A Hilbert-Kunz function with a periodic term that has a given period

TL;DR

This paper demonstrates that for any positive integer period, there exists a one-dimensional local ring whose Hilbert-Kunz function's periodic term has exactly that period, expanding understanding of periodicity in algebraic functions.

Contribution

The paper constructs examples of rings with Hilbert-Kunz functions exhibiting any prescribed immediate periodicity, generalizing previous specific cases.

Findings

Existence of rings with Hilbert-Kunz periodic term of any given period

Explicit construction of rings with specified periodicity

Extension of known periodicity results to all positive integers

Abstract

A result of Monsky states that the Hilbert-Kunz function of a one-dimensional local ring of prime characteristic has a term $\phi$ that is eventually periodic. For example, in the case of a power series ring in one variable over a prime-characteristic field, $\phi$ is the zero function and is therefore immediately periodic with period 1. In additional examples produced by Kunz and Monsky, $\phi$ is immediately periodic with period 2. We show that, for every positive integer $\pi$, there exists a ring for which $\phi$ is immediately periodic with period $\pi$.

Equations

$$H{K}_{\mathfrak{m},R}(e)=e_{HK}(\mathfrak{m},R)\cdot p^e-{\varphi }_{\mathfrak{m},R}(e)$$

$$x^q-x^b{y}^{q-b}=(x^{q-n}{y}^0+x^{q-2n}{y}^n+\cdots +x^b{y}^{q-b-n})(x^n-y^n)\in I\cap (G)$$

$$S(x^b{y}^{q-b},y^q)=y^b(x^b{y}^{q-b})-x^b(y^q)=0$$

$$S(x^b{y}^{q-b},x^n-y^n)=x^{n-b}(x^b{y}^{q-b})-y^{q-b}(x^n-y^n)=y^{q+n-b}$$

$$S(y^q,x^n-y^n)=x^n(y^q)-y^q(x^n-y^n)=y^{q+n}$$

$$\varphi (e+(\omega /2))=(n-c)[n-(n-c)]=c(n-c)=\varphi (e).$$

$$H{K}_{\mathfrak{m},R}(e)=\{\begin{array}{ll}5p^e-4& \text{if e is even;}\\ 5p^e-6& \text{if e is odd.}\end{array}$$

$$H{K}_{\mathfrak{m},R}(e)=\{\begin{array}{ll}7p^e-6& \text{if e≡0 (mod 3);}\\ 7p^e-10& \text{if e≡1 (mod 3);}\\ 7p^e-12& \text{if e≡2 (mod 3).}\end{array}$$

$$H{K}_{\mathfrak{m},R}(e)=\{\begin{array}{ll}15p^e-14& \text{if e≡0 (mod 4);}\\ 15p^e-26& \text{if e≡1 (mod 4);}\\ 15p^e-44& \text{if e≡2 (mod 4);}\\ 15p^e-56& \text{if e≡3 (mod 4).}\end{array}$$

Full text

The author would like to thank Florian Enescu for sharing the interesting example of Monsky that inspired this paper. The author would also like to thank Yongwei Yao, Felipe Pérez, Siang Ng, and Irina Ilioaea for helpful comments during a presentation of these results at Georgia State University.

Let $(R,\mathfrak{m})$ be a one-dimensional commutative Noetherian local ring of prime characteristic $p$. In this note, we study the Hilbert–Kunz function $HK_{\mathfrak{m},R}$, which sends every nonnegative integer $e$ to the length of $R/\mathfrak{m}^{[p^{e}]}$ over $R$, where $\mathfrak{m}^{[p^{e}]}:=(r^{p^{e}}:r\in\mathfrak{m})$.

The general form of $HK_{\mathfrak{m},R}$ is known: Monsky has proved that there exist a positive integer $\operatorname{e}_{HK}(\mathfrak{m},R)$, called the Hilbert–Kunz multiplicity of $R$, and an eventually periodic function $\phi_{\mathfrak{m},R}$ such that

[TABLE]

for every nonnegative integer $e$ [4, Theorem 3.10].

Also, there are algorithms for specialized settings: Kreuzer has covered the case in which $R=\hat{S}^{M}$, where $S$ is a standard graded algebra over a field and $M$ is the maximal homogeneous ideal of $S$ [2, Algorithm 5.4]. Upadhyay has obtained a formula valid in the event that $S$ is a polynomial ring over a field modulo a binomial with zero constant term (and $R=\hat{S}^{M}$ with $M$ the maximal homogeneous ideal of $S$ once again) [5, Theorem 26].

Despite these advances, explicit values of $HK_{\mathfrak{m},R}$ are still scarce. Moreover, when values are available, $\phi_{\mathfrak{m},R}$ does not differ much from one example to the next: For instance, in the six computations achieved by Kreuzer [2, Examples 6.1–6.6], the function $\phi_{\mathfrak{m},R}$ is eventually constant in four cases, and in the remaining two examples, which are attributable to Kunz [3, Example 4.6b] and Monsky [4, page 46], the function $\phi_{\mathfrak{m},R}$ is immediately periodic with period 2.

