The variety defined by the matrix of diagonals is $F$-pure

TL;DR

This paper proves that the algebraic variety defined by the determinant of the matrix of diagonals is $F$-pure across all matrix sizes and prime characteristics, and also identifies a system of parameters for it.

Contribution

It establishes the $F$-purity of the variety defined by the determinant of the matrix of diagonals for all sizes and characteristics, and finds a system of parameters.

Findings

The variety is $F$-pure in all positive prime characteristics.

A system of parameters for the variety is explicitly constructed.

The result applies universally to matrices of all sizes.

Abstract

We prove that the variety defined by the determinant of the matrix of diagonals is $F$-pure for matrices of all sizes and in all positive prime characteristics. Moreover, we find a system of parameters for it.

Equations

$$X=\left[\begin{array}[]{cccc}x_{11}&x_{12}&\ldots&x_{1n}\\ &\ldots&\ldots&\\ x_{n1}&x_{n2}&\ldots&x_{nn}\end{array}\right]$$

$$\tilde{X}=\left[\begin{array}[]{c|c}X_{0}&0\\ \hline\cr 0&x_{nn}\end{array}\right]$$

$$X_{0}=\left[\begin{array}[]{cccc}x_{11}&\ldots&x_{1,n-1}\\ &\ldots&\\ x_{n-1,1}&\ldots&x_{n-1,n-1}\end{array}\right].$$

$$\det \left[\text{diag}(I)\text{diag}(A)\dots \text{diag}(A^{n-1})\right]=\pm 1\text{ if and only if}$$

$$\text{diag}(I),\text{diag}(A),\dots ,\text{diag}(A^{n-1})\text{ span }{\mathbb{Z}}^n\text{ if and only if}$$

$$A=\left[\begin{array}[]{ccccc}1&1&\ldots&1&1\\ 1&1&\ldots&1&0\\ &\ldots&\ldots&&\\ 1&0&\ldots&0&0\\ \end{array}\right].$$

$$B=A^{-1}=\left[\begin{array}[]{ccccc}0&0&\ldots&0&1\\ 0&0&\ldots&1&-1\\ &\ldots&\ldots&&\\ 1&-1&\ldots&0&0\\ \end{array}\right].$$

$$B^{2}=\left[\begin{array}[]{cccccc}1&-1&0&\ldots&0&0\\ -1&2&-1&\ldots&0&0\\ &\ldots&\ldots&&\\ 0&0&0&\ldots&2&-1\\ 0&0&0&\ldots&-1&2\\ \end{array}\right].$$

$$\left[\begin{array}[]{c|c|c}0&0\ldots 0&1\\ \hline\cr 0&&-1\\ 0&\raisebox{-15.0pt}{{\large\mbox{{$B_{0}$}}}}&0\\[-15.49994pt] \vdots&&\vdots\\ 0&&0\\ \hline\cr 1&-1\,0\ldots 0&0\\ \end{array}\right]$$

$$B^{2}=\left[\begin{array}[]{c|c|c}1&-1\quad 0\ldots\ldots\ldots 0\quad 0&0\\ \hline\cr-1&&0\\ 0&\raisebox{-15.0pt}{{\large\mbox{{$B_{0}^{2}+\left[\begin{array}[]{cccc}1&0&\ldots&0\\ 0&0&\ldots&0\\ 0&0&\ldots&0\\ \end{array}\right]$}}}}&0\\[-15.49994pt] \vdots&&\vdots\\ 0&&-1\\ \hline\cr 0&0\quad 0\ldots\ldots\ldots 0\,\,-1&2\\ \end{array}\right].$$

$$(B^{2j-1})_{kl}=\left\{\begin{array}[]{cc}(-1)^{j+1}&\text{ if }k+l=n-j+2,\\ (-1)^{j}&\text{ if }k+l=n+j+1,\\ 0&\text{ if }k+l\leq n-j+1\text{ or }k+l\geq n+j+2.\end{array}\right.$$

$$B^{2(j+1)-1}=B^{2j+1}=B^2{B}^{2j-1}.$$

$$(B^{2j+1}{)}_{kl}=(B^2{B}^{2j-1}{)}_{kl}=\sum _{s=1}^n(B^2{)}_{ks}(B^{2j-1}{)}_{sl}=$$

