Subalgebra and Khovanskii bases equivalence

Subalgebra bases (sometimes also called canonical bases or SAGBI bases) were originally introduced as analogues to Gröbner bases for polynomial algebras independently by Kapur and Madlener [12] and Robbiano and Sweedler [15]. This concept was further generalized to quotient polynomial rings by Stillman and Tsai [16] and to Khovanskii bases of valued algebras by Kaveh and Manon [14].

There are several existing implementations of subalgebra bases for polynomial algebras in computer algebra systems: two implementations [11, 4] using Singular [8], a forthcoming implementation [3] in CoCoA [1], and an implementation [5, 6] in Macaulay2 [10] by several of the authors of this paper. Among these implementations, we note that the recent work in  [4] reports impressive runtimes compared to the alternatives on a test suite of challenging examples. On the other hand, the package [5], described in [6], also handles subalgebra bases for quotient rings. This level of generality is needed for the computations in the present paper.

Showcasing the generality of Khovanskii bases, [14, Example 7.7] constructs finite Khovanskii bases for the standard invariant ring of the alternating group $A=k[x,y,z]^{A_{3}}$. Viewed as a subalgebra $A\subseteq k[x,y,z]$, a finite subalgebra basis for $A$ does not exist [9]. However, in  Example 3.10, we show that there is more to the story: if we present $A$ as the quotient of a polynomial ring, the Khovanskii bases in question are also subalgebra bases in the sense of [16].

Our main goal is to establish explicit connections between the two previously-mentioned notions of Khovanskii bases and subalgebra bases for quotient rings, with an eye towards leveraging existing implementations. We show in Theorem 3.8 that the most common cases of Khovanskii bases, namely those arising from valuations that satisfy our standing hypotheses in Section 3, can also be realized as subalgebra bases of quotients of polynomial rings. We note that the same hypotheses are satisfied by the valuations constructed using tropical geometry in [14, Theorem 1].

As a partial converse, Corollary 4.4 gives sufficient conditions under which subalgebra bases for quotient rings are also Khovanskii bases. Note that this result holds unconditionally in the original setting of polynomial rings. Section 5 contains further discussion of the relationship between the monomial orders and valuations appearing in our constructions. Finally, in Section 6, we apply our results to computing Newton-Okounkov bodies.

Fix a field $k$ and let $R\vcentcolon=k[x_{1},\dots,x_{n}]$ be a polynomial ring with a monomial order $<$. We fix the convention that monomial orders use the maximum convention, that is, $\operatorname{lt}(f+g)\leq\max\{\operatorname{lt}(f),\operatorname{lt}(g)\}$, provided none of $f$, $g$, and $f+g$ are zero.

Let $A\vcentcolon=k[f_{i}\colon i\in\mathcal{I}]$ be a subalgebra of $R$, where $\mathcal{I}$ is some index set. The initial algebra of $A$ is the algebra generated by the leading terms of $A$, that is,

Here, we use the notation $\operatorname{lt}(A)$ instead of the more common notation $\operatorname{in}_{<}(A)$, as we reserve the latter notation for valuations and Khovanskii bases, cf. [17].

One of the most significant differences between subalgebra bases and Gröbner bases is that subalgebra bases do not receive the benefits of the Noetherian property. Therefore, a finitely generated algebra may not have a finite subalgebra basis under any term ordering, see, for example, [15, Example 1.20].

Many of the standard algorithms from Gröbner basis theory have analogues in subalgebra basis theory. For instance, polynomial long division is replaced by subduction. The subduction algorithm provides a rewriting of a polynomial $f\in R$ with respect to a subalgebra basis $\{g_{j}\}_{j\in\mathcal{J}}$. The result of the algorithm is a finite sum

with the following properties: (1) $\alpha_{j}$ is zero for all but finitely many indices, (2) $c_{\alpha}$ is zero for all but finitely many values of $\alpha$, (3) if $r\not=0$, then $\operatorname{lt}(r)\leq\operatorname{lt}(f)$, (4) the nonzero terms of $r$ are not in the initial algebra $\operatorname{lt}(A)$, (5) for every $\alpha$ with $c_{\alpha}\not=0$, $\operatorname{lt}(f)\geq\operatorname{lt}(g^{\alpha})$, and (6) the values of $\operatorname{lm}(g^{\alpha})$ with $c_{\alpha}\not=0$ are distinct. The finiteness in these definitions comes about since monomials orders are well-orders, even when the subalgebra basis itself has infinitely many polynomials. Subduction has many standard properties of other polynomial rewriting procedures: for example, the remainder $r$ is zero if and only if $f\in A$. In fact, the remainder $r$ is independent of the choice of $c_{\alpha}$. Therefore, when $f\in A$, the remainder is zero and exactly one $\alpha$ has both $c_{\alpha}\not=0$ and $\operatorname{lm}(f)=\operatorname{lm}(g^{\alpha})$.

