Lemma 115.5.1. Let $R$ be a ring and let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. If $t$ is integral over $R[x]$, then there exists an $\ell \geq 0$ such that for every $a \in R$ the element $\varphi (a)^\ell t$ is integral over $\varphi _ a : R[y] \to S$, defined by $y \mapsto \varphi (ax)$ and $r \mapsto \varphi (r)$ for $r\in R$.
115.5 Lemmas related to ZMT
The lemmas in this section were originally used in the proof of the (algebraic version of) Zariski's Main Theorem, Algebra, Theorem 10.123.12.
Proof. Say $t^ d + \sum _{i < d} \varphi (f_ i)t^ i = 0$ with $f_ i \in R[x]$. Let $\ell $ be the maximum degree in $x$ of all the $f_ i$. Multiply the equation by $\varphi (a)^\ell $ to get $\varphi (a)^\ell t^ d + \sum _{i < d} \varphi (a^\ell f_ i)t^ i = 0$. Note that each $\varphi (a^\ell f_ i)$ is in the image of $\varphi _ a$. The result follows from Algebra, Lemma 10.123.1. $\square$
Lemma 115.5.2. Let $\varphi : R \to S$ be a ring map. Suppose $t \in S$ satisfies the relation $\varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_ n) t^ n = 0$. Set $u_ n = \varphi (a_ n)$, $u_{n-1} = u_ n t + \varphi (a_{n-1})$, and so on till $u_1 = u_2 t + \varphi (a_1)$. Then all of $u_ n, u_{n-1}, \ldots , u_1$ and $u_ nt, u_{n-1}t, \ldots , u_1t$ are integral over $R$, and the ideals $(\varphi (a_0), \ldots , \varphi (a_ n))$ and $(u_ n, \ldots , u_1)$ of $S$ are equal.
Proof. We prove this by induction on $n$. As $u_ n = \varphi (a_ n)$ we conclude from Algebra, Lemma 10.123.1 that $u_ nt$ is integral over $R$. Of course $u_ n = \varphi (a_ n)$ is integral over $R$. Then $u_{n - 1} = u_ n t + \varphi (a_{n - 1})$ is integral over $R$ (see Algebra, Lemma 10.36.7) and we have
\[ \varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_{n - 1})t^{n - 1} + u_{n - 1}t^{n - 1} = 0. \]
Hence by the induction hypothesis applied to the map $S' \to S$ where $S'$ is the integral closure of $R$ in $S$ and the displayed equation we see that $u_{n-1}, \ldots , u_1$ and $u_{n-1}t, \ldots , u_1t$ are all in $S'$ too. The statement on the ideals is immediate from the shape of the elements and the fact that $u_1t + \varphi (a_0) = 0$. $\square$
Lemma 115.5.3. Let $\varphi : R \to S$ be a ring map. Suppose $t \in S$ satisfies the relation $\varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_ n) t^ n = 0$. Let $J \subset S$ be an ideal such that for at least one $i$ we have $\varphi (a_ i) \not\in J$. Then there exists a $u \in S$, $u \not\in J$ such that both $u$ and $ut$ are integral over $R$.
Proof. This is immediate from Lemma 115.5.2 since one of the elements $u_ i$ will not be in $J$. $\square$
The following two lemmas are a way of describing closed subschemes of $\mathbf{P}^1_ R$ cut out by one (nondegenerate) equation.
Lemma 115.5.4. Let $R$ be a ring. Let $F(X, Y) \in R[X, Y]$ be homogeneous of degree $d$. Assume that for every prime $\mathfrak p$ of $R$ at least one coefficient of $F$ is not in $\mathfrak p$. Let $S = R[X, Y]/(F)$ as a graded ring. Then for all $n \geq d$ the $R$-module $S_ n$ is finite locally free of rank $d$.
Proof. The $R$-module $S_ n$ has a presentation
\[ R[X, Y]_{n-d} \to R[X, Y]_ n \to S_ n \to 0. \]
Thus by Algebra, Lemma 10.79.4 it is enough to show that multiplication by $F$ induces an injective map $\kappa (\mathfrak p)[X, Y] \to \kappa (\mathfrak p)[X, Y]$ for all primes $\mathfrak p$. This is clear from the assumption that $F$ does not map to the zero polynomial mod $\mathfrak p$. The assertion on ranks is clear from this as well. $\square$
Lemma 115.5.5. Let $k$ be a field. Let $F, G \in k[X, Y]$ be homogeneous of degrees $d, e$. Assume $F, G$ relatively prime. Then multiplication by $G$ is injective on $S = k[X, Y]/(F)$.
Proof. This is one way to define “relatively prime”. If you have another definition, then you can show it is equivalent to this one. $\square$
Lemma 115.5.6. Let $R$ be a ring. Let $F(X, Y) \in R[X, Y]$ be homogeneous of degree $d$. Let $S = R[X, Y]/(F)$ as a graded ring. Let $\mathfrak p \subset R$ be a prime such that some coefficient of $F$ is not in $\mathfrak p$. There exists an $f \in R$ $f \not\in \mathfrak p$, an integer $e$, and a $G \in R[X, Y]_ e$ such that multiplication by $G$ induces isomorphisms $(S_ n)_ f \to (S_{n + e})_ f$ for all $n \geq d$.
