Proof. Consider an element $x{\kern-1pt}\in{\kern-1pt} {(\mathfrak {a}{\kern-1pt}+{\kern-1pt}\mathfrak {b})^r}/{\mathfrak {b}}$ . Then, $x{\kern-1pt}={\kern-1pt}\sum x_i\cdots x_r{\kern-1pt}+{\kern-1pt}b$ , where ${x_i{\kern-1pt}\in{\kern-1pt} \mathfrak {a}+\mathfrak {b}}$ and $b\in \mathfrak {b}$ . Let $x_i=a_i+b_i$ and $x_j=a_j+b_j$ be elements in $\mathfrak {a}+\mathfrak {b}$ . We note that $x_i\cdot x_j=~a_1a_2+b_{ij}$ , so that $x_ix_j\in \mathfrak {a}^r+\mathfrak {b}$ . Thus, $\sum x_1\cdots x_r\in \mathfrak {a}^r+\mathfrak {b}$ and, therefore, $x\in ~{(\mathfrak {a}^r+\mathfrak {b})}/{\mathfrak {b}}$ .

On the other hand, if $x\in {(\mathfrak {a}^r+\mathfrak {b})}/{\mathfrak {b}}$ , then $x=(\sum a_1\cdots a_r+b)+\overline {b}$ , where $a_i\in \mathfrak {a}$ and $b,\overline {b}\in \mathfrak {b}$ . Since $\mathfrak {a}\subset \mathfrak {a}+\mathfrak {b}$ , the $a_i$ are elements in $\mathfrak {a}+\mathfrak {b}$ . Therefore, $\sum a_i\cdots a_r\in (\mathfrak {a}+\mathfrak {b})^r$ giving the other inclusion.

Proof. Consider the exact sequence

$$ \begin{align*} 0\to\mathfrak{a}^r\to A\to A/\mathfrak{a}^r\to 0. \end{align*} $$

Tensoring by $A/\mathfrak {b}$ ,

$$ \begin{align*} A/\mathfrak{b}\otimes \mathfrak{a}^r\to A/\mathfrak{b}\to A/\mathfrak{a}^r\otimes A/\mathfrak{b}\to 0. \end{align*} $$

The kernel of the first morphism is $(\mathfrak {a}^r+\mathfrak {b})/\mathfrak {b}$ , so Lemma 2.1 gives the result.

Proof. The inclusion $h\mathfrak {p}^r\subseteq (h)\cap \mathfrak {p}^r$ is always true. We need to show the other inclusion.

Let x be an element in $(h)\cap \mathfrak {p}^r$ . Since $x\in (h)$ , we have $x=ha$ for some $a\in A$ . As $h\not \in \mathfrak {p}$ , then $h\not \in \mathfrak {p}^r$ . Now, $\mathfrak {p}^r$ is primary by [Reference Hochster3, Theorem 2.1] and as $x=ha\in \mathfrak {p}^r$ , we have $a\in \mathfrak {p}^r$ . Therefore, $x=ha\in h\mathfrak {p}^r$ and we obtain the result.

Proof. The assertion follows from the isomorphisms

$$ \begin{align*} H^0(X,\mathcal{A}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\otimes\mathcal{O}_X/\mathcal{I}_{X'})\simeq H^0(X',(\mathcal{A}\otimes\mathcal{O}_X/\mathcal{I}_Z^r)|_{X'})\simeq H^0(X',\mathcal{A}|_{X'}\otimes\mathcal{O}_{X'}/\mathcal{I}_{Z'}^r), \end{align*} $$

where the second isomorphism follows from Lemma 2.2.

Proof. By [Reference Hartshorne2, Theorem 8.18, II], we can choose an element $f\in H^0(X,\mathcal {A})$ such that the zero locus of f, denoted by $X'$ , is a smooth and irreducible subvariety of X, and such that if $d\geq 2$ , then $Z'=Z\cap X'$ is irreducible.

