Hartshorne exercise II.3.22 (c); Dimension of Fibers Let $f:X \to Y$ be a dominant morphism of integral schemes of finite type over a field. Show there is an open dense subset $U  \subseteq X$ s.t. $\dim {U _y} = \dim X - \dim Y$ for all $y$ in the image of $U$. 
Been working on this for awhile now and can't figure it out. 
 A: Here is a proof using only the Noether normalization lemma. Let me know if something is unclear!
Proof. Choose an affine open $\operatorname{Spec} A \subseteq Y$ and an affine open $\operatorname{Spec} B \subseteq X$ such that $f(\operatorname{Spec}B) \subseteq \operatorname{Spec} A$. Then, since $f\rvert_{\operatorname{Spec}B}$ is dominant, the ring homomorphism $A \hookrightarrow B$ is a finite type extension. Letting $S = A \smallsetminus \{0\}$, we also have that $S^{-1}A \hookrightarrow S^{-1}B$ is of finite type. Then, by the Noether normalization lemma, we have a factorization
$$S^{-1}A \hookrightarrow S^{-1}A[x_1,x_2,\ldots,x_n] \hookrightarrow S^{-1}B$$
where the $x_i$ in the second ring are algebraically independent, and the second inclusion is integral.
Now let $z_1,z_2,\ldots,z_m$ generate $B$ as an $A$-algebra. Then, each $z_j$ (as an element of $S^{-1}B$) is integral over $S^{-1}A[x_1,x_2,\ldots,x_n]$. Now let $s \in A$ be the product of all denominators appearing in the integral polynomials that the $z_j$ satisfy, and the denominators appearing in the $x_i$. Then, letting $y_i = sx_i \in B$, we have a factorization
$$A_s \hookrightarrow A_s[y_1,y_2,\ldots,y_n] \hookrightarrow B_s$$
where the second arrow is integral, since $B_s$ is generated by the $sz_j$ over $A_s$.
Letting $U = \operatorname{Spec} B_s$ and $V = \operatorname{Spec} A_s$, this sequence of ring homomorphisms corresponds to the sequence
$$U \to \mathbf{A}^n_V \to V$$
of morphisms of schemes whose composition is $f\rvert_U$, and such that the first morphism is integral (in fact, it is finite and surjective, since it is integral, of finite type, and dominant). For all $v \in V$, we have that
$$\dim U_v = \dim \mathbf{A}^n_{\kappa(v)} = n = \dim \mathbf{A}^n_V - \dim V = \dim U - \dim V,$$
where the first and last equalities are by the fact that integral ring extensions preserve dimension (this follows from Going Up). Finally, since both $X$ and $Y$ are integral and of finite type over $k$, we have that $\dim X = \dim U$ and $\dim Y = \dim V$ by Prop. I.1.10. $\blacksquare$
We actually showed a stronger version of the Noether normalization lemma in the course of the proof, which we state scheme-theoretically:
Relative Noether normalization lemma (see [Atiyah–Macdonald, Exer. 5.20]). Let $f\colon X \to Y$ be a dominant, finite type morphism of schemes, where $Y$ is integral. Then, there exists a non-empty open subset $V \subseteq Y$, an open subset $U \subseteq X$ mapping to $V$, and a factorization
$$U \to \mathbf{A}^n_V \to V$$
of $f\rvert_{U}$ where the first morphism is finite and surjective.
With this version of the Noether normalization lemma, setting $U = f^{-1}(V)$ immediately gives the statement you are looking for. One can also prove this stronger version of the Noether normalization lemma directly, without appealing to the version when $Y = \operatorname{Spec} k$; see [Hochster].
