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\title{Smith reduction and sublattices of finite rank\\with an application to toric varietes}
\author{DBW}
\begin{document}
\maketitle
\abstract{In this note I give a proof of the existence of a Smith normal form for matrices with integer entries. The existence of a good basis for a lattice with a finite index sublattice is a consequence of the Smith normal form. I conclude with an easy application to toric varieties.}
\begin{thm}
Let $A$ be a matrix with integer entries. Then there exist integer-values invertible matrices $C$ and $B$ such that $A = CDB$, where $D$ is a diagonal integer-valued matrix.
\end{thm}
\begin{proof}
We call an elementary operation on a matrix the addition of an integer multiple of one row (or column) to another row (or column). By means of a combination of elementary operations one can exchange two rows or columns, at the cost of a minus sign. An elementary operation on a matrix $A$ corresponds to the multiplication from the left or the right of $A$ with an elementary matrix, which is a matrix that has $1$ along the diagonal and vanishing entries off the diagonal, except at one off-diagonal entry, where it is integer-valued. Thus elemetary matrices are invertible and have unit determinant; in particular, a product of a finite number of elementary matrices is an integer-valued invertible matrix.
Let now $A = (a_{ij})_{1\leq i,j\leq n}$ be an integer-valued matrix. By exchanging rows and or columns, we may assume that $a_{11}\neq 0$. The greatest common divisor of the elements in the first row and the first column (i.e. all $a_{ij}$ with $i$ or $j$ equal to $1$) is an integer combination of the latter; hence we may arrive at the situation (by using the Euclidean algorithm) where $a_{11}$ divides all elements of the first column and of the first row, by only using elemtary operations on $A$. Since now $a_{11}$ divides all elements in the first row and first collumn, we can sweep the first row and column clean, and arrive at the following form of the matrix $A$:
\begin{equation}\label{form1}
A = L A' M\,,~~~~A'=\begin{pmatrix} d & 0 \\ 0 & \tilde{A}\\ \end{pmatrix}\,.
\end{equation}
But then the proof is finished by induction, as we can now focus on $\tilde{A}$.
\end{proof}
\begin{rem}
The previous theorem states that for any integer-valued matrix, there exists a Smith normal form. This normal form is not unique however. The process described in the proof to obtain the Smith normal form is known as Smith reduction.
\end{rem}
\begin{rem}
In the literature the above theorem is stated and proved in greater generality: For any integer-valued $n\times n$-matrix $A$ there exist matrices $C,B\in {\rm SL}_{n}(\Z)$ and integers $d_1,\ldots,d_r$ such that $d_i$ divides $d_j$ for $i\leq j$ and $A = CDB$ with $D={\rm diag}(0,\ldots,0,d_1,d_2,\ldots,d_r)$. The proof is similar; if at the step where one obtains the desired form (\ref{form1}) one $a_{ij}$ cannot be divided by $a_{11}$, one adds the $i$th column to the first and repeats the Euclidean algorithm to the first column, then after a finite number of steps $a_{11}$ divides $a_{ij}$.
\end{rem}
\begin{cor}[Existence of a good basis]
Let $\Lambda$ be a lattice inside $\Z^n$, such that the quotient $\Lambda/\Z^n$ is a finite abelian group; thus $\Lambda$ is a sublattice of finite index. Then there exists a basis $\{ e_1,\ldots,e_n\}$ of $\Z^n$ and nonzero integers $k_1$, \ldots, $k_n$ such that the elements $k_1e_1$, \ldots $k_ne_n$ form a basis of $\Lambda$.
\end{cor}
\begin{proof}
Take any basis $\{m_1,\ldots, m_n\}$ of $\Lambda$. Note, that any basis of $\Lambda$ must have $n$ elements, since there exist at most $n$ linearly independent elements in $\Z^n$ and since $\Lambda\otimes_\Z \mathbb{R}$ must be a real vector space of dimension $n$. Write $A$ for the matrix whose columns are the basis vectors $m_1$, \ldots, $m_n$. Then we can write $A = BKC$ with $B$ and $C$ invertible integer-valued matrices and $K$ is a diagonal integer-valued matrix $K={\rm diag}(k_1,\ldots,k_n)$. The matrix $AC^{-1}$ is a matrix whose columns are a basis for $\Lambda$. The matrix $B$ is a matrix whose columns are a basis of $\Z^n$. Let us write $b_1$, \ldots, $b_n$ for the columns of $B$. Then the columns of $BK$ are the vectors $b_1 k_1$, \ldots, $b_nk_n$. This proves the corollary.