Consider now the exact sequence

$$ \begin{align*} 0\to\mathcal{I}_{X'}\to\mathcal{O}_X\to\mathcal{O}_{X}/\mathcal{I}_{X'}\to 0. \end{align*} $$

Tensoring by $\mathcal {O}_X/\mathcal {I}_Z^r$ , we obtain

(3.1) $$ \begin{align} \cdots\to \mathrm{Tor}^{\mathcal{O}_X}(\mathcal{O}_X/\mathcal{I}_{X'}~;\mathcal{O}_X/\mathcal{I}_Z^r)\to\mathcal{I}_{X'}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{O}_X/\mathcal{I}_{X'}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to0. \end{align} $$

Lemma 2.3 implies $\mathrm {Tor}^{\mathcal {O}_X}(\mathcal {O}_X/\mathcal {I}_{X'}~;\mathcal {O}_X/\mathcal {I}_Z^r)$ is trivial, so the sequence is

$$ \begin{align*} 0\to\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{I}_{X'}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{O}_X/\mathcal{I}_{X'}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to0. \end{align*} $$

Since $\mathcal {A}$ is flat,

$$ \begin{align*} 0\to\mathcal{A}\otimes\mathcal{I}_{X'}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{A}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{A}\otimes\mathcal{O}_X/\mathcal{I}_{X'}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to0. \end{align*} $$

Finally, by the choice of f, we have $\mathcal {I}_{X'}=\mathcal {A}^\vee $ . Thus, the final sequence is

$$ \begin{align*} 0\to\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{A}\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to\mathcal{A}\otimes\mathcal{O}_X/\mathcal{I}_X'\otimes\mathcal{O}_X/\mathcal{I}_Z^r\to0. \end{align*} $$

Passing to global sections, we have the equality

$$ \begin{align*} h^0(X,\mathcal{A}\otimes\mathcal{O}_X/\mathcal{I}_Z^r) &= h^0(X,\mathcal{A}\otimes\mathcal{O}/\mathcal{I}_Z^r\otimes\mathcal{O}_X/\mathcal{I}_{X'})+h^0(X,\mathcal{O}_X/\mathcal{I}_Z^r) \\ &= h^0(X',\mathcal{A}|_{X'}\otimes\mathcal{O}_{X'}/\mathcal{I}_{Z'}^r)+h^0(X,\mathcal{O}_X/\mathcal{I}_Z^r) \quad\text{(by Lemma }{3.1}). \end{align*} $$

Now, consider $f'\in H^0(X',\mathcal {A}|_{X'})$ with the same properties as f and denote by $X^{\prime }_1$ the zero locus of $f'$ and $Z^{\prime }_1=Z'\cap X^{\prime }_1$ . Then,

$$ \begin{align*} h^0(X',\mathcal{A}|_{X'}\otimes\mathcal{O}_{X'}/\mathcal{I}_{Z'}^r)=h^0(X_1',\mathcal{A}|_{X_1'}\otimes\mathcal{O}_{X_1'}/\mathcal{I}_{Z_1'}^r)+h^0(X',\mathcal{O}_{X'}/\mathcal{I}_{Z_1'}^r). \end{align*} $$

Sequence (3.1) implies $h^0(X,\mathcal {O}_X/\mathcal {I}_Z^r)\geq h^0(X',\mathcal {O}_{X'}/\mathcal {I}_{Z_1'}^r)$ .

We can repeat this argument until we obtain a curve Y and a set of points W obtained by the intersection of X and Z, respectively, with hyperplanes that come from $\mathcal {A}$ as described before. Then, we need to compute $h^0(Y,\mathcal {O}_Y/\mathcal {I}_W^r)$ . This dimension is equal to $\#W\cdot \binom {c+r-1}{c}$ , that is, $(\mathcal {A}^d\cdot Z)\cdot \binom {c+r-1}{c}$ . For r sufficiently large, the last binomial coefficient is equal to ${r^c}/{c!}+O(r^{c-1})$ . Thus, inductively,

$$ \begin{align*} h^0(X,\mathcal{A}\otimes\mathcal{O}_X/\mathcal{I}_Z^r)\leq \frac{r^c}{c!}(\mathcal{A}^d\cdot Z)+d\cdot h^0(X,\mathcal{O}_X/\mathcal{I}_Z^r)+O((\mathcal{A}^d\cdot Z)r^{c-1}). \end{align*} $$

By the assumption on $\mathcal {N}_{Z/X}$ , [Reference Lazarsfeld4, Proposition 6.3.13] implies $h^0(X,\mathcal {O}_X/\mathcal {I}_Z^r)\leq 1$ , which leads to the desired inequality.