2.1. If $I$ and $J$ are ideals of a ring $R$ , then the saturation of  $I$  with respect to  $J$ is $I:J^{\infty }=\bigcup _{i=1}^{\infty }(I:J^{i})$ . Recall that $I:J^{\infty }$ is obtained from $I$ by removing all primary components whose radical contain $J$ . We write $\text{gcd}$ to mean greatest common divisor. If $I$ is a homogeneous ideal in $k[x,y]$ , then we denote the $\text{gcd}$ of a generating set of $I$ by $\text{gcd}(I)$ ; notice that this polynomial generates the saturation $I:(x,y)^{\infty }$ .

2.2. If $R$ is a ring, then we write $\operatorname{Quot}(R)$ for the total ring of quotients of $R$ ; that is,

$$\begin{eqnarray}\operatorname{Quot}(R)=U^{-1}R,\end{eqnarray}$$

where $U$ is the set of nonzerodivisors on $R$ . If $R$ is a domain, then the total ring of quotients of $R$ is usually called the quotient field of $R$ . An extension of domains $A\subset B$ is called birational if $A$ and $B$ have the same quotient field.

If $A\subset B$ is an extension of domains so that $\operatorname{Quot}(B)$ is algebraic over $\operatorname{Quot}(A)$ , then we use the three notations

$$\begin{eqnarray}[B:A],\quad \operatorname{rank}_{A}B,\quad \text{and}\quad [\operatorname{Quot}(B):\operatorname{Quot}(A)]\end{eqnarray}$$

interchangeably. Indeed, in this situation,

(2.2.1) $$\begin{eqnarray}\operatorname{Quot}(A)\otimes _{A}B=\operatorname{Quot}(B).\end{eqnarray}$$

The rank of $B$ as an $A$ -module, denoted $\operatorname{rank}_{A}B$ , is defined to be the dimension of the left hand side of (2.2.1) as a vector space over $\operatorname{Quot}(A)$ . The dimension of the right side of (2.2.1), as a vector space over $\operatorname{Quot}(A)$ , is denoted $[\operatorname{Quot}(B):\operatorname{Quot}(A)]$ .

2.3. If $R$ is a Noetherian ring, $M$ is a finitely generated $R$ -module of Krull dimension $s$ , and $\mathfrak{a}$ is an ideal of $R$ with the Krull dimension of $R/\mathfrak{a}$ equal to zero, then the multiplicity of the  $R$ -module  $M$  with respect to the ideal  $\mathfrak{a}$ is

$$\begin{eqnarray}e_{\mathfrak{a}}(M)=s!\lim _{n\rightarrow \infty }\frac{\unicode[STIX]{x1D706}_{R}(M/\mathfrak{a}^{n}M)}{n^{s}},\end{eqnarray}$$

where $\unicode[STIX]{x1D706}_{R}(\underline{\phantom{x}})$ represents the length of an $R$ -module. If $R$ is local with maximal ideal $\mathfrak{m}$ , then we often write $e(M)$ or $e_{R}(M)$ in place of $e_{\mathfrak{m}}(M)$ . Similarly, if $R$ is a standard-graded algebra over a field (see 2.4) with maximal homogeneous ideal $\mathfrak{m}$ , and $M$ is a graded $R$ -module, then we often write $e(M)$ or $e_{R}(M)$ in place of $e_{\mathfrak{m}}(M)$ . In either case, we call the common value $e(M)=e_{R}(M)=e_{\mathfrak{m}}(M)$ the multiplicity of the  $R$ -module  $M$ ; in particular, $e(R)$ means $e_{R}(R)$ .

2.4. Let $R=\bigoplus _{0\leqslant i}R_{i}$ be a Noetherian graded algebra and $M$ be a finitely generated graded $R$ -module. It follows that

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{R_{0}}(M_{n})=\unicode[STIX]{x1D706}_{R}\left(\frac{\underset{n\leqslant i}{\bigoplus }M_{i}}{\underset{n<i}{\bigoplus }M_{i}}\right).\end{eqnarray}$$

This nonnegative integer is the value of the Hilbert function of the $R$ -module $M$ at $n$ , denoted $\operatorname{HF}_{M}(n)$ .

Let $k$ be a field. The graded algebra $R=\bigoplus _{0\leqslant i}R_{i}$ is a standard-graded  $k$ -algebra if $R_{0}$ is equal to $k$ , $R_{1}$ is a finitely generated $R_{0}$ -module, and $R$ is generated by $R_{1}$ as an algebra over $R_{0}$ . If $R$ is a standard-graded $k$ -algebra and $M$ is a finitely generated, graded $R$ -module with positive Krull dimension $s$ , then the multiplicity of $M$ may be expressed in terms of Hilbert functions:

$$\begin{eqnarray}e_{R}(M)=(s-1)!\lim _{n\rightarrow \infty }\frac{\operatorname{HF}_{M}(n)}{n^{s-1}}.\end{eqnarray}$$

2.5. If $R$ is a standard-graded $k$ -algebra with maximal homogeneous ideal $\mathfrak{m}$ , then the multiplicity $e(R)$ of the standard-graded $k$ -algebra $R$ is equal to the multiplicity $e(R_{\mathfrak{m}})$ of the local ring $R_{\mathfrak{m}}$ .

2.6. One important application of multiplicity is the following well-known theorem of Nagata [Reference Nagata41, 40.6]; see also [Reference Swanson and Huneke51, Exercise 11.8, Example 11.1.11]. If $C$ is a Noetherian formally equidimensional local ring, then $e(C)=1$ if and only if $C$ is a regular local ring. When we apply this result, $C$ is a one-dimensional Noetherian local domain and the hypothesis that $C$ be formally equidimensional is automatically satisfied.

2.7. We use the associativity formula for multiplicity; see for example, [Reference Bruns and Herzog5, Corollary 4.6.8] or [Reference Swanson and Huneke51, Theorem 11.2.4]. Let $R$ be a Noetherian ring, $\mathfrak{a}$ be an ideal of $R$ with the Krull dimension of $R/\mathfrak{a}$ equal to zero, and $M$ be a finitely generated $R$ -module. Then

$$\begin{eqnarray}e_{\mathfrak{a}}(M)=\mathop{\sum }_{P}\unicode[STIX]{x1D706}_{R_{P}}(M_{P})e_{\mathfrak{a}}(R/P),\end{eqnarray}$$

where $P$ varies over the prime ideals in the support of $M$ with the property that the Krull dimension of $R/P$ is equal to the Krull dimension of $M$ .The proof makes use of a filtration of $M$ whose factors are cyclic modules defined by prime ideals of $R$ .

Observation 2.8. Assume $A\subseteq B$ is a module-finite extension of standard-graded $k$ -algebras which also are domains. Then $e_{B}(B)=e_{A}(A)\operatorname{rank}_{A}B$ .

Proof. One has $e_{B}(B)=e_{A}(B)=e_{A}(A)\operatorname{rank}_{A}B$ . The first equality holds because $B$ has the same Hilbert function independent of whether $B$ is viewed as an $A$ -module or a $B$ -module. The second equality is the associativity formula for multiplicity.◻

2.9. A dominant rational map  of projective varieties is birational if the induced map of function fields $K(Y){\hookrightarrow}K(X)$ is an isomorphism. In general, the degree of the rational map $\unicode[STIX]{x1D6F9}$ is the dimension of the field extension $[K(X):K(\operatorname{Im}\unicode[STIX]{x1D6F9})]$ .

Proof. According to 2.9, the degree of $\operatorname{Proj}(\unicode[STIX]{x1D713})$ is defined to be $[B_{(0)}:(\unicode[STIX]{x1D713}(A))_{(0)}]$ , where $B_{(0)}$ and $(\unicode[STIX]{x1D713}(A))_{(0)}$ are the subfields of $\operatorname{Quot}(B)$ and $\operatorname{Quot}(\unicode[STIX]{x1D713}(A))$ , respectively, which consist of the homogeneous elements of degree zero. Let $a_{1}$ be an element of $A_{1}$ with $\unicode[STIX]{x1D713}(a_{1})\neq 0$ . It is easy to see that $\unicode[STIX]{x1D713}(a_{1})$ is transcendental over $B_{(0)}$ , $\operatorname{Quot}(B)=B_{(0)}(\unicode[STIX]{x1D713}(a_{1}))$ , and $\operatorname{Quot}(\unicode[STIX]{x1D713}(A))=(\unicode[STIX]{x1D713}(A))_{(0)}(\unicode[STIX]{x1D713}(a_{1}))$ . It follows that

$$\begin{eqnarray}[B_{(0)}:(\unicode[STIX]{x1D713}(A))_{(0)}]=[\operatorname{Quot}(B):\operatorname{Quot}(\unicode[STIX]{x1D713}(A))].\square\end{eqnarray}$$

2.11. If $R=\bigoplus _{0\leqslant i}R_{i}$ is a graded ring and $s$ is a positive integer, then the $s$ th Veronese ring of $R$ is equal to $R^{(s)}=\bigoplus _{0\leqslant i}R_{is}$ . One regrades the Veronese ring in order to have the component of $R^{(s)}$ in degree $i$ be $R_{is}$ . The $s$ th Veronese of a graded module $M=\bigoplus M_{i}$ is formed in a similar manner: $M^{(s)}=\bigoplus M_{is}$ , with $M_{is}$ having degree  $i$ .

