## 33.18 Variants of Noether normalization

Noether normalization is the statement that if $k$ is a field and $A$ is a finite type $k$ algebra of dimension $d$, then there exists a finite injective $k$-algebra homomorphism $k[x_1, \ldots , x_ d] \to A$. See Algebra, Lemma 10.115.4. Geometrically this means there is a finite surjective morphism $\mathop{\mathrm{Spec}}(A) \to \mathbf{A}^ d_ k$ over $\mathop{\mathrm{Spec}}(k)$.

Lemma 33.18.1. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ with image $s \in S$. Let $V \subset S$ be an affine open neighbourhood of $s$. If $f$ is locally of finite type and $\dim _ x(X_ s) = d$, then there exists an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ and a factorization

$U \xrightarrow {\pi } \mathbf{A}^ d_ V \to V$

of $f|_ U : U \to V$ such that $\pi$ is quasi-finite.

Proof. This follows from Algebra, Lemma 10.125.2. $\square$

Lemma 33.18.2. Let $f : X \to S$ be a finite type morphism of affine schemes. Let $s \in S$. If $\dim (X_ s) = d$, then there exists a factorization

$X \xrightarrow {\pi } \mathbf{A}^ d_ S \to S$

of $f$ such that the morphism $\pi _ s : X_ s \to \mathbf{A}^ d_{\kappa (s)}$ of fibres over $s$ is finite.

Proof. Write $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$ and let $A \to B$ be the ring map corresponding to $f$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $s$. We can choose a surjection $A[x_1, \ldots , x_ r] \to B$. By Algebra, Lemma 10.115.4 there exist elements $y_1, \ldots , y_ d \in A$ in the $\mathbf{Z}$-subalgebra of $A$ generated by $x_1, \ldots , x_ r$ such that the $A$-algebra homomorphism $A[t_1, \ldots , t_ d] \to B$ sending $t_ i$ to $y_ i$ induces a finite $\kappa (\mathfrak p)$-algebra homomorphism $\kappa (\mathfrak p)[t_1, \ldots , t_ d] \to B \otimes _ A \kappa (\mathfrak p)$. This proves the lemma. $\square$

Lemma 33.18.3. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. Let $V = \mathop{\mathrm{Spec}}(A)$ be an affine open neighbourhood of $f(x)$ in $S$. If $f$ is unramified at $x$, then there exist exists an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ such that we have a commutative diagram

$\xymatrix{ X \ar[d] & U \ar[l] \ar[rd] \ar[r]^-j & \mathop{\mathrm{Spec}}(A[t]_{g'}/(g)) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(A[t]) = \mathbf{A}^1_ V \ar[ld] \\ Y & & V \ar[ll] }$

where $j$ is an immersion, $g \in A[t]$ is a monic polynomial, and $g'$ is the derivative of $g$ with respect to $t$. If $f$ is étale at $x$, then we may choose the diagram such that $j$ is an open immersion.

Proof. The unramified case is a translation of Algebra, Proposition 10.152.1. In the étale case this is a translation of Algebra, Proposition 10.144.4 or equivalently it follows from Morphisms, Lemma 29.36.14 although the statements differ slightly. $\square$

Lemma 33.18.4. Let $f : X \to S$ be a finite type morphism of affine schemes. Let $x \in X$ with image $s \in S$. Let

$r = \dim _{\kappa (x)} \Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x) = \dim _{\kappa (x)} \Omega _{X_ s/s, x} \otimes _{\mathcal{O}_{X_ s, x}} \kappa (x) = \dim _{\kappa (x)} T_{X/S, x}$

Then there exists a factorization

$X \xrightarrow {\pi } \mathbf{A}^ r_ S \to S$

of $f$ such that $\pi$ is unramified at $x$.

Proof. By Morphisms, Lemma 29.32.12 the first dimension is finite. The first equality follows as the restriction of $\Omega _{X/S}$ to the fibre is the module of differentials from Morphisms, Lemma 29.32.10. The last equality follows from Lemma 33.16.4. Thus we see that the statement makes sense.

To prove the lemma write $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$ and let $A \to B$ be the ring map corresponding to $f$. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $x$. Choose a surjection of $A$-algebras $A[x_1, \ldots , x_ t] \to B$. Since $\Omega _{B/A}$ is generated by $\text{d}x_1, \ldots , \text{d}x_ t$ we see that their images in $\Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x)$ generate this as a $\kappa (x)$-vector space. After renumbering we may assume that $\text{d}x_1, \ldots , \text{d}x_ r$ map to a basis of $\Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x)$. We claim that $P = A[x_1, \ldots , x_ r] \to B$ is unramified at $\mathfrak q$. To see this it suffices to show that $\Omega _{B/P, \mathfrak q} = 0$ (Algebra, Lemma 10.151.3). Note that $\Omega _{B/P}$ is the quotient of $\Omega _{B/A}$ by the submodule generated by $\text{d}x_1, \ldots , \text{d}x_ r$. Hence $\Omega _{B/P, \mathfrak q} \otimes _{B_\mathfrak q} \kappa (\mathfrak q) = 0$ by our choice of $x_1, \ldots , x_ r$. By Nakayama's lemma, more precisely Algebra, Lemma 10.20.1 part (2) which applies as $\Omega _{B/P}$ is finite (see reference above), we conclude that $\Omega _{B/P, \mathfrak q} = 0$. $\square$

Lemma 33.18.5. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ with image $s \in S$. Let $V \subset S$ be an affine open neighbourhood of $s$. If $f$ is locally of finite type and

$r = \dim _{\kappa (x)} \Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x) = \dim _{\kappa (x)} \Omega _{X_ s/s, x} \otimes _{\mathcal{O}_{X_ s, x}} \kappa (x) = \dim _{\kappa (x)} T_{X/S, x}$

then there exist

1. an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ and a factorization

$U \xrightarrow {j} \mathbf{A}^{r + 1}_ V \to V$

of $f|_ U$ such that $j$ is an immersion, or

2. an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ and a factorization

$U \xrightarrow {j} D \to V$

of $f|_ U$ such that $j$ is a closed immersion and $D \to V$ is smooth of relative dimension $r$.

Proof. Pick any affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$. Apply Lemma 33.18.4 to $U \to V$ to get $U \to \mathbf{A}^ r_ V \to V$ as in the statement of that lemma. By Lemma 33.18.3 we get a factorization

$U \xrightarrow {j} D \xrightarrow {j'} \mathbf{A}^{r + 1}_ V \xrightarrow {p} \mathbf{A}^ r_ V \to V$

where $j$ and $j'$ are immersions, $p$ is the projection, and $p \circ j'$ is standard étale. Thus we see in particular that (1) and (2) hold. $\square$

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