Annihilators in $\mathbb{N}^k$-graded and $\mathbb{Z}^k$-graded rings

TL;DR

This paper explores properties of annihilators in graded rings, showing that non-trivial right ideals have homogeneous annihilators in certain graded structures, extending previous results to more general settings.

Contribution

It generalizes McCoy's result by demonstrating that in $ abla$-graded rings, right annihilators contain homogeneous elements, and under certain conditions, degree 0 annihilators can be found.

Findings

Non-trivial right ideals in $ abla$-graded rings have homogeneous annihilators.

In subrings of $bZ^k$-graded rings with non-annihilation properties, degree 0 annihilators exist.

Extension of annihilator properties from polynomial rings to more general graded rings.

Abstract

It has been shown by McCoy that a right ideal of a polynomial ring with several indeterminates has a non-trivial homogeneous right annihilator of degree 0 provided its right annihilator is non-trivial to begin with. In this note, it is documented that any $\mathbb{N}$-graded ring $R$ has a slightly weaker property: the right annihilator of a right ideal contains a homogeneous non-zero element, if it is non-trivial to begin with. If $R$ is a subring of a $\mathbb{Z}^k$ -graded ring $S$ satisfying a certain non-annihilation property (which is the case if $S$ is strongly graded, for example), then it is possible to find annihilators of degree 0.

Equations

$$A^{\mathrm{r}}=\{r\in R|\forall a\in A:ar=0\}.$$

$$r=\sum _{g\in G}{r}_g$$

$$z=\sum _{j=\ell }^u{z}_j$$

$$0=xy=\sum _{i=m}^n{x}_{i}y=\sum _{i=m}^p{x}_{i}y.$$

$${\mathbb{N}}^{k+1}={\mathbb{N}}^k\oplus \mathbb{N}\text{rTo}\mathbb{N};$$

$$T_j=\underset{v\in {\tau }^{-1}(j)}{⨁}{R}_v.$$

$$x=\sum _{i=m}^n{x}_i,$$

$${\mathbb{N}}^{k+1}={\mathbb{N}}^k\oplus \mathbb{N}\text{rTo}{\mathbb{N}}^k;$$

$$S_s=\underset{v\in {\sigma }^{-1}(s)}{⨁}{R}_v.$$

$$z=z_{v_1}+z_{v_2}+\dots +z_{v_{\ell }},$$

$$\tau (v_1)<\tau (v_2)<\dots <\tau (v_{\ell }).$$

$$x\in S_v$$

$$a_w{S}_{-w}=⟨a,S_{-w}⟩\subseteq ⟨a,S⟩=\{0\}$$

$$a_v=a_v\cdot 1=\sum _j(a_v{y}_j)z_j=0$$

$$S_t=\underset{j}{⨁}K\cdot e_{j,j+t},$$

$$S=K[X_1,X_1^{-1},X_2,X_2^{-1},\cdots ,X_k,X_k^{-1}]$$

$$f_{j}rg=0\text{for all j and all r∈R.}$$

$$(y_i{f}_j)rg=y_i(f_{j}rg)=y_i\cdot 0=0$$

$$f_{j}rh=\sum _i{x}_i(y_i{f}_{j}rh)=0$$

Full text

Work on this paper commenced during a research visit of the author to the Beijing Institute of Technology in January 2017. Their hospitality and financial support are gratefully acknowledged.

1. Introduction

In 1942, McCoy proved the following remarkable result concerning zero divisors in polynomial rings:

Theorem 1.1** (McCoy [McC42, Theorems 2 and 3]).**

Let $K$ be a commutative unital ring, and let $f\in K[x_{1},\,x_{2},\,\cdots,\,x_{k}]$ be a non-zero polynomial which is a zero divisor in $K[x_{1},\,x_{2},\,\cdots,\,x_{k}]$. Then there exists a non-zero $c\in K$ such that $f\cdot c=0$.