2.12. Recall that the Rees algebra ${\mathcal{R}}(I)$ of an ideal $I$ in a commutative ring $R$ is the graded subalgebra $R[It]$ of the polynomial ring $R[t]$ . If $R$ has a distinguished maximal ideal $\mathfrak{m}$ (that is, if $R$ is graded with maximal homogeneous ideal $\mathfrak{m}$ or if $R$ is local with maximal ideal $\mathfrak{m}$ ), then the special fiber ring of $I$ is ${\mathcal{R}}(I)\otimes _{R}R/\mathfrak{m}$ .

In the typical situation in the present paper, the ring $R$ will be a standard-graded polynomial ring over a field $k$ , and $I$ will be a homogeneous ideal of $R$ generated by homogeneous forms of the same degree $d$ , and, after regrading, the $k$ -subalgebra $k[I_{d}]$ of $R$ will be the coordinate ring of a projective variety. In this case, there is a $k$ -algebra isomorphism from the projective coordinate ring $k[I_{d}]$ , to the special fiber ring ${\mathcal{F}}(I)$ . Indeed,

where $\mathfrak{m}$ is the maximal homogeneous ideal of $R$ .

2.13. Let $J\subset I$ be ideals in a commutative Noetherian ring $R$ . The following conditions are equivalent:

1. (a) there exists a nonnegative integer $m$ with $JI^{m}=I^{m+1}$ , and

2. (b) the Rees algebra ${\mathcal{R}}(I)$ is finitely generated as a module over ${\mathcal{R}}(J)$ .

When these conditions occur, one says that $J$ is a reduction of $I$ or $I$ is integral over $J$ . (For details, see, for example, [Reference Swanson and Huneke51, 8.21, 1.25, 1.1.1, 1.2.1].)

2.14. If $R$ is a Noetherian domain, then the ring $S$ is an $S_{2}$ -ification of  $R$ if $R\subset S\subset \operatorname{Quot}(R)$ , $S$ is module-finite over $R$ , $S$ satisfies Serre’s condition $(S_{2})$ as an $R$ -module, and for each $s\in S$ , the ideal $R:_{R}s$ of $R$ has height at least $2$ . It follows from [Reference Hochster and Huneke23, 2.3] that the $S_{2}$ -fication of $R$ is unique and from [Reference Hochster and Huneke23, 2.7] that if $R$ has a canonical module $\unicode[STIX]{x1D714}_{R}$ then $\operatorname{End}_{R}(\unicode[STIX]{x1D714}_{R})$ is the $S_{2}$ -ification of $R$ .

2.15. The concept of “generalized row ideals” appears widely in the literature; see, for example, [Reference Eisenbud, Huneke and Ulrich18, Reference Eisenbud and Ulrich19, Reference Green21]. Let $M$ be a matrix with entries in a $k$ -algebra and $p$ be a nonzero row vector with entries from $k$ , where $k$ is a field. A generalized row of $M$ is the product $pM$ . If $pM$ is a generalized row of $M$ , then the ideal $I_{1}(pM)$ is called a generalized row ideal of $R$ .

2.16. Let $X$ be a topological space. We say that a general point of  $X$ has a certain property if there exists a dense, open subset $U$ of $X$ so that every point of $U$ has the property.

Data 3.1. Let $k$ be an infinite field and  be a rational map defined by $k$ -linearly independent homogeneous forms $g_{1},\ldots ,g_{n}$ of degree $d$ in $R=k[x_{1},\ldots ,x_{s}]$ , a standard-graded polynomial ring over $k$ in $s$ variables with maximal homogeneous ideal $\mathfrak{m}$ , $I$ be the homogeneous ideal $(g_{1},\ldots ,g_{n})$ in $R$ , and $\unicode[STIX]{x1D711}$ be a homogeneous syzygy matrix of $[g_{1},\ldots ,g_{n}]$ . Let $S=k[T_{1},\ldots ,T_{n}]$ be a standard-graded polynomial ring over $k$ in $n$ variables. The map $\unicode[STIX]{x1D6F9}$ corresponds to the $k$ -algebra homomorphism $\unicode[STIX]{x1D713}:S\rightarrow R$ , which sends $T_{i}$ to $g_{i}$ . Let $A$ be the image $k[I_{d}]$ of this homomorphism and $B$ be the Veronese ring $k[R_{d}]$ . After regrading, we view $A\subset B$ as standard-graded $k$ -algebras. Notice that $A$ is the homogeneous coordinate ring of the image of $\unicode[STIX]{x1D6F9}$ . Notice also that, according to Remark 2.10, $[B:A]$ is the degree of the rational map $\unicode[STIX]{x1D6F9}$ .

Observation 3.2. Adopt the data of 3.1. If $I$ is $\mathfrak{m}$ -primary then $d^{s-1}=e(A)[B:A]$ .

Proof. The hypothesis that the ideal $I$ is $\mathfrak{m}$ -primary forces the ring extension $A\subset B$ to be module-finite; since $\mathfrak{m}^{dm}\subset I$ for some m, hence $B_{m}=A_{1}B_{m-1}$ . Observation 2.8 yields $e(A)[B:A]=e(B)$ . On the other hand from the Hilbert function of $B$ one sees $e(B)=d^{s-1}$ .◻

Remarks 3.4. (1) Definition 3.3 gives the correct notion of fiber as a set. Indeed, since the ideal $I$ is the extension to $R$ of the homogeneous maximal ideal of $S$ , the prime ideals of $\operatorname{Proj}\,(R/(\mathfrak{P}R:_{R}I^{\infty }))$ correspond to the primes of $\operatorname{Proj}\,(R)$ that contract to the prime $\mathfrak{P}$ of $\operatorname{Proj}\,(S)$ . In particular, the rational closed points of $\operatorname{Proj}\,(R/(\mathfrak{P}R:_{R}I^{\infty }))$ correspond to the points in the domain of $\unicode[STIX]{x1D6F9}$ that map to the point $p$ of $\mathbb{P}_{k}^{n-1}$ .

(2) If $p$ is the point $[\unicode[STIX]{x1D6FC}_{1}:\cdots :\unicode[STIX]{x1D6FC}_{n}]$ of $\mathbb{P}_{k}^{n-1}$ , then $\mathfrak{P}$ is the prime ideal

(3.4.1) $$\begin{eqnarray}\mathfrak{P}=I_{2}\left(\left[\begin{array}{@{}ccc@{}}\unicode[STIX]{x1D6FC}_{1} & \ldots \, & \unicode[STIX]{x1D6FC}_{n}\\ T_{1} & \ldots \, & T_{n}\end{array}\right]\right)\end{eqnarray}$$

of $S$ , and $\mathfrak{P}R$ and $\mathfrak{P}B$ are the extensions of $\mathfrak{P}$ to the rings $R$ and $B$ , respectively, under the ring homomorphisms:

where $\text{incl}$ is the inclusion map. So, in particular, the ideals $\mathfrak{P}R$ and $\mathfrak{P}B$ both are generated by the $2\times 2$ minors of

(3.4.2) $$\begin{eqnarray}I_{2}\left(\left[\begin{array}{@{}ccc@{}}\unicode[STIX]{x1D6FC}_{1} & \ldots \, & \unicode[STIX]{x1D6FC}_{n}\\ g_{1} & \ldots \, & g_{n}\end{array}\right]\right).\end{eqnarray}$$

Proof.