In response to the lack of diversity in concrete examples of $\phi_{\mathfrak{m},R}$, we show here that, for every positive integer $\pi$, there exists a ring $(R,\mathfrak{m})$ for which $\phi_{\mathfrak{m},R}$ is immediately periodic with period $\pi$. To accomplish our goal, we first provide an infinite family of rings whose Hilbert–Kunz functions we can compute explicitly. To illustrate the content of our result, we recover Monsky’s example with period 2 and give new examples with periods 3 and 4. We then appeal to a classical observation of Dirichlet to prove our main theorem.

1. (1)

Let $e$ be a nonnegative integer, and let $b$ be an integer satisfying $p^{e}\equiv b$ (mod $n$) and $1\leqslant b\leqslant n-1$. Then $HK_{\mathfrak{m},R}(e)=np^{e}-b(n-b)$. 2. (2)

$\phi_{\mathfrak{m},R}$* is immediately periodic with $\pi\in\{\omega/2,\omega\}$.* 3. (3)

$\pi=\omega/2$* if and only if $\omega$ is even and $p^{\omega/2}\equiv-1$ (mod $n$).*

If $q<n$, then $I=(x^{q},y^{q})$, and so $\dim_{k}(T)=q^{2}=nq-b(n-b)$, completing the proof. Of course, $q\neq n$ since, by hypothesis, $n$ is not divisible by $p$. Therefore, we may assume that $q>n$.

Next, we establish that $G:=\{x^{b}y^{q-b},y^{q},x^{n}-y^{n}\}$ is a Gröbner basis for $I$ with respect to lex($x>y$), the lexicographic monomial order induced by the relation $x>y$. Certainly, $I$ is equal to the ideal of $k[x,y]$ generated by $G$ since

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and since $y^{q},x^{n}-y^{n}\in I\cap(G)$ by definition. To verify that $G$ satisfies Buchberger’s Criterion, we note that

[TABLE]

and that

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and

[TABLE]

are divisible by $y^{q}$ and no other member of $G$. Hence $G$ is a Gröbner basis for $I$ with respect to lex($x>y$).

Now let $A$ be the set of monomials in $k[x,y]$ not divisible by any of the leading terms of the members of $G$, and let - denote the natural projection of $k[x,y]$ onto $T$. Then $\{\overline{f}:f\in A\}$ is a basis for $T$ over $k$. Hence $\dim_{k}(T)=|A|=nq-b(n-b)$.

(2) Write $\phi:=\phi_{\mathfrak{m},R}$. Then Part (1) implies that $\phi$ is immediately periodic with $\pi$ dividing $\omega$. Hence $\phi(\pi)=\phi(0)$. Part (1) now gives us $p^{\pi}(n-p^{\pi})\equiv 1(n-1)$ (mod $n$), which implies that $p^{2\pi}\equiv 1$ (mod $n$), which in turn implies that $\omega$ divides $2\pi$. Hence $\pi\in\{\omega/2,\omega\}$.

(3) Let $\phi:=\phi_{\mathfrak{m},R}$, and suppose that $\pi=\omega/2$. Since $\pi$ is an integer, $\omega$ must be even. Let $b$ be an integer such that $p^{\pi}\equiv b$ (mod $n$) and $1\leqslant b\leqslant n-1$. Since Part (2) yields $\phi(0)=\phi(\pi)$, we get $1(n-1)=b(n-b)$. Solving for $b$, we get $b\in\{1,n-1\}$. If $b=1$, then $p^{\pi}\equiv b\equiv 1$ (mod $n$), and so $\omega$ divides $\pi=\omega/2$, a contradiction. Hence $b=n-1$, and so $p^{\omega/2}=p^{\pi}\equiv b\equiv-1$ (mod $n$).

Conversely, suppose that $\omega$ is even and that $p^{\omega/2}\equiv-1$ (mod $n$). Let $e$ be an integer with $0\leqslant e\leqslant(\omega/2)-1$, and let $c$ be an integer satisfying $p^{e}\equiv c$ (mod $n$) and $1\leqslant c\leqslant n-1$. Since $p^{\omega/2}\equiv-1$ (mod $n$) by hypothesis, $p^{[e+(\omega/2)]}\equiv-c\equiv n-c$ (mod $n$). So, by Part (1),

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We have thus shown that $\pi$ divides $\omega/2$. Now, by Part (2), we get $\pi=\omega/2$. ∎

Theorem 1 recovers the following example of Monsky in which $\pi=\omega/2$:

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Hence $\pi=2=\omega/2$.

The next example presents a case in which $\pi=3=\omega$.

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Hence $\pi=3=\omega$.

Our final example illustrates that the congruence condition in Part (3) of Theorem 1 is not redundant:

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Hence $\pi=4=\omega$.

We now recall a result of Dirichlet and prove our main theorem.

In light of our observations, the following question naturally arises: