Noether Normalization and rational points over finite fields Let $k$ be a finite field with $q$ elements, $U$ an integral scheme of finite type over $k$ and $f: U \to \mathbb{A}_k^d$ a finite dominant morphism (obtained by Noether normalization, $d=\dim U$). If $K$ is an extension of $k$ of degree $n$, how exactly does one get the inequality $|U(K)| \leq deg(f)\cdot q^{n\cdot d}$?
This seems intuitive to me but I failed trying to explain it to myself. 
$U(K)$ is the set of $K$-rational points of $U$.
Thanks in advance for your help!
 A: As noted in the comments (and, as I understand, what caused the OP to post this question in the first place), there is a bit of a problem with the obvious approach of bounding the number of $K$-points in the fibres.
However, for the real question you're interested in (the proof of Lemma 1.4 of these notes), we're allowed to pass to a dense open by the same induction argument. Thus, by generic freeness (Tag 051S), we can restrict to an open $V \subseteq \mathbb A^n_k$ such that $f \colon f^{-1}(V) \to V$ is finite flat. For such morphisms, the (scheme-theoretic) degree is constant, thus
$$|f^{-1}(V)(K)| \leq \deg f \cdot |V(K)| \leq \deg f \cdot q^{nd}.$$
The same induction now applies.
Remark. This is a little unsatisfying because it does not answer the literal question that the OP posted. I would be interested to know if there exists a counterexample, or whether the result is somehow true regardless of the failure of the argument suggested above.
A: This is an answer to the question posed in Remy's answer (I don't have the points to leave a comment). If you have a finite (or quasi-finite) dominant morphism $f : X \to Y$ of varieties over a field $k$ and $Y$ is normal, then each fibre $f^{-1}(y)$ has at most $\deg(f)$ points, where $\deg(f)$ is defined as the degree of the extension $k(Y) \subset k(X)$ of function fields.
In the case of a finite morphism $f : X \to \mathbf{A}^d_k$ with $k$ infinite and $y$ a $k$-rational point you can prove it as follows. Choose a general line $L \subset \mathbf{A}^d_k$ through $y$. Then $L$ meets the locus where $f$ is flat hence $C = f^{-1}(L) \to L$ is generically flat of the same degree. By going down for integral over normal, we see that every  point of $f^{-1}(y)$ is the specialization of a point of $C$ mapping to the generic point of $L$. Hence if $C' \subset C$ is the closed subscheme cut out by the torsion (sections supported in dim $0$) in $\mathcal{O}_C$, then $C' \to L$ and $C \to L$ have the same set of points over $y$. OK, and now the morphism $C' \to L$ is flat (as $L$ is a smooth curve) and this is the case you professed to be happy with. (Warning: $C'$ need not be irreducible and need not even be reduced, but it doesn't matter for the argument --- you can replace $C'$ by its reduction and even by its normalization which would only increase the number of points over $y$.)
If $k$ is finite, then you have to replace the line in the argument above by a higher degree curve to make sure it meets the flat locus.
This argument doesn't prove the general normal case because we don't know we can find a smooth curve through every point. To prove it in general I suggest using completions.