1. (1) It suffices to show that $(\mathfrak{Q}\cap A)R$ and $(\mathfrak{Q}\cap A^{(t)})R$ are equal locally at any homogeneous relevant prime ideal of $R$ that contains either ideal. Any such prime ideal contracts to $\mathfrak{Q}\cap A$ in $A$ , and $\mathfrak{Q}\cap A$ is a relevant prime ideal of $A$ because $q$ is in the domain of $\unicode[STIX]{x1D6F9}$ . Therefore, the two ideals $\mathfrak{Q}\cap A$ and $(\mathfrak{Q}\cap A^{(t)})A$ of $A$ coincide locally at $\mathfrak{Q}\cap A$ .

2. (2) In a similar manner it suffices to show that the two ideals $\mathfrak{Q}\cap C$ and $(\mathfrak{Q}\cap A)C$ are equal locally at the prime ideal $\mathfrak{Q}\cap A$ of $A$ . To see this, notice that $A_{\mathfrak{Q}\cap A}=C_{\mathfrak{Q}\cap A}$ because the extension $A\subset C$ is birational and the point $q$ is general. ◻

Proof. No harm is done if we assume that $k$ is algebraically closed because the hypotheses and conclusions remain unchanged under this change of base.

The rational map  is defined at general points $q$ and their images $p$ , which correspond to prime ideals $\mathfrak{p}$ , are general in $\operatorname{Proj}(A)$ . Since $A$ and $B$ have the same Krull dimension, the field extension $\text{Quot}(A)\subset \operatorname{Quot}(B)$ is algebraic and therefore finite. Since $\mathfrak{p}$ is general, the Generic Freeness Lemma implies that $B_{\mathfrak{p}}$ is free as an $A_{\mathfrak{p}}$ -module. Therefore, $B_{\mathfrak{p}}$ is a finitely generated $A_{\mathfrak{p}}$ -module and

$$\begin{eqnarray}[\operatorname{Quot}(B):\operatorname{Quot}(A)]=\unicode[STIX]{x1D707}_{A_{\mathfrak{p}}}(B_{\mathfrak{p}})=\unicode[STIX]{x1D706}_{A_{\mathfrak{p}}}(B\otimes _{A}k(\mathfrak{p})).\end{eqnarray}$$

The ring $B\otimes _{A}k(\mathfrak{p})$ is Artinian and therefore $B$ is equal to the direct product $\displaystyle \times _{\mathfrak{q}^{\prime }}B_{\mathfrak{q}^{\prime }}/\mathfrak{p}B_{\mathfrak{q}^{\prime }}$ , where $\mathfrak{q}^{\prime }$ varies over all primes in $\operatorname{Proj}(B)$ with $\mathfrak{q}^{\prime }\cap A=\mathfrak{p}$ . Therefore,

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{A_{\mathfrak{p}}}(B\otimes _{A}k(\mathfrak{p}))=\mathop{\sum }_{\mathfrak{q}^{\prime }}\unicode[STIX]{x1D706}_{A_{\mathfrak{p}}}(B_{\mathfrak{q}^{\prime }}/\mathfrak{p}B_{\mathfrak{q}^{\prime }}).\end{eqnarray}$$

Since $B_{\mathfrak{p}}$ is a finitely generated $A_{\mathfrak{p}}$ -module, the field extension $\text{Quot}(A/\mathfrak{p})\subset \operatorname{Quot}(B/\mathfrak{q}^{\prime })$ is algebraic, and therefore $A/\mathfrak{p}$ and $B/\mathfrak{q}^{\prime }$ have the same Krull dimension. Thus the rings $A/\mathfrak{p}$ and $B/\mathfrak{q}^{\prime }$ are standard-graded one-dimensional domains over the algebraically closed field $k$ , hence both are polynomial rings in one variable. Since the inclusion $A/\mathfrak{p}\subset B/\mathfrak{q}^{\prime }$ is homogeneous, it then follows that it is actually an equality. Therefore, $k(\mathfrak{p})=k(\mathfrak{q}^{\prime })$ and we obtain

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{A_{\mathfrak{p}}}(B_{\mathfrak{q}^{\prime }}/\mathfrak{p}B_{\mathfrak{q}^{\prime }})=\unicode[STIX]{x1D706}_{B_{\mathfrak{q}^{\prime }}}(B_{\mathfrak{q}^{\prime }}/\mathfrak{p}B_{\mathfrak{q}^{\prime }}).\end{eqnarray}$$

Since $\operatorname{Proj}(B/\mathfrak{p}B)\simeq \operatorname{Proj}(R/\mathfrak{p}R)$ , there exists a unique $\mathfrak{Q}^{\prime }\in \operatorname{Proj}(R)$ with $\mathfrak{Q}^{\prime }\cap A=\mathfrak{p}$ corresponding to each $\mathfrak{q}^{\prime }$ and furthermore

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{B_{\mathfrak{q}^{\prime }}}(B_{\mathfrak{q}^{\prime }}/\mathfrak{p}B_{\mathfrak{q}^{\prime }})=\unicode[STIX]{x1D706}_{R_{\mathfrak{Q}^{\prime }}}(R_{\mathfrak{Q}^{\prime }}/\mathfrak{p}R_{\mathfrak{Q}^{\prime }}).\end{eqnarray}$$

In summary we obtain

$$\begin{eqnarray}[\operatorname{Quot}(B):\operatorname{Quot}(A)]=\mathop{\sum }_{\mathfrak{q}^{\prime }}\unicode[STIX]{x1D706}_{A_{\mathfrak{p}}}(B_{\mathfrak{q}^{\prime }}/\mathfrak{p}B_{\mathfrak{q}^{\prime }})=\mathop{\sum }_{\mathfrak{Q}^{\prime }}\unicode[STIX]{x1D706}_{R_{\mathfrak{Q}^{\prime }}}(R_{\mathfrak{Q}^{\prime }}/\mathfrak{p}R_{\mathfrak{Q}^{\prime }}).\end{eqnarray}$$

Let $\mathfrak{m}_{A}$ denote the homogeneous maximal ideal of $A$ . Notice that $\mathfrak{m}_{A}R=I$ and recall that $\text{dim}\,A/\mathfrak{p}$ is one. It follows that

$$\begin{eqnarray}\displaystyle \{\mathfrak{Q}^{\prime }\mid \mathfrak{Q}^{\prime }\in \operatorname{Proj}(R),\mathfrak{Q}^{\prime }\cap A=\mathfrak{p}\} & = & \displaystyle \{\mathfrak{Q}^{\prime }\mid \mathfrak{Q}^{\prime }\in \operatorname{Proj}(R),\mathfrak{p}\subset \mathfrak{Q}^{\prime },\mathfrak{m}_{A}\not \subset \mathfrak{Q}^{\prime }\}\nonumber\\ \displaystyle & = & \displaystyle \{\mathfrak{Q}^{\prime }\mid \mathfrak{Q}^{\prime }\in \operatorname{Proj}(R),\mathfrak{p}R\subset \mathfrak{Q}^{\prime },I\not \subset \mathfrak{Q}^{\prime }\}\nonumber\\ \displaystyle & = & \displaystyle \operatorname{Proj}(R)\cap V(\mathfrak{p}R:_{R}I^{\infty }).\nonumber\end{eqnarray}$$

The rings $R/\mathfrak{Q}^{\prime }$ are polynomial rings in one variable for every prime ideal $\mathfrak{Q}^{\prime }$ in the above set. In particular, these prime ideals correspond to the minimal primes of maximal dimension of the ring $R/(\mathfrak{p}R:_{R}I^{\infty })$ . Now the associativity formula for multiplicity implies that

$$\begin{eqnarray}\mathop{\sum }_{\mathfrak{Q}^{\prime }}\unicode[STIX]{x1D706}_{R_{\mathfrak{Q}^{\prime }}}(R_{\mathfrak{Q}^{\prime }}/\mathfrak{p}R_{\mathfrak{Q}^{\prime }})=e(R/(\mathfrak{p}R:_{R}I^{\infty })),\end{eqnarray}$$

which completes the proof. ◻

Proof. Let $f_{1},\ldots ,f_{s}$ be general $k$ -linear combinations of homogeneous minimal generators of $I$ . Let $C=k[f_{1},\ldots ,f_{s}]$ . Since $A$ has dimension $s$ , the homogeneous ring extension $C\subset A$ is module-finite and $C$ is a polynomial ring. Hence $[A:C]=e(A)$ by Observation 2.8, for instance.