$$(B^2{)}_{kk}(B^{2j-1}{)}_{kl}+(B^2{)}_{k,k+1}(B^{2j-1}{)}_{k+1,l}+(B^2{)}_{k,k-1}(B^{2j-1}{)}_{k-1,l}.$$

$$(B^{2j+1}{)}_{kl}=(B^2{)}_{kk}\cdot 0+(-1)(-1{)}^{j+1}+(-1)\cdot 0=(-1{)}^{j+2}.$$

$$(B^{2j+1}{)}_{kl}=(B^2{)}_{kk}\cdot 0+(-1)\cdot 0+(-1)(-1{)}^j=(-1{)}^{j+1}.$$

$$(B^{2j+1}{)}_{kl}=(B^2{)}_{kk}\cdot 0+(-1)\cdot 0+(-1)\cdot 0=0.$$

$$\left[\text{diag}(B)\text{diag}(B^3)\dots \text{diag}(B^{2j-1})\dots \text{diag}(B^{2n-1})\right]=$$

$$\left[\begin{array}[]{cccccccccccc}0&0&0&0&0&&0&0&0&1\\ 0&0&0&0&0&&0&1&*&*\\ 0&0&0&0&0&&*&*&*&*\\ &&\ldots&&&&&\ldots\\ 0&0&0&0&1&&*&*&*&*\\ 0&0&1&*&*&&*&*&*&*\\ 1&&*&*&*&\ldots&*&*&*&*\\ 0&1&*&*&*&&*&*&*&*\\ 0&0&0&1&*&&*&*&*&*\\ &&\ldots&&&&&\quad\ldots\\ 0&0&0&0&0&&*&*&*&*\\ 0&0&0&0&0&&1&*&*&*\\ 0&0&0&0&0&&0&*&1&*\\ \end{array}\right],$$

$$\left[\begin{array}[]{cccccccccccc}0&0&0&0&&0&0&0&-1\\ 0&0&0&0&&0&-1&*&*\\ 0&0&0&0&&*&*&*&*\\ &&\ldots&&&&&\ldots\\ 0&0&0&-1&&*&*&*&*\\ 0&-1&*&*&&*&*&*&*\\ -1&*&*&*&&*&*&*&*\\ 0&0&-1&*&&*&*&*&*\\ \ldots&&&&&\quad\ldots\\ 0&0&0&0&&*&*&*&*\\ 0&0&0&0&&-1&*&*&*\\ 0&0&0&0&&0&*&-1&*\\ \end{array}\right].$$

$$\Lambda =\left\{(i,j)|1\le i\le n,n+1-i\le j\le n\right\}$$

$$\Omega =\left\{(k,l)|1\le k\le n-1,1\le l\le n-k,(k,l)\ne (1,1)\right\}.$$

$$\mathcal{S}=\left\{x_{ij},x_{11}-x_{kl}|(i,j)\in \Lambda ,(k,l)\in \Omega \right\}$$

$$X=\left[\begin{array}[]{cc}x_{11}&0\\ 0&0\end{array}\right]$$

$$\mathcal{P}(X)=\det\left[\begin{array}[]{cc}1&x_{11}\\ 1&0\end{array}\right]=-x_{11}.$$

$$X=\left[\begin{array}[]{ccc}x_{11}&x_{11}&0\\ x_{11}&0&0\\ 0&0&0\end{array}\right],\,X^{2}=\left[\begin{array}[]{ccc}2x_{11}^{2}&x_{11}^{2}&0\\ x_{11}^{2}&x_{11}^{2}&0\\ 0&0&0\end{array}\right]$$

$$\mathcal{P}(X)=\det\left[\begin{array}[]{ccc}1&x_{11}&2x_{11}^{2}\\ 1&0&x_{11}^{2}\\ 1&0&0\end{array}\right]=x_{11}^{3}.$$