The result is not valid for non-commutative $K$. Indeed, consider the ring $R=\mathrm{Mat}_{2}(\mathbb{Z})$ of square matrices of size $2$ over $\mathbb{Z}$, and write $E_{ij}$ for the usual matrix units. The polynomials $f=E_{21}+E_{11}x+E_{22}x^{2}+E_{12}x^{3}$ and $g=E_{11}-E_{12}x$ satisfy $g\cdot f=0$, but no $M\in R$ annihilates $f$. Nevertheless, an ideal-theoretic variation of Theorem 1.1 holds:

Theorem 1.2** (McCoy [McC57, Theorem]).**

Let $K$ be a unital ring. If the right ideal $A$ of $K[x_{1},\,x_{2},\,\cdots,\,x_{k}]$ has non-trivial right annihilator, then there exists a non-zero $c\in K$ with $A\cdot c=\{0\}$.

This can be re-phrased in terms of a grading on the polynomial ring: If the right ideal $A$ has non-trivial right annihilator, then there exists a non-zero annihilator which is homogeneous of degree [math]. In particular, the right annihilator of $A$ contains a non-zero graded ideal of $K[x_{1},\,x_{2},\,\cdots,\,x_{k}]$ which intersects $K$ non-trivially.

Let $G$ be an additive monoid with zero element [math]. We say that a $G$-graded ring $T$ has the graded right McCoy property for elements if for all non-zero $f,g\in T$ with $f\cdot g=0$ there exists a homogeneous element $h\in T$ of degree [math] such that $f\cdot h=0$. We say that $T$ has the graded right McCoy property for right ideals if for all right ideals $A\unlhd T$ with non-trivial right annihilator $A^{\mathrm{r}}$ there exists a non-zero homogeneous element of degree [math] in $A^{\mathrm{r}}$.

Thus a polynomial ring $T=K[X_{1},\,X_{2},\,\cdots,\,X_{k}]$, for $K$ a unital ring, has the graded right McCoy property for right ideals by Theorem 1.2, and if $K$ is unital and commutative, it has the graded right McCoy property for elements by Theorem 1.1.

It is the purpose of this note to re-visit these results strictly from the point of view of graded algebra, and provide a different perspective on the special place polynomial rings occupy in the theory. We show that every $\mathbb{N}^{k}$-graded ring $T$ has the weak graded right McCoy property for right ideals (Theorem 4.1): if the right ideal $A\unlhd T$ has non-trivial right annihilator $A^{\mathrm{r}}$ there exists a non-zero homogeneous element, of possibly non-zero degree, in $A^{\mathrm{r}}$. If $T$ arises as the $\mathbb{N}^{k}$-graded subring of a strongly $\mathbb{Z}^{k}$-graded ring, or more generally of a $\mathbb{Z}^{k}$-graded ring satisfying a certain non-annihilation condition, then $T$ actually possesses the graded right McCoy property for right ideals (Theorem 5.4); this applies, for example, to polynomial rings.

For semi-commutative $T$ the (weak) graded McCoy property for right ideals implies the (weak) graded McCoy property for elements, which is recorded in Corollaries 4.3 and 5.8.

2. Notation and conventions

Given a right ideal $A$ of a (possibly non-unital) ring $R$ we write $A^{\mathrm{r}}$ for the right annihilator of $A$ in $R$, that is,

[TABLE]

The set $A^{\mathrm{r}}$ is a two-sided ideal of $R$.

Rings graded by a monoid

Given a monoid $G$, additively written, a $G$-graded ring is a ring $R$ together with a decomposition $R=\bigoplus_{g\in G}R_{g}$ into abelian groups such that $R_{g}R_{h}\subseteq R_{g+h}$ for all $g,h\in G$. Elements of $R_{g}$ are called homogeneous of degree $g$; we may say $R$-homogeneous of degree $g$ if we want to emphasise the ring $R$ and its grading.

Every element $r$ of a $G$-graded ring $R$ can be uniquely written as a sum

[TABLE]

where $r_{g}\in R_{g}$, with almost all $r_{g}$ zero. The set $\mathrm{supp}(r)=\{g\in G\,|\,r_{g}\neq 0\}$ is the support of $r$. — The following elementary Lemma is central:

Lemma 2.2**.**

Suppose $G$ is an additively written monoid with neutral element [math] (so that $a+0=a=0+a$ for all $a\in G$). Let $T$ be a $G$-graded ring. Let $r=\sum_{g\in G}r_{g}\in T$ as in (2.1), and let $s\in T$ be homogeneous.