We have the inclusion of domains

$$\begin{eqnarray}C\subset A\subset B.\end{eqnarray}$$

We compute the field degree $[B:C]$ in two different ways. First,

$$\begin{eqnarray}[B:C]=[B:A]\cdot e(A).\end{eqnarray}$$

Next, we wish to apply Proposition 3.6 to express $[B:C]$ . By the general choice of $f_{1},\ldots ,f_{s}$ , we may assume that $[0:\ldots :0:1]$ is a general point of $\operatorname{Proj}(C)\simeq \mathbb{P}_{k}^{s-1}$ . The homogeneous prime ideal of $C$ corresponding to this point is $\mathfrak{p}=(f_{1},\ldots ,f_{s-1})C$ . Therefore, Proposition 3.6 shows that

$$\begin{eqnarray}[B:C]=e\left(\frac{R}{(f_{1},\ldots ,f_{s-1}):_{R}(f_{s})^{\infty }}\right).\end{eqnarray}$$

On the other hand,

$$\begin{eqnarray}(f_{1},\ldots ,f_{s-1}):_{R}(f_{s})^{\infty }=(f_{1},\ldots ,f_{s-1}):_{R}I^{\infty }\end{eqnarray}$$

because $f_{s}$ is a general $k$ -linear combination of the homogeneous minimal generators of $I$ . The proof is complete.◻

Proof. According to [Reference Xie53, 2.5] (see also [Reference Achilles and Manaresi1, 3.8] and [Reference Nishida and Ulrich42, 3.6])

$$\begin{eqnarray}j(I)=\unicode[STIX]{x1D706}_{R}\left(\frac{R}{((f_{1},\ldots ,f_{s-1}):_{R}I^{\infty },f_{s})}\right),\end{eqnarray}$$

in the language of Corollary 3.7. On the other hand, $f_{s}$ is regular on $R/(f_{1},\ldots ,f_{s-1}):_{R}I^{\infty }$ , and $f_{s}$ is a homogeneous element of $R$ of degree $d$ ; so,

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{R}\left(\frac{R}{((f_{1},\ldots ,f_{s-1}):_{R}I^{\infty },f_{s})}\right)=d\cdot e\left(\frac{R}{(f_{1},\ldots ,f_{s-1}):_{R}I^{\infty }}\right),\end{eqnarray}$$

and the assertion follows from Corollary 3.7. ◻

Observation 3.9. Adopt Data 3.1. If $\mathfrak{P}$ is the homogeneous prime ideal in $S$ which corresponds to the rational point $p$ in $\mathbb{P}_{k}^{n-1}$ , then the ideals $\mathfrak{P}R:I^{\infty }$ and $I_{1}(p\unicode[STIX]{x1D711}):I^{\infty }$ of $R$ are equal.

Proof. Let $p=[\unicode[STIX]{x1D6FC}_{1}:\cdots :\unicode[STIX]{x1D6FC}_{n}]$ with $\unicode[STIX]{x1D6FC}_{i}$ in $k$ . Recall the generating sets given in (3.4.1) and (3.4.2) for the ideals $\mathfrak{P}$ and $\mathfrak{P}R$ of $S$ and $R$ , respectively. Let $\unicode[STIX]{x1D712}$ be an invertible matrix with entries in $k$ and $(0,\ldots ,0,1)=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{n})\unicode[STIX]{x1D712}$ . Define $[g_{1}^{\prime },\ldots ,g_{n}^{\prime }]=\pmb{g}^{\prime }$ by

(3.9.1) $$\begin{eqnarray}\pmb{g}^{\prime }=[g_{1},\ldots ,g_{n}]\unicode[STIX]{x1D712}.\end{eqnarray}$$

The entries of $\pmb{g}^{\prime }$ generate $I$ and $\unicode[STIX]{x1D711}^{\prime }=\unicode[STIX]{x1D712}^{-1}\unicode[STIX]{x1D711}$ is a homogeneous syzygy matrix for $\pmb{g}^{\prime }$ . One consequence of this last statement is the fact that the bottom row of $\unicode[STIX]{x1D711}^{\prime }$ generates the ideal $(g_{1}^{\prime },\ldots ,g_{n-1}^{\prime }):g_{n}^{\prime }$ . On the other hand,

$$\begin{eqnarray}\left[\begin{array}{@{}ccc@{}}\unicode[STIX]{x1D6FC}_{1} & \ldots \, & \unicode[STIX]{x1D6FC}_{n}\\ g_{1} & \ldots \, & g_{n}\end{array}\right]\unicode[STIX]{x1D712}=\left[\begin{array}{@{}cccc@{}}0 & \ldots \, & 0 & 1\\ g_{1}^{\prime } & \ldots \, & g_{n-1}^{\prime } & g_{n}^{\prime }\end{array}\right];\end{eqnarray}$$

and the ideal of $2\times 2$ minors of a matrix is unchanged by row and column operations; therefore, $\mathfrak{P}R$ is generated by $(g_{1}^{\prime },\ldots ,g_{n-1}^{\prime })$ . We now see that

(3.9.2) $$\begin{eqnarray}I_{1}(p\unicode[STIX]{x1D711})=I_{1}([0,\ldots ,0,1]\unicode[STIX]{x1D711}^{\prime })=(g_{1}^{\prime },\ldots ,g_{n-1}^{\prime }):g_{n}^{\prime }=\mathfrak{P}R:I;\end{eqnarray}$$

hence, $I_{1}(p\unicode[STIX]{x1D711}):I^{\infty }=(\mathfrak{P}R:I):I^{\infty }=\mathfrak{P}R:I^{\infty }$ .◻

Proof. For part (1) we use Observation 3.9, Remark 3.4(1) and the fact that the ideal $I_{1}(p\unicode[STIX]{x1D711}):I^{\infty }$ is not the unit ideal if and only it has dimension at least one. Item (2) follows from Observation 3.9. To prove item (3) we apply Observation 3.9 and Proposition 3.6.◻

Proof. Items (1)–(3) follow from Corollary 3.10. As for item (4), notice that each column of $\unicode[STIX]{x1D711}$ contains a nonzero element, hence the linear space

$$\begin{eqnarray}V=\{[\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n}]\in k^{n}\mid \sum \unicode[STIX]{x1D6FD}_{i}\unicode[STIX]{x1D711}_{ij}=0\}\end{eqnarray}$$

is a proper linear subspace of $\mathbb{P}_{k}^{n-1}$ . Because $g_{1},\ldots ,g_{n}$ are $k$ -linearly independent the image of $\unicode[STIX]{x1D6F9}$ cannot be contained in $V$ . Thus if $q$ is a general point in $\mathbb{P}_{k}^{1}$ every entry in the product $p\unicode[STIX]{x1D711}=\unicode[STIX]{x1D6F9}(q)\unicode[STIX]{x1D711}$ is nonzero. Now item (4) follows from (3).◻

Data 4.1. (1) Let $R=k[x_{1},x_{2}]$ be a standard-graded polynomial ring over an infinite perfect field $k$ with maximal homogeneous ideal $\mathfrak{m}$ , $I$ be a height two ideal of $R$ which is generated by homogeneous forms of the same degree $d$ , $A$ and $B$ be the $k$ -subalgebras $A=k[I_{d}]$ and $B=k[R_{d}]$ of $R$ , and $r$ be the degree of the field extension $\operatorname{Quot}(A)\subset \operatorname{Quot}(B)$ . View $A$ and $B$ as standard-graded $k$ -algebras with maximal homogeneous ideals $\mathfrak{m}_{A}$ and $\mathfrak{m}_{B}$ , respectively. Let $e=e(A)$ .

(2) Let $g_{1},\ldots ,g_{n}$ be homogeneous forms in $R$ of degree $d$ which minimally generate the ideal $I$ and consider the morphism  which is defined by these forms; that is, $\unicode[STIX]{x1D6F9}$ sends the point $q$ of $\mathbb{P}_{k}^{1}$ to the point $[g_{1}(q):\cdots :g_{n}(q)]\in \mathbb{P}_{k}^{n-1}$ . The corresponding homomorphism of $k$ -algebras is $\unicode[STIX]{x1D713}:S=k[T_{1},\ldots ,T_{n}]\rightarrow R$ , which sends $T_{i}$ to $g_{i}$ , for each $i$ .

(3) Let $\unicode[STIX]{x1D711}$ be a minimal homogeneous Hilbert–Burch matrix for the row vector $[g_{1},\ldots ,g_{n}]$ . In other words, there exist shifts $D_{j}$ and a unit $u$ in $k$ so that

(4.1.1)

is a minimal homogeneous resolution of $R/I$ and $g_{i}$ is equal to $u(-1)^{i}$ times the determinant of $\unicode[STIX]{x1D711}$ with row $i$ deleted, for $1\leqslant i\leqslant n$ .

Proof. Apply Proposition 4.3. In the present situation, $A$ and $B$ are normal of dimension $2$ and therefore Cohen–Macaulay, and $B$ has minimal multiplicity since $d=e(B)=\unicode[STIX]{x1D707}(\mathfrak{m}_{B})-1$ . It follows that $A$ has minimal multiplicity as well, namely,

$$\begin{eqnarray}e=\unicode[STIX]{x1D707}(\mathfrak{m}_{A})-1=\unicode[STIX]{x1D707}(I)-1=n-1.\end{eqnarray}$$

We conclude that the syzygy matrix $\unicode[STIX]{x1D711}$ has size $e+1\times e$ . Each entry in column $j$ of $\unicode[STIX]{x1D711}$ is a form of degree $D_{j}$ , in the language of (4.1.1), and, according to Corollary 3.11(4), $D_{j}\geqslant r$ . The $g_{i}$ have degree $d$ and each one of them is the determinant of an $e\times e$ minor of $\unicode[STIX]{x1D711}$ ; thus, $d=\sum _{j=1}^{e}D_{j}\geqslant er=d$ , where the last equality follows from Observation 3.2. We conclude that each entry of $\unicode[STIX]{x1D711}$ is a homogeneous form of degree  $r$ .

Let $q_{1},\ldots ,q_{e+1}$ be general points in $\mathbb{P}_{k}^{1}$ . Hence the points $\unicode[STIX]{x1D6F9}(q_{1}),\ldots ,\unicode[STIX]{x1D6F9}(q_{e+1})$ are general on the curve ${\mathcal{C}}=\text{Im}\,\unicode[STIX]{x1D6F9}$ , and since ${\mathcal{C}}$ is nondegenerate in $\mathbb{P}^{e}$ , it follows that $\unicode[STIX]{x1D6F9}(q_{1}),\ldots ,\unicode[STIX]{x1D6F9}(q_{e+1})$ span $\mathbb{P}^{e}$ . In other words, the $(e+1)\times (e+1)$ scalar matrix $\unicode[STIX]{x1D6E4}$ , whose rows are $\unicode[STIX]{x1D6F9}(q_{i})$ , is invertible. We apply Corollary 3.11(3) to see that each $\text{gcd}(I_{1}(\unicode[STIX]{x1D6F9}(q_{i})\unicode[STIX]{x1D711}))$ is a form in $R$ of degree $r$ . Write $f_{i}=\text{gcd}(I_{1}(\unicode[STIX]{x1D6F9}(q_{i})\unicode[STIX]{x1D711}))$ . Degree considerations show that

(4.4.1) $$\begin{eqnarray}\unicode[STIX]{x1D6F9}(q_{i})\unicode[STIX]{x1D711}=f_{i}\cdot \unicode[STIX]{x1D706}_{i},\end{eqnarray}$$

where $\unicode[STIX]{x1D706}_{i}$ is a vector of scalars. It follows that

(4.4.2) $$\begin{eqnarray}\unicode[STIX]{x1D6E4}\unicode[STIX]{x1D711}=D\unicode[STIX]{x1D6EC},\end{eqnarray}$$

where $D$ is the diagonal matrix whose diagonal entries are the forms $f_{1},\ldots ,f_{e+1}$ , and $\unicode[STIX]{x1D6EC}$ is the $(e+1)\times e$ scalar matrix whose rows are the $\unicode[STIX]{x1D706}_{i}$ . We notice that $\unicode[STIX]{x1D6EC}$ has maximal rank $e$ because $\unicode[STIX]{x1D711}$ does.

Let $q$ be another general point in $\mathbb{P}^{1}$ . Write

(4.4.3) $$\begin{eqnarray}\unicode[STIX]{x1D6F9}(q)\unicode[STIX]{x1D711}=f\unicode[STIX]{x1D706},\end{eqnarray}$$

as in (4.4.1), where $f$ is a form of degree $r$ in $R$ and $\unicode[STIX]{x1D706}$ is a row vector of scalars from $k$ . There is an $e\times (e-1)$ matrix of scalars $\unicode[STIX]{x1D6E5}$ so that $\unicode[STIX]{x1D706}\unicode[STIX]{x1D6E5}=0$ and $\unicode[STIX]{x1D6E5}$ has full rank $e-1$ . Combine (4.4.3), the definition of $\unicode[STIX]{x1D6F9}$ as given in 4.1(2), and (4.4.2) to see that

$$\begin{eqnarray}0=f\unicode[STIX]{x1D706}\unicode[STIX]{x1D6E5}=\unicode[STIX]{x1D6F9}(q)\unicode[STIX]{x1D711}\unicode[STIX]{x1D6E5}=[g_{1}(q),\ldots ,g_{e+1}(q)]\unicode[STIX]{x1D6E4}^{-1}D\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D6E5}.\end{eqnarray}$$

The point $q$ in $\mathbb{P}^{1}$ is general; so the product $[g_{1}(q),\ldots ,g_{e+1}(q)]\unicode[STIX]{x1D6E4}^{-1}$ is a row vector of nonzero scalars and the product $[g_{1}(q),\ldots ,g_{e+1}(q)]\unicode[STIX]{x1D6E4}^{-1}D$ is equal to $[f_{1},\ldots ,f_{e+1}]D^{\prime }$ , where $D^{\prime }$ is an invertible diagonal matrix of scalars. Thus,

$$\begin{eqnarray}0=[f_{1},\ldots ,f_{e+1}]D^{\prime }\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D6E5},\end{eqnarray}$$

where $D^{\prime }\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D6E5}$ is an $(e+1)\times (e-1)$ matrix of scalars of rank $e-1$ . We conclude that the vector space $V$ spanned by the homogeneous forms $f_{1},\ldots ,f_{e+1}$ has dimension at most two. Select a generating set $f_{i},f_{j}$ for $V$ . Every entry of the Hilbert–Burch matrix $\unicode[STIX]{x1D711}$ is an element of the ring $k[f_{i},f_{j}]$ (as can be seen from (4.4.2)); hence $A=k[I_{d}]\subset k[f_{i},f_{j}]$ and the proof of Claim 4.4 is complete.◻

Proof. We prove the theorem by reducing to the normal case. The integral closure $\overline{A}$ of $A$ in $\operatorname{Quot}(A)$ is a graded $k$ -subalgebra of $B$ , though not necessarily standard-graded. However, $\overline{A}$ is a finitely generated graded module over the standard-graded $k$ -algebra $A$ and hence there exists a positive integer $s$ such that the $t$ -Veronese subring of the integral closure of $A$ , denoted $\overline{A}^{(t)}$ , is standard-graded for all $t\geqslant s$ . We observe that $\overline{A}^{(t)}=\overline{A^{(t)}}$ .

We apply Claim 4.4 to $\overline{A^{(s)}}$ . For general points $q_{1}$ and $q_{2}$ in $\mathbb{P}_{k}^{1}$ we obtain forms $\widetilde{f_{1}}$ and $\widetilde{f_{2}}$ such that $\overline{A^{(s)}}\subset k[\widetilde{f_{1}},\widetilde{f_{2}}]$ . Let $\mathfrak{Q_{i}}$ be the prime ideals in $R$ corresponding to $q_{i}$ . We have

$$\begin{eqnarray}\displaystyle \widetilde{f_{i}}R & = & \displaystyle ((\mathfrak{Q_{i}}\cap \overline{A^{(s)}})R):\mathfrak{m}^{\infty }\nonumber\\ \displaystyle & = & \displaystyle ((\mathfrak{Q_{i}}\cap A^{(s)})R):\mathfrak{m}^{\infty }=((\mathfrak{Q_{i}}\cap A)R):\mathfrak{m}^{\infty }=f_{i}R,\nonumber\end{eqnarray}$$

where the first and last equality follow from Observation 3.9 (and 2.1) and the second and third equality hold due to Remark 3.5. We conclude that $\widetilde{f_{i}}k=f_{i}k$ and therefore $\overline{A^{(s)}}\subset k[f_{1},f_{2}]$ . In the same manner we obtain $\overline{A^{(s+1)}}\subset k[f_{1},f_{2}]$ . Hence for every nonzero homogeneous element $\unicode[STIX]{x1D6FC}$ in $A$ we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\frac{\unicode[STIX]{x1D6FC}^{s+1}}{\unicode[STIX]{x1D6FC}^{s}}\in \text{Quot}(k[f_{1},f_{2}]).\end{eqnarray}$$

On the other hand, $\unicode[STIX]{x1D6FC}$ is integral over $A^{(s)}$ , hence integral over $k[f_{1},f_{2}]$ . We conclude that $\unicode[STIX]{x1D6FC}\in k[f_{1},f_{2}]$ since $k[f_{1},f_{2}]$ is normal.◻

Proof. The forms $f_{1}$ and $f_{2}$ have degree $r$ according to Corollary 3.11(3). From Theorem 4.2 we obtain $I_{d}\subset R^{\prime }=k[f_{1},f_{2}]$ . In particular, $I\subset (f_{1},f_{2})R$ because $I=I_{d}R$ . It follows that $f_{1}$ , $f_{2}$ form a regular sequence in $R$ and hence $R$ is a free module over $R^{\prime }=k[f_{1},f_{2}]$ , necessarily of rank $r^{2}$ . Parts (1) and (5) are established. At this point, items (2), (3), and (6) are clear.