$$X=\left[\begin{array}[]{cccc}x_{11}&x_{11}&x_{11}&0\\ x_{11}&x_{11}&0&0\\ x_{11}&0&0&0\\ 0&0&0&0\end{array}\right],\,X^{2}=\left[\begin{array}[]{cccc}3x_{11}^{2}&2x_{11}^{2}&x_{11}^{2}&0\\ 2x_{11}^{2}&2x_{11}^{2}&x_{11}^{2}&0\\ x_{11}^{2}&x_{11}^{2}&x_{11}^{2}&0\\ 0&0&0&0\end{array}\right],$$

$$X^{3}=\left[\begin{array}[]{cccc}6x_{11}^{3}&5x_{11}^{3}&3x_{11}^{3}&0\\ 5x_{11}^{3}&4x_{11}^{3}&2x_{11}^{3}&0\\ 3x_{11}^{3}&2x_{11}^{3}&x_{11}^{3}&0\\ 0&0&0&0\end{array}\right],\text{ and }\mathcal{P}(X)=\det\left[\begin{array}[]{cccc}1&x_{11}&3x_{11}^{2}&6x_{11}^{3}\\ 1&x_{11}&2x_{11}^{2}&4x_{11}^{3}\\ 1&0&x_{11}^{2}&x_{11}^{3}\\ 1&0&0&0\end{array}\right]=-x_{11}^{6}.$$

$$\{x_{ij},x_{11}-x_{kl}|(i,j)\in \Lambda ,(k,l)\in \Omega \}$$

Full text

*Keywords: * Frobenius, singularities, $F$-purity, system of parameters.

1 Introduction and preliminaries

Let $X=(x_{ij})_{1\leq i,\,j\leq n}$ be a square matrix of size $n$ with indeterminate entries over a field $K$ and $R=K[X]$ be the polynomial ring over $K$ in $\{x_{ij}\,|\,1\leq i,\,j\leq n\}$. Let $D(X)$ be an $n\times n$ matrix whose $j$th column consists of the diagonal entries of the matrix $X^{j-1}$ written from left to right with the convention that $X^{0}$ is the identity matrix of size $n$. Define $\mathcal{P}(X)=\det(D(X))$. In [You11] and [You21], H-W.Young studies the varieties of nearly commuting matrices and derives their important properties such as the decomposition into irreducible components through the use of the polynomial $\mathcal{P}(X)$. We hope to understand better their Frobenius singularities and as it can be seen in [Kad18] this is also closely related to the singularities of the variety defined by $\mathcal{P}(X)$. In this paper we prove that the latter is $F$-pure and find a homogeneous system of parameters for it. Let us first define the necessary preliminaries. The following notation is fixed for the rest of the paper. Let

[TABLE]

and

[TABLE]

where

[TABLE]

That is, $\tilde{X}$ is the matrix obtained from $X$ by setting all the entries of its last column and the last row to 0 except for the entry $x_{nn}$. Let $X_{0}$ be the matrix obtained from $X$ by deleting its last column and last row.

Lemma 1.1** ([You21], Lemma 1.3).**

$\mathcal{P}(X)$* is an irreducible polynomial.*

Lemma 1.2** ([You21], Proof of Lemma 1.3).**

$\mathcal{P}(\tilde{X})=\mathcal{P}(X_{0})c_{X_{0}}(x_{nn})$* where $c_{X_{0}}(t)$ is the characteristic polynomial of $X_{0}$.*

Remark 1*.*

The above lemma is also true when we annihilate either only the last column or only the last row of $X$ with the exception of the entry $x_{nn}$.

2 A system of parameters

First, we find a homogeneous system of parameters on $R/(\mathcal{P}(X))$. To do so we prove several useful lemmas.

Lemma 2.1**.**

Let $A$ be a square matrix of size $n$ with integer entries and with $\det(A)=\pm 1$. Then the following are equivalent

1. (a)

$\mathcal{P}(A)=\pm 1$. 2. (b)

The diagonals of $I,\,A,\,A^{2},\ldots,A^{n-1}$ span $\mathbb{Z}^{n}$. 3. (c)

The diagonals of the elements of $\mathbb{Z}[A,A^{-1}]$ span $\mathbb{Z}^{n}$. 4. (d)

There exist $n$ integer powers of $A$ with the property that their diagonals span $\mathbb{Z}^{n}$.

Proof.