1. (1)

If $G$ is right cancellative so that $a+c=b+c$ implies $a=b$, then $rs=0$ implies $r_{g}s=0$ for all $g\in G$. 2. (2)

If $G$ is left cancellative so that $c+a=c+b$ implies $a=b$, then $sr=0$ implies $sr_{g}=0$ for all $g\in G$.

Proof.

We prove (1) only. The elements $r_{g}s$ is homogeneous of degree $g+h$. Now $g+h=g^{\prime}+h$ if and only if $g=g^{\prime}$ as $G$ is right cancellative, so $rs=\sum_{g\in G}r_{g}s$ is the unique decomposition of $rs$ into homogeneous elements of distinct degree. Thus $rs=0$ entails $r_{g}s=0$ for all $g\in G$ as claimed. ∎

One can also show that if $G$ is right cancellative or left cancellative, and if $T$ has a unity, then the unit element $1\in T$ is homogeneous of degree [math]. Indeed, supposing that $G$ is right cancellative write $r=1=\sum_{g\in G}r_{g}$ as in (2.1). For any $h\in G$ and any homogeneous element $s\in T_{h}$ we find $s=1\cdot s=\sum_{g\in G}r_{g}s$, with $r_{g}s\in T_{g+h}$. By uniqueness of the representation, this implies $r_{g}s=0$ whenever $h\neq g+h$, i.e., whenever $g\neq 0$ (recall that $G$ is right cancellative). By distributivity, we have $r_{g}z=0$ for any $z\in T$ and $g\neq 0$. Applying this to $z=1$ yields $1=z=1\cdot z=\sum_{g\in G}r_{g}z=r_{0}z=r_{0}\cdot 1=r_{0}$ which shows that the unit is homogeneous of degree [math] as claimed. — For $G$ a group this is an observation of Dade [Dad80, Proposition 1.4].

Rings graded by $\mathbb{N}$

Suppose that $T=\bigoplus_{j\geq 0}T_{j}$ is an $\mathbb{N}$-graded ring. Every non-zero element $z\in T$ can be written uniquely in the form

[TABLE]

with $z_{j}\in T_{j}$, and both $z_{\ell}$ and $z_{u}$ non-zero. We call the expression (2.3) the canonical form of $z$. We say that $z$ has lower degree $\ell$ and upper degree $u$; the quantity $a=u-\ell\geq 0$ is called the amplitude of $z$. The element $z$ is homogeneous if and only if it is of amplitude [math]. We remark that if $z$ has amplitude $a$ and $x$ is a homogeneous element, then $zx$ has amplitude not exceeding $a$. Indeed, if $\ell$ and $u$ denote the lower and upper degree of $z$, respectively, and $x$ has degree $d$, then the lower degree of $zx$ is at least $\ell+d$ while the upper degree is at most $u+d$. Inequality occurs if and only if $z_{\ell}x=0$ or $z_{u}x=0$.

3. Annihilators in $\mathbb{N}$-graded rings

Theorem 3.1**.**

Suppose that $T$ is an $\mathbb{N}$-graded ring. Let $A$ be a right ideal in $T$ such that $A^{\mathrm{r}}$, its right annihilator, is non-zero. If $A^{\mathrm{r}}$ contains a non-zero element $y$ of positive amplitude $a$, then $A^{\mathrm{r}}$ also contains a non-zero element $z$ of amplitude less than $a$. In particular, $A^{\mathrm{r}}$ contains a non-zero homogeneous element.

A version of this Theorem for commutative polynomial rings, referring to degree rather than amplitude, was given by Forsythe [For43, Theorem A].

Proof.

Write $y$ in canonical form $y=\sum_{\ell}^{u}y_{j}$; note that $a=u-\ell>0$ by hypothesis. If $y_{u}\in A^{\mathrm{r}}$ then $z=y_{u}$ has amplitude [math], and is the desired element of amplitude less than $a$.