To prove (4) we use the following diagram, where the relevant field degrees are displayed:

The diagram shows $[k[R_{d/r}^{\prime }]:A]=1$ .◻

Proof. Items (1) and (2) follow from items (1)–(6) of Corollary 4.5. Item (3) is now clear. Item (4) is [Reference Cox, Kustin, Polini and Ulrich16, 0.11]. ◻

Observation 4.7. Adopt the data of 4.1 and assume that $I$ is a monomial ideal. Then

$$\begin{eqnarray}[B:A]=\text{gcd}(\text{column degrees of }\unicode[STIX]{x1D711}).\end{eqnarray}$$

Proof. We may assume that $g_{i}=x^{a_{i}}y^{d-a_{i}}$ with $0=a_{1}<a_{2}<\cdots <a_{n}=d$ . Hence in the language of (4.1.1),

$$\begin{eqnarray}D_{j}=a_{j+1}-a_{j}\quad \text{for }1\leqslant j\leqslant n-1.\end{eqnarray}$$

On the other hand, as is well known,

$$\begin{eqnarray}[R:A]\mathbb{Z}=I_{2}\left(\begin{array}{@{}cccc@{}}a_{1} & a_{2} & \ldots & a_{n}\\ d-a_{1} & d-a_{2} & \ldots & d-a_{n}\end{array}\right).\end{eqnarray}$$

After row and column operations on the matrix, this determinantal ideal becomes

$$\begin{eqnarray}d\cdot I_{2}\left(\begin{array}{@{}cccc@{}}0 & a_{2}-a_{1} & \ldots & a_{n}-a_{n-1}\\ 1 & 0 & \ldots & 0\end{array}\right)=d\cdot \text{gcd}(D_{1},\ldots ,D_{n-1})\mathbb{Z}.\end{eqnarray}$$

It follows that

$$\begin{eqnarray}[B:A]=\frac{[R:A]}{d}=\text{gcd}(D_{1},\ldots ,D_{n-1})=\text{gcd}(\text{column degrees of }\unicode[STIX]{x1D711}).\square\end{eqnarray}$$

Proof. Item (1) is Observation 3.2; (2) follows from (1) and the definition of birationality; and (3) is Theorem 4.2 and Corollary 4.5. ◻

Data 5.1. Let $R$ be a standard-graded domain over a field $k$ , of dimension $s$ , with maximal homogeneous ideal $\mathfrak{m}$ . Let $I$ be a homogeneous ideal in $R$ generated by forms of degree $d$ , $A$ be the ring $k[I_{d}]$ , and $B$ be the Veronese ring $k[R_{d}]$ . After regrading, we view $A\subset B$ as standard-graded $k$ -algebras.

Proof. Inside the Rees ring ${\mathcal{R}}:={\mathcal{R}}(I)=R[It]\subset R[t]$ we consider the $k$ -subalgebra $A^{\prime }=k[I_{d}t]$ , which is isomorphic to $A$ . Write $K$ for the quotient field of $A^{\prime }$ . We notice that ${\mathcal{R}}\otimes _{A^{\prime }}K$ is a standard-graded $K$ -algebra with maximal homogeneous ideal $\mathfrak{m}({\mathcal{R}}\otimes _{A^{\prime }}K)$ and $({\mathcal{R}}\otimes _{A^{\prime }}K)_{\mathfrak{m}({\mathcal{R}}\otimes _{A^{\prime }}K)}$ is equal to ${\mathcal{R}}_{\mathfrak{m}{\mathcal{R}}}$ . Therefore, the local ring ${\mathcal{R}}_{\mathfrak{m}{\mathcal{R}}}$ and the standard-graded $K$ -algebra ${\mathcal{R}}\otimes _{A^{\prime }}K$ have the same multiplicity and the same Krull dimension. This dimension is one by Remark 5.2.

Let $g$ be a nonzero homogeneous element of $I_{d}$ . Notice that $g$ is a homogeneous nonzerodivisor of degree $d$ in the one-dimensional standard-graded $K$ -algebra ${\mathcal{R}}\otimes _{A^{\prime }}K$ . We obtain

$$\begin{eqnarray}\displaystyle e({\mathcal{R}}\otimes _{A^{\prime }}K)=e(({\mathcal{R}}\otimes _{A^{\prime }}K)^{(d)}) & = & \displaystyle [\vphantom{E_{E_{E}}^{E_{E}}}({\mathcal{R}}\otimes _{A^{\prime }}K)^{(d)}:K[g]\vphantom{E_{E_{E}}^{E_{E}}}]\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{d}[({\mathcal{R}}\otimes _{A^{\prime }}K):K[g]\vphantom{E_{E_{E}}^{E_{E}}}]\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{d}[R[t]:A[t]\vphantom{E_{E_{E}}^{E_{E}}}].\nonumber\end{eqnarray}$$

The last equality holds because

$$\begin{eqnarray}\text{Quot}(K[g])=\text{Quot}(k[I_{d}t,g])=\text{Quot}(k[I_{d},t])=\text{Quot}(A[t])\end{eqnarray}$$

and $\text{Quot}({\mathcal{R}}\otimes _{A^{\prime }}K)=\text{Quot}(R[t])$ . Now we conclude that

$$\begin{eqnarray}e({\mathcal{R}}_{~\mathfrak{m}{\mathcal{R}}})=e({\mathcal{R}}\otimes _{A^{\prime }}K)=\frac{1}{d}[R:A]=[B:A].\end{eqnarray}$$

This concludes the proof of the first equality.

The second equality follows from [Reference Simis, Ulrich and Vasconcelos49, 6.1(a) and 3.4]. Here we give a self-contained proof. We write ${\mathcal{G}}:=\text{gr}_{I}(R)={\mathcal{R}}/I{\mathcal{R}}$ for the associated graded ring of $I$ . Notice that ${\mathcal{G}}/\mathfrak{m}{\mathcal{G}}\cong {\mathcal{R}}/\mathfrak{m}{\mathcal{R}}\cong A$ . In particular, $\mathfrak{m}{\mathcal{G}}$ is a prime ideal in ${\mathcal{G}}$ and ${\mathcal{G}}_{\mathfrak{m}{\mathcal{G}}}$ is Artinian, because $A$ is a domain and $\text{dim}\,A=s=\text{dim}\,{\mathcal{G}}$ .

The $j$ -multiplicity of $I$ is

$$\begin{eqnarray}j(I)=e(0:_{{\mathcal{G}}}\mathfrak{m}^{\infty }).\end{eqnarray}$$

Since $\text{Supp}_{{\mathcal{G}}}(0:_{{\mathcal{G}}}\mathfrak{m}^{\infty })=V(\mathfrak{m}{\mathcal{G}})$ , the associativity formula for multiplicity 2.7 gives

$$\begin{eqnarray}e(0:_{{\mathcal{G}}}\mathfrak{m}^{\infty })=\unicode[STIX]{x1D706}_{{\mathcal{G}}_{\mathfrak{ m}{\mathcal{G}}}}((0:_{{\mathcal{G}}}\mathfrak{m}^{\infty })_{\mathfrak{ m}{\mathcal{G}}})\cdot e({\mathcal{G}}/\mathfrak{m}{\mathcal{G}})=\unicode[STIX]{x1D706}({\mathcal{G}}_{\mathfrak{m}{\mathcal{G}}})\cdot e(A).\end{eqnarray}$$

We conclude that

$$\begin{eqnarray}j(I)=\unicode[STIX]{x1D706}({\mathcal{G}}_{\mathfrak{m}{\mathcal{G}}})\cdot e(A).\end{eqnarray}$$

(The same equality was proved in [Reference Jeffries, Montaño and Varbaro32, the proof of 3.1].)