The term order induces a filtration on $A$ by $k$-vector spaces as follows: For any $\alpha\in{\mathbb{N}}^{n}$,

$F_{<\alpha}(A)$ is defined similarly, by replacing $\leq$ by $<$ in Equation (3). Then the associated graded algebra is

Let $A$ be a finitely generated $k$-algebra and domain, as in the definition of Khovanskii bases. Suppose that $\{g_{1},\dots,g_{m}\}$ is a finite Khovanskii basis for $A$. Our goal is to find a subalgebra basis in the sense of [16] that reflects the structure and properties of $A$.

We maintain the following standing hypotheses on a valuation in order to simplify theorem statements.

Let $R\vcentcolon=k[x_{1},\dots,x_{m}]$ be a presentation ring for $A$ corresponding to our Khovanskii basis, meaning that there is a presentation map $\pi:R\rightarrow A$ defined by $\pi(x_{i})=g_{i}$. Let $I$ be the kernel of this map. Using the valuation $\nu,$ we define a monomial order on $R$.

We verify that $<$ defines a monomial order. It is a total ordering since both the image of $\nu$ is totally ordered and $<^{\prime}$ induces a total ordering on ${\mathbb{N}}^{m}$. Also, $<$ refines the partial order given by divisibility due to the property $\nu(fg)=\nu(f)+\nu(g).$ Finally, we note that $\nu$ lifts the trivial valuation on $k$ and its image is well-ordered. Thus, there cannot be any $f\in A\setminus k$ such that $\nu(f)\succ 0$. Since $<^{\prime}$ satisfies $1\leq^{\prime}x^{\alpha}$ for all $\alpha$, we conclude that $1\leq x^{\alpha}$ for any $\alpha\in{\mathbb{N}}^{m}$.

We now consider $R/I$ as a subalgebra of itself. For a Khovanskii basis $\{g_{1},\ldots,g_{m}\},$ we observe that $\{[x_{1}],\dots,[x_{m}]\}$ form a subalgebra basis for $R/I$. To verify this, we note that if $\operatorname{lt}([x_{i}])\not=x_{i}+\operatorname{lt}(I)$, then $x_{i}$ is not a standard monomial. Therefore, for all $f\in R$, $x_{i}$ does not appear in $\operatorname{lt}(\tilde{f})$, and so $\operatorname{lt}([f])$ can be represented without $x_{i}$. Therefore, $\{[x_{1}],\dots,[x_{m}]\}$ and $\{[x_{1}],\dots,\widehat{[x_{i}]},\dots[x_{m}]\}$ generate the same subalgebra of $\operatorname{gr}_{>}(R/I)$. Continuing recursively, we find that for every $f$ in $R$, $\operatorname{lt}([f])$ can be represented by a monomial in the variables $x_{i}$ such that $\operatorname{lt}([x_{i}])=x_{i}+\operatorname{lt}(I)$. Hence, these variables generate $\operatorname{gr}_{>}(R/I)$ and form a subalgebra basis.

Therefore, not only are $R/I$ and $A$ isomorphic as $k$-algebras, but the presentation map $\pi$ also takes the subalgebra basis $\{[x_{1}],\dots,[x_{m}]\}$ for $R/I$ to the Khovanskii basis $\{g_{1},\dots,g_{m}\}$ for $A$.

Next, we show that the grading and filtration given by this monomial order match that of the valuation.

In the previous proof, we note that generally $\tilde{f}\neq\overline{f}$. We can only conclude that the images of their leading terms under $\pi$ must have the same valuation. This relationship can be seen more precisely in the following lemma, which shows that the filtrations of $R/I$ and $A$ are compatible with the quotient $\pi$.

By construction, the elements of $R/I$ are in bijective correspondence with those of $A$. Hence, the elements of $F_{\leq\alpha}(R/I)$ are in bijective correspondence with the elements of $F_{\succeq\nu(\pi(x^{\alpha}))}(A)$. By the previous result, we see that if $\nu(\pi(x^{\alpha}))=\nu(\pi(x^{\beta}))$, then $F_{\leq\alpha}(R/I)=F_{\leq\beta}(R/I)$ since the right-hand-side of the equality is the same for both $\alpha$ and $\beta$.