The equivalence of (a) and (b) is clear as the columns of the matrix $D(A)$ are the diagonals of $I,\,A,\,A^{2},\ldots,A^{n-1}$.

By Cayley-Hamilton’s theorem we have that $A^{n}\in\sum_{i=0}^{n-1}\mathbb{Z}A^{i}$ and hence $I\in\sum_{i=1}^{n}\mathbb{Z}A^{i}$. Therefore, since $A$ is invertible, we also have that $A^{-1}\in\sum_{i=0}^{n-1}\mathbb{Z}A^{i}$ as well as $A^{-h}$ for all integers $h$. Thus $\mathbb{Z}[A]=\mathbb{Z}[A,A^{-1}]$. Finally,

[TABLE]

[TABLE]

the diagonals of all the integer powers of $A$ span $\mathbb{Z}^{n}$ if and only if there exist $n$ integer powers of $A$ with the property that their diagonals span $\mathbb{Z}^{n}$.

∎

Lemma 2.2**.**

Let

[TABLE]

be a square matrix of size $n\geq 1$ with the property that all the entries strictly below the main anti-diagonal are 0 and the rest are equal to 1. Then $\mathcal{P}(A)$ is equal to either 1 or -1.

Proof.

First observe that $\det(A)=\pm 1$. Hence the matrix is invertible and it can be shown that

[TABLE]

$B$ has two non-zero anti-diagonals and the rest of the entries are equal to 0. To show that $\mathcal{P}(A)=\pm 1$ it is necessary and sufficient to show that there exist $n$ powers of $B$ so that their diagonals span $\mathbb{Z}^{n}$, see Lemma 2.1. We claim that for this purpose it is sufficient to take $n$ odd powers of $B$.

Claim 2.1**.**

[TABLE]

We show this by induction with the induction step equal to 2. Cases $n=2$ and $n=3$ can be easily verified.

Write $B$ as

[TABLE]

then

[TABLE]

Claim 2.2**.**

For all $1\leq j\leq n-1$, we have that

[TABLE]

We prove the claim by induction on $j$. When $j=1$, it is true. Suppose that the claim is true for all integers less than or equal to $j$. Then

[TABLE]

Therefore,

[TABLE]

For now assume that $k>1$.

[TABLE]

Suppose first that $k+l=n-j+1$. Then

[TABLE]

Next, suppose that $k+l=n+j+2$

[TABLE]

Finally, consider the case when $k+l\leq n-j\text{ or }k+l\geq n+j+3$. We have that

[TABLE]

Thus the formula for $B^{2j-1}$ is true.

Remark 2*.*

The case $k=1$ goes along the same lines as above with the exception when we have that $1+l=n+j+2$. Then $l=n+j+1>n$ and this is not possible.

Now we are ready to finish the proof of the lemma. Consider the matrix whose columns are the diagonals of the odd powers of $B$ written from left to right. Observe that in each odd power the main diagonal meets only one of the sub-anti-diagonals that we highlighted above. Therefore, we have that

[TABLE]

for matrices of odd sizes

[TABLE]

and for matrices of even sizes

[TABLE]

Since every row of the matrices has a pivot position, we conclude that the determinants of these matrices are not zero and thus the columns of each of them span $\mathbb{Z}^{n}$. This finishes the proof of the lemma. ∎

Theorem 2.3**.**

Let $X=(x_{ij})_{1\leq i,\,j\leq n}$ be a square matrix of size $n$ with indeterminate entries over a field $K$ and $R=K[X]$ be the polynomial ring over $K$ in $\{x_{ij}\,|\,1\leq i,\,j\leq n\}$. Let

[TABLE]

and

[TABLE]

Then

[TABLE]

is a homogeneous system of parameters and hence a regular sequence on $R/(\mathcal{P}(X))$.

Proof.

We prove the lemma by induction on $n$.

Consider the first few small cases.

Let $n=2$. In this case modulo $(\mathcal{S})$ we have that

[TABLE]

and

[TABLE]

Let $n=3$. In this case

[TABLE]

and

[TABLE]

Let $n=4$. In this case

[TABLE]

[TABLE]

Claim 2.3**.**

$\mathcal{P}(X)=\pm x_{11}^{n(n-1)/2}$ in the quotient ring of $R$ by the ideal generated by

[TABLE]

Remark 3*.*

In our proof of the claim we do not establish exactly the sign of the image of $\mathcal{P}(X)$. We only show that it is either 1 or -1.

Consider the matrices $X$, $X_{0}$, $\tilde{X}$, and set the elements strictly below the main anti-diagonal of $X_{0}$ and $x_{nn}$ to 0, that is,

[TABLE]

Since $P(\tilde{X})=P(X_{0})c_{X_{0}}(x_{nn})$ where $c_{X_{0}}$ is the characteristic polynomial of $X_{0}$ (see Lemma 1.2), we have that $P(\tilde{\tilde{X}})=P(\tilde{X_{0}})(-1)^{(n-2)(n-1)/2}\prod_{i=1}^{n-1}x_{i,n-i}$. Kill the elements $x_{11}-x_{kl}$ for all $(k,l)\in\Omega$. Then $P(\tilde{X_{0}})=x_{11}^{(n-1)(n-2)/2}P(Y_{0})$, where

[TABLE]

Hence

[TABLE]

By Lemma 2.2 we have that $P(Y_{0})=\pm 1$, which finishes the proof of the claim.

Finally, we have that $R/(\mathcal{P}(X),\mathcal{S})\cong K[x_{11}]/(x_{11}^{n(n-1)/2})$, which has Krull dimension 0. Hence, $\mathcal{S}$ is indeed a system of parameters on $R/(\mathcal{P}(X))$ and, since the ring is a complete intersection, it is also a regular sequence. ∎

3 The variety defined by $\mathcal{P}(X)$ is $F$-pure

Definition 3.1**.**

Let $S$ be a ring with positive prime characteristic $p$. Then $S$ is called $F$-pure if the Frobenius endomorphism $F:S\rightarrow S$ with $F(s)=s^{p}$ is pure, that is, for every $S$-module $M$ we have that $F\otimes_{S}\text{id}_{M}:S\otimes M\rightarrow S\otimes M$ is injective.

Next is Fedder’s criterion specialized for hypersurfaces and we use it to prove $F$-purity of $R/(\mathcal{P}(X))$.

Lemma 3.1**.**

(Fedder’s criterion, [Fed83], Proposition 2.1) Let $S$ be a polynomial ring over a field $K$ of positive prime characteristic $p$ and let $f\in S$ be a homogeneous polynomial. Then $S/(f)$ is $F$-pure if the polynomial $f^{p-1}$ has a non-zero monomial term in which every indeterminate has degree at most $p-1$.

We also need the fact that $F$-purity deforms for Gorenstein rings.

Lemma 3.2**.**

([Fed83], Theorem 3.4(2)) Let $S$ be a Gorenstein ring and let $x\in S$ be a non-zero divisor. Then if $S/xS$ is $F$-pure, then so is $S$.

For more examples of $F$-pure rings and other related notions the reader may refer to [Fed87], [FW89], [HR76], [BH98]. Computer algebra system Macaulay2, [GS], is a great tool in studying rings and their properties.

Let $\mathcal{S}_{\Lambda}=\{x_{ij}:1\leq i\leq n,\,n-i+1\leq j\leq n\}\subseteq\mathcal{S}$. These are the entries of the matrix $X$ on and below the main anti-diagonal. It has been shown in the previous section that $\mathcal{S}_{\Lambda}$ is part of a homogeneous system of parameters and a regular sequence on $R/(\mathcal{P}(X))$.

Theorem 3.3**.**

Let $K$ be a field of positive prime characteristic $p$. Then $R/(\mathcal{P}(X))$ is $F$-pure for all $n\geq 1$.

Proof.

We use the fact that $F$-purity deforms for Gorenstein rings, [Fed83], and we have that $R/(\mathcal{P}(X))$ is a complete intersection and hence is Gorenstein. Therefore, it is sufficient to show that we have an $F$-pure ring once we take the quotient of the ring $R/(\mathcal{P}(X))$ by the ideal generated by the regular sequence $\mathcal{S}_{\Lambda}$.

Let us first take a look at what we have for few small values of $n$.

Case $n=1$ is trivial. $\mathcal{P}(X)=1$ and $R/(\mathcal{P}(X))=0$.

Case $n=2$: \mathcal{P}(X)=\det\left[\begin{array}[]{cc}1&x_{11}\\ 1&x_{22}\end{array}\right]=x_{22}-x_{11}. Then $R/(\mathcal{P}(X))\cong K[x_{11},x_{12},x_{21}]$ is regular and hence $F$-pure.

Now we are ready to prove the general statement. We do it by induction on $n$.

First observe the following: if our statement is true for a fixed $n$, that is, if $R/(\mathcal{P}(X),\mathcal{S}_{\Lambda})$ is $F$-pure, then so is $R/(\mathcal{P}(X),\mathcal{S}_{\Lambda}^{0})$, where

[TABLE]

that is, the entries strictly below the main anti-diagonal. Moreover, if a particular monomial term of $\mathcal{P}(X)$ has a nonzero coefficient in $R/(\mathcal{S}_{\Lambda})$, then it is a monomial term of $\mathcal{P}(X)$ with a nonzero coefficient in $R/(\mathcal{S}_{\Lambda}^{0})$. A partial converse is also true. If $\mathcal{P}(X)$ in $R/(\mathcal{S}_{\Lambda}^{0})$ has a nonzero monomial term in the entries of $X$ which are strictly above the main anti-diagonal, then so does $\mathcal{P}(X)$ in $R/(\mathcal{S}_{\Lambda})$.

The first non-trivial case is $n=3$. Let $\hat{X}$ be the matrix $X$ modulo the elements of $\mathcal{S}_{\Lambda}^{0}$, that is,

[TABLE]

and

[TABLE]

[TABLE]

Since $P(\hat{X})$ has a monomial term $x_{11}x_{12}x_{21}$ with a coefficient $1$ modulo $p$, so does $\mathcal{P}(X)$ in $R/(\mathcal{S}_{\Lambda})$.

Here is our induction hypothesis: for all $k<n$ we have that $\mathcal{P}(X)^{p-1}$ in $R/(\mathcal{S}_{\Lambda})$ and in $R/(\mathcal{S}_{\Lambda}^{0})$ has a monomial term $\prod_{i=1}^{k-1}\prod_{j=1}^{k-i}x_{ij}^{p-1}$ with coefficient $\pm 1$ modulo $p$. In other words, this monomial term is the $(p-1)^{\text{st}}$ power of the product of the entries of $X$ strictly above the main anti-diagonal. The basis of the induction is verified above.

Now consider the matrices $X_{0}$, $\tilde{X}$ and $\tilde{\tilde{X}}$. As in the Theorem 1, since $P(\tilde{X})=P(X_{0})c_{X_{0}}(x_{nn})$ where $c_{X_{0}}$ is the characteristic polynomial of $X_{0}$ (see Lemma 1.2) we have that $P(\tilde{\tilde{X}})=P(\tilde{X_{0}})(-1)^{(n-2)(n-1)/2}\prod_{i=1}^{n-1}x_{i,n-i}$. By induction hypothesis $P(\tilde{X_{0}})^{p-1}$ has a monomial term $\prod_{i=1}^{n-2}\prod_{j=1}^{n-1-i}x_{ij}^{p-1}$ with coefficient $\pm 1$.

Hence $P(\tilde{\tilde{X}})^{p-1}$ has a monomial term $\prod_{i=1}^{n-1}\prod_{j=1}^{n-i}x_{ij}^{p-1}$ with coefficient $\pm 1$. Therefore, by Fedder’s criterion we have that $R/(\mathcal{P}(X),\mathcal{S}_{\Lambda})$ is $F$-pure and thus so is $R/(\mathcal{P}(X))$. ∎

The next natural question that one can consider is whether the variety defined by the polynomial $\mathcal{P}(X)$ is $F$-regular. It is certainly true when $n=1$ and $n=2$, but is unknown for larger values of $n$.

Conjecture 3.1**.**

Let $K$ be a field of positive prime characteristic $p$. Then $R/(\mathcal{P}(X))$ is $F$-regular for all $n\geq 1$.

4 Acknowledgement

The author is grateful to Mel Hochster for valuable discussions and comments and thanks the referee for useful suggestions on improving the paper.