-1} + \frac{b'_{n-2, i}}{b_{n-2, i}}a_{i}^{n_{i}-2} + \cdots + \frac{b'_{1,i}}{b_{1, i}}a_{i} + \frac{b'_{0,i}}{b_{0,i}} = 0 $$ If we define $b_{i}$ to be the product of all the denominators in $P_{i}$, then we can define the polynomials $$ b_{i}P_{i} \left( \frac{a_{i}}{1} \right) := Q_{i}(a_{i}) = 0 $$ with coefficients now in $B$ and leading coefficient $b_{i}$. Further still, define $b$ to be the product of all the (finitely many) leading coefficients $b_{i}$. Now in the localizations $B_{b}$ and $A_{b}$, the polynomials $Q_{i}$ are sent to monic (up to a factor of a unit) vanishing polynomials. So we have shown that the finite-type morphism of rings $$ \phi_{b}: B_{b} \longrightarrow A_{b} $$ is integral, and since finite-type, integral morphisms are finite, we have that this is finite. We know also that finite morphisms are affine local on the target, and we know that $\text{Spec } B_{b}$ and $\text{Spec } A_{b}$ are dense open sets (since $X$ and $Y$ are irreducible). Therefore, we can recover our desired finite morphism as required.
I should add that there are two points I am a little uncomfortable with. The first: Is it true that in the localization $A_{b}$ and $B_{b}$ that every element will be integral over $A_{b}$? I think that the generators (as an algebra) being integral is sufficient for this, right? Secondly: Is my very last comment, that we can find a morphism $f^{-1}(U) \longrightarrow U$ because of affine localness of finite morphisms actually a valid deduction?