Suppose that $\operatorname{lm}(\tilde{f})=x^{\alpha}$ and $\nu(\pi(x^{\alpha}))=\nu(\pi(x^{\beta}))$. Since $[f]\in F_{\leq\alpha}(R/I)=F_{\leq\beta}(R/I),$ it follows that $x^{\alpha}\leq x^{\beta}$. In other words, $x^{\alpha}$ is the smallest monomial with respect to $<$ whose valuation is $\nu(\pi(x^{\alpha}))$.

On the other hand, suppose that $x^{\beta}$ is not the smallest monomial with respect to $<$ with valuation $\nu(\pi(x^{\beta}))$. The observation above implies that if $\nu(\pi(f))=\nu(x^{\beta})$, then $\operatorname{lt}(\tilde{f})<x^{\beta}$. Therefore, $[f]\in F_{<\beta}(R/I)$. Putting this together, we have the following conclusions:

In the case of Corollary 3.6, we see that $F_{\leq\beta}(R/I)/F_{<\beta}(R/I)$ is trivial. On the other hand, in the case of Corollary 3.7, we conclude that

These two observations directly imply that the associated graded algebras for both $R/I$ under $<$ and $A$ under $\succ$ are equal. Collecting these results, we have the following:

Two remarks further illustrate the relationship between $\nu$ and $<$.

The construction of Theorem 3.8 assumes it is known a priori that $\{g_{1},\dots,g_{m}\}$ forms a Khovanskii basis. Without this assumption, the construction is merely existential. We next state a criterion that can identify when $\{g_{1},\dots,g_{m}\}$ form a Khovanskii basis.

A Khovanskii or subalgebra basis is minimal if none of its proper subsets form a Khovanskii or subalgebra basis for the same algebra. Our previous constructions respect minimality.

Let $R\vcentcolon=k[x_{1},\dots,x_{m}]$. Suppose that $R/I$ is a finitely generated $k$-algebra and domain with a monomial order $<$. In this case, we can apply the theory of Khovanskii bases directly to $R/I$, provided we can find a suitable valuation on $R/I$.

A motivating attempt would be to use $\tilde{\mu}:R/I\setminus\{[0]\}\rightarrow{\mathbb{Z}}^{m}$ defined as $[f]\mapsto-\operatorname{exp}(\operatorname{lm}(\tilde{f})),$ where $\operatorname{exp}$ denotes the exponent of the input monomial. In many cases, however, this is not a valuation. In particular, suppose that $x^{\alpha}$ and $x^{\beta}$ are standard monomials, but their product $x^{\alpha+\beta}$ is not a standard monomial. In this case, $\tilde{\mu}([x^{\alpha}])+\tilde{\mu}([x^{\beta}])=-(\alpha+\beta),$ but this does not equal $\tilde{\mu}([x^{\alpha+\beta}])$ since $\operatorname{lm}(\widetilde{x^{\alpha+\beta}})\not=x^{\alpha+\beta}$. We proceed to fix this deficiency.

The key object in our construction is the following lattice:

We then define the torsion-free portion of $\mathbb{Z}^{m}/K$ as

From this, we define the map $\mu:R/I\setminus\{[0]\}\rightarrow\operatorname{torfree}(\mathbb{Z}^{m}/K)$ where $\mu([f])$ maps to the image of $\tilde{\mu}([f])$ in this quotient. We now define an order on the image of $\mu$. In particular, suppose that $\mathfrak{a}$ and $\mathfrak{b}$ are in the image of $\mu$. We say that $\mathfrak{a}\prec\mathfrak{b}$ if the smallest monomial $x^{\alpha}$ with $\mu([x^{\alpha}])=\mathfrak{a}$ is greater than the smallest monomial $x^{\beta}$ with $\mu([x^{\beta}])=\mathfrak{b}$.

We observe that the monomial $x^{\alpha}$, as defined above, is a standard monomial. In particular, for any $[f]\in\mu^{-1}(\mathfrak{a})$, it follows that $[\operatorname{lt}(\tilde{f})]\in\mu^{-1}(\mathfrak{a})$. We note that $\operatorname{lt}(\tilde{f})$ is both a standard monomial and smaller than $\operatorname{lt}(f)$.