\end{proof}
\begin{prop}
Any finitely generated abelian group is of the form $\Z^m\times \Z_{r_1}\times\cdots \times \Z_{r_k}$.
\end{prop}
\begin{proof}
If $G$ is a finitely generated group, there is an epimorphism $\Z^n\to G$. The kernel of this morphism is a sublattice $\Lambda$ in $\Z^n$. Hence $G\cong \Z^n/\Lambda$. Let $e_1$, \ldots, $e_n$ be the standard basis vectors of $\Z^n$ and let $b_1,\ldots, b_r$ be any set of generators of $\Lambda$, where $r = {\rm dim}_\mathbb{R}\Lambda\otimes_Z \mathbb{R}$. Then we can adjoin $m=n-r$ elements of the $e_i$ such that these together with the $b_j$ form a basis for $\Z^n\otimes_\Z \mathbb{R}$. Thus we can decompose $\Z^n = \Z^m\times \Z^{k}$, with $k=n-m$ and where $\Lambda$ lies in the second factor and is of finite index in $\Z^{k}$. We can find a basis $f_1,\ldots,f_k$ of $\Z^{k}$ and integers $r_1,\ldots, r_k$ such that $r_1f_1,\ldots, r_kf_k$ form a basis of $\Lambda$. But then $\Z^n/\Lambda = \Z^m\times (\Z^k/\Lambda)$ and in the obtained basis
it is obvious to see that $\Z^k/\Lambda = \Z_{r_1}\times\cdots \times \Z_{r_k}$.
\end{proof}
We now discuss a simple application of the Smith normal form to toric varieties; we show the one-to-one correspondences between integral commutative semigroups and affine toric varieties. All our semigroups are commutative and we will work over an algebraically closed field $k$, with arbitrary characteristic though.
If $S$ is a semigroup, we write $k[S]$ for the $k$-algebra generated by all elements $x^s$ where $s$ runs over all elements of $S$. We will restrict to finitely generated semigroups, so that the $k$-algebras $k[S]$ will always be finitely generated, hence noetherian. Even more, we will assume our semigroups are integral, by which we mean that they can be embedded into $\Z^n$. Then automatically $k[S]$ will always be finitely generated and admits an embedding $k[S]\to k[Z^n] =k[X_1,X_{1}^{-1},\ldots,X_n,X_{n}^{-1}]$. It follows that $k[S]$ is a domain, thus $X_S = {\rm Spec}(k[S])$ is an integral scheme over $k$.
\begin{defn}
For any integral semigroup $S$ we define the universal enveloping group of $S$ to be a group $G(S)$ with an injective morphism $i_S:S\to G(S)$ such that for any morphism of semigroups $f:S\to H$, where $H$ is a group, there exists a unique morphism $j:G(S)\to H$ such that $f = j\circ i_S$.
\end{defn}
\begin{thm}
For any integral semigroup $S$ the universal enveloping group exists and is unique up to isomorphism.
\end{thm}
\begin{proof}
Uniqueness of $G(S)$ is obvious by the requirement of the uniqueness of the morphism $i_S:S\to G(S)$ announced in the definition, hence existence is all that is required to prove. We first fix some embedding $S\to \Z^n$ and consider then the subgroup $G(S)$ in $\Z^n$ generated by all elements of $S$. That is, $G(S)$ consists of all elements $s - s'$, where $s,s'\in S$. If $f:S\to H$ is some morphism of semigroups with $H$ a group, then $j(s-s')$ has to be $j(s) - j(s')$, which is unambiguously defined as the image of $S$ necessarily lies in the centre of $H$. Thus uniqueness is proved.
\end{proof}
Hence for any semigroup $S$ we have a canonical morphism of affine $k$-schemes ${\rm Spec}(k[G(S)])\to {\rm Spec}(k[S]$. The object ${\rm Spec}(k[G(S)])$ is a torus.
\begin{lem}
Let $S$ be an integral semigroup. The morphism of $k$-schemes ${\rm Spec}(k[G(S)])\to {\rm Spec}(k[S]$ is an open embedding.
\end{lem}
\begin{proof}
We have to prove that the image of ${\rm Spec}(k[G(S)])\to {\rm Spec}(k[S]$ is an open subset in $X_S = {\rm Spec}(k[S]$. If $S$ is generated by elements $s_1$, \ldots, $s_n$, then $G(S)$ is as a group generated by elements $s_1$, \ldots, $s_n$ and $y=-(s_1+\ldots+s_n)$. Hence $k[G(S)] = k[S]_{y}$, which is a localization at $y$, and hence ${rm Spec}(k[G(S)])$ is a principal open subset of $X_S$.