Proof. During the course of the proof we may replace $R$ by $R_ f$ for $f\in R$, $f\not\in \mathfrak p$ (finitely often). As a first step we do such a replacement such that some coefficient of $F$ is invertible in $R$. In particular the modules $S_ n$ are now locally free of rank $d$ for $n \geq d$ by Lemma 115.5.4. Pick any $G \in R[X, Y]_ e$ such that the image of $G$ in $\kappa (\mathfrak p)[X, Y]$ is relatively prime to the image of $F(X, Y)$ (this is possible for some $e$). Apply Algebra, Lemma 10.79.4 to the map induced by multiplication by $G$ from $S_ d \to S_{d + e}$. By our choice of $G$ and Lemma 115.5.5 we see $S_ d \otimes \kappa (\mathfrak p) \to S_{d + e} \otimes \kappa (\mathfrak p)$ is bijective. Thus, after replacing $R$ by $R_ f$ for a suitable $f$ we may assume that $G : S_ d \to S_{d + e}$ is bijective. This in turn implies that the image of $G$ in $\kappa (\mathfrak p')[X, Y]$ is relatively prime to the image of $F$ for all primes $\mathfrak p'$ of $R$. And then by Algebra, Lemma 10.79.4 again we see that all the maps $G : S_ d \to S_{d + e}$, $n \geq d$ are isomorphisms. $\square$
Remark 115.5.7. Let $R$ be a ring. Suppose that we have $F \in R[X, Y]_ d$ and $G \in R[X, Y]_ e$ such that, setting $S = R[X, Y]/(F)$ we have (1) $S_ n$ is finite locally free of rank $d$ for all $n \geq d$, and (2) multiplication by $G$ defines isomorphisms $S_ n \to S_{n + e}$ for all $n \geq d$. In this case we may define a finite, locally free $R$-algebra $A$ as follows:
as an $R$-module $A = S_{ed}$, and
multiplication $A \times A \to A$ is given by the rule that $H_1 H_2 = H_3$ if and only if $G^ d H_3 = H_1 H_2$ in $S_{2ed}$.
This makes sense because multiplication by $G^ d$ induces a bijective map $S_{de} \to S_{2de}$. It is easy to see that this defines a ring structure. Note the confusing fact that the element $G^ d$ defines the unit element of the ring $A$.
Lemma 115.5.8. Let $R$ be a ring, let $f \in R$. Suppose we have $S$, $S'$ and the solid arrows forming the following commutative diagram of rings Assume that $R_ f \to S'$ is finite. Then we can find a finite ring map $R \to S''$ and dotted arrows as in the diagram such that $S' = (S'')_ f$.
\[ \xymatrix{ & S'' \ar@{-->}[rd] \ar@{-->}[dd] & \\ R \ar[rr] \ar@{-->}[ru] \ar[d] & & S \ar[d] \\ R_ f \ar[r] & S' \ar[r] & S_ f } \]
Proof. Namely, suppose that $S'$ is generated by $x_ i$ over $R_ f$, $i = 1, \ldots , w$. Let $P_ i(t) \in R_ f[t]$ be a monic polynomial such that $P_ i(x_ i) = 0$. Say $P_ i$ has degree $d_ i > 0$. Write $P_ i(t) = t^{d_ i} + \sum _{j < d_ i} (a_{ij}/f^ n) t^ j$ for some uniform $n$. Also write the image of $x_ i$ in $S_ f$ as $g_ i / f^ n$ for suitable $g_ i \in S$. Then we know that the element $\xi _ i = f^{nd_ i} g_ i^{d_ i} + \sum _{j < d_ i} f^{n(d_ i - j)} a_{ij} g_ i^ j$ of $S$ is killed by a power of $f$. Hence upon increasing $n$ to $n'$, which replaces $g_ i$ by $f^{n' - n}g_ i$ we may assume $\xi _ i = 0$. Then $S'$ is generated by the elements $f^ n x_ i$, each of which is a zero of the monic polynomial $Q_ i(t) = t^{d_ i} + \sum _{j < d_ i} f^{n(d_ i - j)} a_{ij} t^ j$ with coefficients in $R$. Also, by construction $Q_ i(f^ ng_ i) = 0$ in $S$. Thus we get a finite $R$-algebra $S'' = R[z_1, \ldots , z_ w]/(Q_1(z_1), \ldots , Q_ w(z_ w))$ which fits into a commutative diagram as above. The map $\alpha : S'' \to S$ maps $z_ i$ to $f^ ng_ i$ and the map $\beta : S'' \to S'$ maps $z_ i$ to $f^ nx_ i$. It may not yet be the case that $\beta $ induces an isomorphism $(S'')_ f \cong S'$. For the moment we only know that this map is surjective. The problem is that there could be elements $h/f^ n \in (S'')_ f$ which map to zero in $S'$ but are not zero. In this case $\beta (h)$ is an element of $S$ such that $f^ N \beta (h) = 0$ for some $N$. Thus $f^ N h$ is an element ot the ideal $J = \{ h \in S'' \mid \alpha (h) = 0 \text{ and } \beta (h) = 0\} $ of $S''$. OK, and it is easy to see that $S''/J$ does the job. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)