If $g\not =0$ is an element of $I_{d}$ as above, then $gt\in {\mathcal{R}}\smallsetminus \mathfrak{m}{\mathcal{R}}$ , hence $I/g=It/gt\subset {\mathcal{R}}$ . It follows that $I{\mathcal{R}}_{~\mathfrak{m}{\mathcal{R}}}=g{\mathcal{R}}_{~\mathfrak{m}{\mathcal{R}}}$ , which gives

$$\begin{eqnarray}{\mathcal{G}}_{\mathfrak{m}{\mathcal{G}}}\cong {\mathcal{R}}_{~\mathfrak{m}{\mathcal{G}}}/(g).\end{eqnarray}$$

Now

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D706}(G_{\mathfrak{m}{\mathcal{G}}}) & = & \displaystyle \unicode[STIX]{x1D706}({\mathcal{R}}_{\mathfrak{m}{\mathcal{G}}}/(g))=e(({\mathcal{R}}\otimes _{A^{\prime }}K)/(g))=d\cdot e({\mathcal{R}}\otimes _{A^{\prime }}K)\nonumber\\ \displaystyle & = & \displaystyle d\cdot e({\mathcal{R}}_{\mathfrak{m}{\mathcal{R}}})=d\cdot [B:A].\nonumber\end{eqnarray}$$

We conclude that $j(I)=d\cdot [B:A]\cdot e(A)$ .◻

Proof. We apply the notation of Data 5.1 to the present setting. The ideal $I$ is generated by forms of degree $d-1$ , and the algebra $A$ is the homogeneous coordinate ring of $X^{\prime }\subset \mathbb{P}_{k}^{n-1^{\prime }}$ . According to [Reference Kleiman, Carrell, Geramita and Russell33, Theorem 4] or [Reference Piene43, Proposition 3.3], the extension $A\subset B$ is birational, that is, $[B:A]=1$ . Now Theorem 5.3 shows that $j(I)=(d-1)\text{deg}\,X^{\prime }$ .◻

Remark.

If $k$ is infinite, then $I$ has a reduction generated by $s$ forms of degree $d$ .

Proof. Notice that $k[J_{d}]$ is a homogeneous Noether normalization of $A=k[I_{d}]$ . Therefore, $e(A)=[k[I_{d}]:k[J_{d}]]$ . On the other hand,

$$\begin{eqnarray}[k[I_{d}]:k[J_{d}]]=\frac{[B:k[J_{d}]]}{[B:k[I_{d}]]}=\frac{e({\mathcal{R}}(J)_{\mathfrak{m}{\mathcal{R}}(J)})}{e({\mathcal{R}}(I)_{\mathfrak{m}{\mathcal{R}}(I)})},\end{eqnarray}$$

where the last equality follows from Theorem 5.3 applied to the two ideals $J$ and $I$ , respectively.◻

Proof. Notice that if the ring extension $A\subset B$ is birational, then $A$ and $B$ have the same transcendence degree over $k$ and therefore, $\text{dim}\,A=\text{dim}\,B=s$ . Furthermore, $\text{dim}\,{\mathcal{R}}(I)_{\mathfrak{m}{\mathcal{R}}(I)}=1$ if and only if $\text{dim}\,A=s$ according to Remark 5.2. Now the asserted equivalences follow from Theorem 5.3 and Corollary 5.5.◻

Proof. Since $I$ is $\mathfrak{m}$ -primary the ring $A$ has Krull dimension $s$ . In particular, $\mathfrak{m}{\mathcal{R}}(I)$ is a prime ideal of height one. If $Q$ is any other height one prime ideal of ${\mathcal{R}}(I)$ , then $Q$ does not contain $\mathfrak{m}$ and hence does not contain $I$ . It follows that ${\mathcal{R}}(I)_{Q}$ is regular. Thus ${\mathcal{R}}(I)$ satisfies Serre’s condition $(R_{1})$ if and only if ${\mathcal{R}}(I)_{\mathfrak{m}{\mathcal{R}}(I)}$ is a discrete valuation ring. Now the equivalence of (1) and (2) follows from Corollary 5.6.

To see the equivalence of (2) and (3) notice that ${\mathcal{R}}(\mathfrak{m}^{d})$ is the integral closure of ${\mathcal{R}}(I)$ . Hence ${\mathcal{R}}(I)$ satisfies Serre’s condition $(R_{1})$ if and only if the ${\mathcal{R}}(I)$ -module ${\mathcal{R}}(\mathfrak{m}^{d})/{\mathcal{R}}(I)$ has codimension at least two, which in turn is equivalent to the equality $\unicode[STIX]{x1D714}_{{\mathcal{R}}(I)}=\unicode[STIX]{x1D714}_{{\mathcal{R}}(\mathfrak{m}^{d})}$ . The second equivalence is obtained by analyzing the long exact sequence of $\text{Ext}_{R\otimes _{k}S}^{\bullet }(-,\unicode[STIX]{x1D714}_{R\otimes _{k}S})$ associated to the short exact sequence

$$\begin{eqnarray}0\longrightarrow {\mathcal{R}}(I)\longrightarrow {\mathcal{R}}(\mathfrak{m}^{d})\longrightarrow {\mathcal{R}}(\mathfrak{m}^{d})/{\mathcal{R}}(I)\longrightarrow 0\end{eqnarray}$$

and bearing in mind that ${\mathcal{R}}(\mathfrak{m}^{d})$ is Cohen–Macaulay. The equivalence of (3) and (4) is explained in Remark 5.7.

To see that (5) is equivalent to the other statements we apply Corollary 5.6. We may assume that $k$ is infinite. It suffices to show the equality $e({\mathcal{R}}(J)_{\mathfrak{m}{\mathcal{R}}(J)})=d^{s-1}$ for the ideal $J$ of Corollary 5.6. Since $I$ is $\mathfrak{m}$ -primary, the ideal $J$ is generated by a regular sequence of $s$ forms of degree  $d$ . Therefore, $[R:k[J_{d}]]=d^{s}$ and then $e({\mathcal{R}}(J)_{\mathfrak{m}{\mathcal{R}}(J)})=[B:k[J_{d}]]=d^{s-1}$ according to Theorem 5.3.

We now show that (2) implies (6). We will repeatedly use the fact from [Reference Polini, Ulrich and Vitulli45, 2.1] that the core of $\mathfrak{m}$ -primary ideals localizes. The first equality follows from [Reference Corso, Polini and Ulrich12, 4.5] as $I$ is generated by forms of the same degree. Recall that the Rees ring $R[It]$ and the extended Rees ring $R[It,t^{-1}]$ have isomorphic projective spectra. Hence if the Rees ring satisfies $R_{1}$ , then so does $\operatorname{Proj}(R[It,t^{-1}])$ . It follows that $R[It,t^{-1}]$ satisfies $R_{1}$ as well because it suffices to consider prime ideals containing $t^{-1}$ and because the ideal $(It,t^{-1})R[It,t^{-1}]$ has height $s+1>1$ . One uses the argument that (2) implies (3) to see that the $R_{1}$ property of the extended Rees ring implies the bigraded isomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D714}_{R[It,t^{-1}]}\cong \unicode[STIX]{x1D714}_{R[\mathfrak{m}^{d}t,t^{-1}]}.\end{eqnarray}$$

By [Reference Polini and Ulrich44, 2.2] and [Reference Fouli, Polini and Ulrich20, 1.2] the core can be recovered from the canonical module of the extended Rees ring. Therefore, $\operatorname{core}(I)=\operatorname{core}(\mathfrak{m}^{d})$ , which is the second equality in (6). The third equality follows from [Reference Corso, Polini and Ulrich13, 4.2]. Finally, one has $\mathfrak{m}^{(d-1)s+1}=\text{adj}(\mathfrak{m}^{ds})$ by [Reference Lipman40, 1.3.2(c)] and $\text{adj}(\mathfrak{m}^{ds})=\text{adj}\,(I^{s})$ because $\mathfrak{m}^{ds}$ is integral over $I^{s}$ .