\end{proof}
\begin{cor}
The dimension of ${\rm Spec}(k[S])$ is the rank of the group $G(S)$.
\end{cor}
\begin{defn}
We define the category of affine toric varieties as the category with objects ${\rm Spec}(k[S])$, where $S$ is an integral semigroup, and where the morphisms are the morphisms of $k$-schemes ${\rm Spec}(k[S']) \to {\rm Spec}(k[S])$induced by a morphism of semigroups $S\to S'$.
\end{defn}
It is not hard to paraphrase the above definition in terms of objects $T\to X$, where $X$ is the spectrum of a $k$-algebra $k[S]$ with $S$ an integral semigroup and with $T$ a torus and with morphisms $(T,X)\to (T',X')$ a commuting diagram
\begin{pspicture}(0,0)(3,3)
\rput(0,0){$X$}\rput(3,0){$X'$}\rput(0,3){$T$}\rput(3,3){$T'$}
\psline{->}(0.5,0)(2.5,0)\psline{->}(0,2.5)(0,0.5)
\psline{->}(0.5,3)(2.5,3)\psline{->}(3,2.5)(3,0.5)
\end{pspicture}
So we have seen that integral semigroups induce a couple $(T,X)$ where $X$ is an affine variety with an open embedding of the torus $T$ in $X$, such that the action of $T$ on itself extends to an action of $T$ on $X$; we have a morphism of $k$-algebras $k[X]\to k[T]\otimes k[X]$ given by $x^s \mapsto x^s\otimes x^s$, for any $s\in S$. Now we proceed to prove that given an affine variety $X$ with an open embedding of a torus $T$ in $X$, such that the action of $T$ on itself extends to an action of $T$ on $X$, there is a semigroup such that $X = {\rm Spec}(k[S])$ and $T = {\rm Spec}(k[G(S)]$. This is where the usage of the Smith normal form will appear.
\begin{thm}
Let $X$ be an affine variety over $k$ and $T\to X$ an open embedding of a torus $T$ in $X$, such that the action of $T$ on itself can be extended to an action of $T$ on $X$. Then there is an integral semigroup $S$ such that $T\cong {\rm Spec}(k[G(S)])$ and $X= {\rm Spec}(k[S])$, and the open embedding $T\to X$ is induced by the morphism of algebras $k[S]\to k[G(S)]$.
\end{thm}
\begin{proof}
We can always write $T = {\rm Spec}(k[\Z^n])$ for some $n$. Since $X$ is affine it is the spectrum of some $k$-algebra $A$. Since the action of the torus extends, we have a commutative diagram
\begin{pspicture}
(0,-.3)(3,3.2)
\rput(0,3){$A$}\rput(3.1,3){$A\otimes_k k[\Z^n]$}\rput(0,0){$k[\Z^n]$}\rput(3.3,0){$k[\Z^n]\otimes_k k[\Z^n]$}
\psline{->}(0.5,0)(2.2,0)\psline{->}(0,2.5)(0,0.5)
\psline{->}(0.5,3)(2.2,3)\psline{->}(3,2.5)(3,0.5)
\end{pspicture}
where the vertical maps are embeddings. Hence $A\subset k[\Z^n]$ and if $\sum a_\alpha x^\alpha$ is in $A$, the morphism $A\to A\otimes k[\Z^n]$ is given by $\sum a_\alpha x^\alpha\mapsto \sum a_\alpha x^\alpha\otimes x^\alpha$. But then it follows that $A = \bigoplus k x^\alpha$, where the sum runs over all $\alpha$, such that $x^\alpha$ is in $A$. Since $A$ is a ring, it follows that $A$ is of the form $k[S]$ with $S$ some sub-semigroup of $\Z^n$.
The proof is thus finished if we can show that $G(S) = \Z^n$. As the dimensions of $T$ and $X$ must match, we see that $S$ must generate a rank $n$ sublattice $\Lambda$. If this is a proper lattice, then by the existence of a good basis we can find a basis $e_1,\ldots, e_n$ of $\Z^n$ such $k_1 e_1,\ldots, k_ne_n$ is a basis of $\Lambda$, where $k_i\geq 1$ are not all equal to one. On the level of coordinates inside some affine space, the morphism $k[S]\to k[\Z^n]$ then corresponds to $x_i \mapsto x_{i}^{k_1}$, which is not an embedding unless all $k_i$ are equal to one. Hence $\Z^n = \Lambda$ and $G(S) = \Z^n $.
\end{proof}
\end{document}