To see (7), notice that ${\mathcal{R}}(I)$ is normal, hence ${\mathcal{R}}(I)={\mathcal{R}}(\mathfrak{m}^{d})$ . The last ring is Cohen–Macaulay because it is a direct summand of ${\mathcal{R}}(\mathfrak{m})$ , which is a Cohen–Macaulay ring and a finitely generated module over ${\mathcal{R}}(\mathfrak{m}^{d})$ .◻

Proof. From the presentation of the Rees ring

$$\begin{eqnarray}{\mathcal{R}}(\mathfrak{m})=\frac{R[T_{1},\ldots ,T_{s}]}{I_{2}\left(\left[\begin{array}{@{}ccc@{}}x_{1} & \ldots & x_{s}\\ T_{1} & \ldots & T_{s}\end{array}\right]\right)}\end{eqnarray}$$

one computes

$$\begin{eqnarray}\unicode[STIX]{x1D714}_{{\mathcal{R}}(\mathfrak{m})}\cong x_{1}^{2}t(x_{1},x_{1}t)^{s-2}{\mathcal{R}}(\mathfrak{m}),\end{eqnarray}$$

where the factor $x_{1}^{2}t$ is required to make the isomorphism bihomogeneous. The Rees ring ${\mathcal{R}}(\mathfrak{m}^{d})$ is obtained by taking the $d$ -Veronese subring of ${\mathcal{R}}(\mathfrak{m})$ with respect to the grading given by the $T$ -variables and then rescaling the grading. Since the Veronese functor commutes with taking canonical modules, we obtain the desired statement.

The natural inclusion ${\mathcal{R}}(m^{d})\subset \operatorname{End}_{{\mathcal{R}}(m^{d})}(\unicode[STIX]{x1D714}_{{\mathcal{R}}(m^{d})})$ is an equality because ${\mathcal{R}}(m^{d})$ is Cohen–Macaulay and $\operatorname{End}_{{\mathcal{R}}(m^{d})}(\unicode[STIX]{x1D714}_{{\mathcal{R}}(m^{d})})$ is its $S_{2}$ -ification.◻

Proof. Let $A^{\prime }$ be the subring $k[I_{d}t]$ of ${\mathcal{R}}(I)=R[I_{d}t]$ and $C^{\prime }$ be the subring $k[K_{d}t]$ of ${\mathcal{R}}(K)=R[K_{d}t]$ . We first prove that the ideals

(5.10.1) $$\begin{eqnarray}\operatorname{rad}({\mathcal{R}}(I):_{{\mathcal{R}}(I)}{\mathcal{R}}(K))\quad \text{and}\quad \operatorname{rad}((\mathfrak{m},A^{\prime }:_{A^{\prime }}C^{\prime }){\mathcal{R}}(I))\end{eqnarray}$$

of ${\mathcal{R}}(I)$ are equal.

We prove the inclusion “ $\supset$ ” for the ideals of (5.10.1). The Rees algebra ${\mathcal{R}}(K)$ is a module-finite extension of ${\mathcal{R}}(I)$ because $I$ is $\mathfrak{m}$ -primary; hence, $I$ is a reduction of $K$ and there is a positive integer $t$ with $K^{t}K^{n}\subset I^{n}$ for all nonnegative integers $n$ . It follows that $K^{t}\subset {\mathcal{R}}(I):_{{\mathcal{R}}(I)}{\mathcal{R}}(K)$ ; and therefore, $\mathfrak{m}\subset \operatorname{rad}({\mathcal{R}}(I):_{{\mathcal{R}}(I)}{\mathcal{R}}(K))$ . Clearly, $A^{\prime }:_{A^{\prime }}C^{\prime }\subset {\mathcal{R}}(I):_{{\mathcal{R}}(I)}{\mathcal{R}}(K)$ because ${\mathcal{R}}(I)=R[A^{\prime }]$ and ${\mathcal{R}}(K)=R[C^{\prime }]$ .

We prove the inclusion “ $\subset$ ” for the ideals of (5.10.1). Let $\unicode[STIX]{x1D6FC}$ be a bihomogeneous element of ${\mathcal{R}}(I):_{{\mathcal{R}}(I)}{\mathcal{R}}(K)$ . In particular, $\unicode[STIX]{x1D6FC}=ft^{j}$ for some homogeneous form $f$ in $I$ and some nonnegative integer $j$ . If the homogeneous form $f$ has degree $i+dj$ as a polynomial in $R=k[x_{1},\ldots ,x_{s}]$ , then the bi-degree of $\unicode[STIX]{x1D6FC}$ in ${\mathcal{R}}(I)$ is $(i,j)$ . If $i$ is positive, then $\unicode[STIX]{x1D6FC}$ is in $\mathfrak{m}{\mathcal{R}}(I)$ . If $i=0$ , then $\unicode[STIX]{x1D6FC}$ is in $A^{\prime }$ and $\unicode[STIX]{x1D6FC}\cdot C^{\prime }\subset \unicode[STIX]{x1D6FC}\cdot {\mathcal{R}}(K)\subset {\mathcal{R}}(I)$ . One may compute $x$ -degree to see that $\unicode[STIX]{x1D6FC}\cdot C^{\prime }$ is actually contained in $A^{\prime }$ . In either case, $\unicode[STIX]{x1D6FC}\in (\mathfrak{m},A^{\prime }:_{A^{\prime }}C^{\prime }){\mathcal{R}}(I)$ .

Now that the assertion of (5.10.1) has been established, we complete the proof of the result by applying the natural quotient homomorphism ${\mathcal{R}}(I)\rightarrow {\mathcal{R}}(I)/\mathfrak{m}{\mathcal{R}}(I)$ to the ideals of (5.10.1). The ring ${\mathcal{R}}(I)$ is a domain and a finitely generated algebra over a field, and the ideal $\mathfrak{m}{\mathcal{R}}(I)$ of ${\mathcal{R}}(I)$ has height $1$ . Thus,

$$\begin{eqnarray}\begin{array}{@{}lllll@{}}\operatorname{ht}({\mathcal{R}}(I):_{{\mathcal{R}}(I)}{\mathcal{R}}(K))-1\, & =\, & \operatorname{ht}\frac{\operatorname{rad}({\mathcal{R}}(I):_{{\mathcal{R}}(I)}{\mathcal{R}}(K))}{\mathfrak{m}{\mathcal{R}}(I)}\, & =\, & \operatorname{ht}(\mathfrak{m},A^{\prime }:_{A^{\prime }}C^{\prime })\frac{{\mathcal{R}}(I)}{\mathfrak{m}{\mathcal{R}}(I)}\\ \, & =\, & \operatorname{ht}({\mathcal{F}}(I):_{{\mathcal{F}}(I)}{\mathcal{F}}(K))\, & =\, & \operatorname{ht}(A:_{A}C).\end{array}\end{eqnarray}$$

Proof. When $i=1$ , the statement is $(1)\;\Longleftrightarrow \;(2)$ from Corollary 5.8. Henceforth, we assume $2\leqslant i$ and that $A\subset B$ is a birational extension. Thus, both extensions ${\mathcal{R}}(I)\subset {\mathcal{R}}(\mathfrak{m}^{d})$ and $A\subset B$ are module-finite and birational.

Moreover, $\dim {\mathcal{R}}(I)=s+1$ and ${\mathcal{R}}(\mathfrak{m}^{d})$ has an isolated singularity. Hence for any prime ideal $Q$ of ${\mathcal{R}}(I)$ with $\dim {\mathcal{R}}(I)_{Q}\leqslant s$ , the ring ${\mathcal{R}}(I)_{Q}$ is regular if and only if ${\mathcal{R}}(I)_{Q}={\mathcal{R}}(\mathfrak{m}^{d})_{Q}$ . This shows that ${\mathcal{R}}(I)$ satisfies $(R_{i})$ if and only if $i<\operatorname{ht}({\mathcal{R}}(I):_{{\mathcal{R}}(I)}{\mathcal{R}}(\mathfrak{m}^{d}))$ .

Similarly, $\dim A=s$ and $B$ has an isolated singularity. Hence again, $A_{q}$ is regular if and only if $A_{q}=B_{q}$ for any prime ideal $q$ of $A$ with $\dim A_{q}\leqslant s-1$ . We conclude that $A$ satisfies $(R_{i-1})$ if and only if $i-1<\operatorname{ht}(A:_{A}B)$ .