Otherwise, if $y_{u}\notin A^{\mathrm{r}}$, we can choose an element $x\in A$, with canonical form $\sum_{i=m}^{n}x_{i}$ , such that $xy_{u}\neq 0$. If $x_{i}y=0$ for all $i$ then we must in particular have $x_{i}y_{u}=0$ for all $i$ by Lemma 2.2 (2). Thus $xy_{u}=\sum_{m}^{n}x_{i}y_{u}=0$ contradicting the choice of $x$. Consequently, there exists a maximal index $p$ with $x_{p}y\neq 0$. But $y\in A^{\mathrm{r}}$ so that

[TABLE]

It follows that $x_{p}y_{u}=0$ so that $z=x_{p}y$ has amplitude less than $a$; indeed, the upper degree of $x_{p}y$ is less than $p+u$, while the lower degree of $x_{p}y$ is at least $p+\ell$. — As $A^{\mathrm{r}}$ is a two-sided ideal, $z=x_{p}y\in A^{\mathrm{r}}$ is the desired non-zero element of amplitude less than $a$.

The last sentence of the Theorem follows by repeated application of what we proved already, yielding non-zero elements in $A^{\mathrm{r}}$ of successively smaller amplitude. The process must stop with an element of amplitude [math].∎

Corollary 3.2**.**

Suppose that $T$ is an $\mathbb{N}$-graded ring. Let $A$ be a right ideal in $T$ such that $A^{\mathrm{r}}$, its right annihilator, is non-zero. Then $A^{\mathrm{r}}$ contains a non-trivial graded ideal of $T$. ∎

The result is best possible. For let $T=K[x]/\langle x^{2}\rangle$, with $K$ a field, graded by $\deg(x)=1$. Let $A=\langle x\rangle$ be the right ideal of all polynomials in $x$ without constant term. It is annihilated by $x\in T_{1}$, but there is no non-zero homogeneous annihilator of degree [math].

Remark 3.3**.**

It is no coincidence that the proof of Theorem 3.1 makes use of the natural ordering on $\mathbb{N}$: One cannot expect McCoy-type results unless the grading monoid lies in an ordered group. For example, let $K$ be a field and consider $R=K[x]/(x^{2}-1)$ as a $\mathbb{Z}/2$-graded ring, with $x$ having degree $1$. This ring contains zero divisors as $(1-x)(1+x)=0$, but $R$ does not contain any non-zero homogeneous zero divisors. On the other hand, if $T$ is a $G$-graded ring with $G$ an ordered group, then any equality $a\cdot b=0$ with non-zero $a$ and $b$ yields, by passing to leading terms with respect to the total order on $G$, to a pair $(x,y)$ of homogeneous non-zero elements with $x\cdot y=0$.

4. Annihilators in $\mathbb{N}^{k}$-graded rings

Theorem 4.1**.**

Suppose that $R$ is an $\mathbb{N}^{k}$-graded ring. Let $A$ be a right ideal in $R$ such that $A^{\mathrm{r}}$, its right annihilator, is non-trivial. Then $A^{\mathrm{r}}$ contains a non-zero homogeneous element.

Proof.

We use induction on $k$, the case $k=1$ being the final sentence of Theorem 3.1 applied to $T=R$.

So suppose that $R$ is $\mathbb{N}^{k+1}$-graded. We let $T$ denote the $\mathbb{N}$-graded ring which is identical to $R$ as a ring, but with grading defined by the last coordinate of $\mathbb{N}^{k+1}$. More explicitly, denote by $\tau\colon\mathbb{N}^{k+1}\rTo\mathbb{N}$ the projection

[TABLE]

writing $R=\bigoplus_{v\in\mathbb{N}^{k+1}}R_{v}$ we let $T=\bigoplus_{j=0}^{\infty}T_{j}$ where

[TABLE]

Now suppose $A$ is a right ideal of the ring $R=T$ which has non-trivial right annihilator $A^{\mathrm{r}}$. By Theorem 3.1 we find a non-zero element $y\in A^{\mathrm{r}}$ which is homogeneous of degree $d$ as an element of $T$, i.e., $y\in A^{\mathrm{r}}\cap T_{d}$.

Write a general element $x\in A\subseteq T$ in canonical form with respect to the ring $T$ (i.e., with respect to the $\mathbb{N}$-grading),

[TABLE]

and let $J$ denote the right ideal of $T=R$ generated by all the resulting elements $x_{i}$ (letting $x$ vary over all of $A$); that is, $J$ is the smallest graded right ideal of $T$ containing $A$. As $x_{i}\in T_{i}$ for all $i$, and as $y\in T_{d}$ is homogeneous, the equality $xy=0$ (true since $y\in A^{\mathrm{r}}$) implies $x_{i}y=0$ for all $i$, by Lemma 2.2 (1). Thus in fact $y\neq 0$ is a right annihilator of $J$, that is, $y\in J^{\mathrm{r}}$

Next, we let $S$ denote the $\mathbb{N}^{k}$-graded ring which is identical to $R$ (and $T$) as a ring, but with grading given by the first $k$ coordinates of $\mathbb{N}^{k+1}=\mathbb{N}^{k}\oplus\mathbb{N}$. More explicitly, denote by $\sigma\colon\mathbb{N}^{k+1}\rTo\mathbb{N}^{k}$ the projection

[TABLE]

writing $R=\bigoplus_{v\in\mathbb{N}^{k+1}}R_{v}$ as before we let $S=\bigoplus_{s\in\mathbb{N}^{k}}S_{s}$ where

[TABLE]

Recall now that $J\unlhd S$ has non-trivial right annihilator as it contains the element $y\neq 0$ constructed above. By our induction hypothesis, applied to the $\mathbb{N}^{k}$-graded ring $S$ and the ideal $J$, we can find a homogeneous non-zero right annihilator $z$ of $J$ in $S$, of degree $s\in\mathbb{N}^{k}$ say. Such an element can uniquely be written as a sum of non-zero elements

[TABLE]

with $z_{v_{j}}\in R_{v_{j}}$ and $\sigma(v_{j})=s$ for all $j$ such that

[TABLE]

We specifically choose $z$ with $\ell$ as small as possible. If $\ell=1$ we are done: the element $z=z_{v_{1}}$ is a homogeneous element of $R$ which annihilates $J$, and thus annihilates $A$; note that $J^{\mathrm{r}}\subseteq A^{\mathrm{r}}$. On the other hand, $\ell>1$ cannot happen. Indeed, if $\ell>1$ then $z_{v_{\ell}}\notin J^{\mathrm{r}}$ by minimality of $\ell$. This means we can find an element $x\in J$ with $xz_{v_{\ell}}\neq 0$. In fact, as $J\unlhd T$ is a graded ideal, as remarked before, we can ensure that $x$ is $T$-homogeneous. But then $xz=0$ implies $xz_{v_{\ell}}=0$ by Lemma 2.2 (2), a contradiction. Thus we must have $\ell=1$, finishing the induction. ∎

Corollary 4.2**.**

Suppose that $R$ is an $\mathbb{N}^{k}$-graded ring. Let $A$ be a right ideal in $R$ such that $A^{\mathrm{r}}$, its right annihilator, is non-zero. Then $A^{\mathrm{r}}$ contains a non-trivial graded ideal of $R$. ∎

Corollary 4.3**.**

Let $R$ be an $\mathbb{N}^{k}$-graded ring, and let $(f_{i})_{i\in I}$ be a family of elements of $R$. Suppose there exists a non-zero element $g\in R$ such that $f_{i}g=0$ for all $i$. Suppose further that $R$ is semi-commutative so that $ab=0$ implies $arb=0$ for all $r\in R$. Then there exists a non-zero homogeneous element $h\in R_{v}$, of possibly non-zero degree $v\in\mathbb{N}^{k}$, such that $f_{i}h=0$ for all $i\in I$.

Proof.

Let $A=\langle f_{i}\,|\,i\in I\rangle$ be the right ideal generated by the elements $f_{i}$. As $R$ is semi-commutative, $g$ is a non-trivial annihilator of $A$. Hence by Theorem 4.1 there exists a homogeneous non-zero element $h\in A^{\mathrm{r}}$; this element $h$ annihilates in particular the specified generators $f_{i}$ of $A$. ∎

5. Positive subrings of $\mathbb{Z}^{k}$-graded rings

Let $S=\bigoplus_{v\in\mathbb{Z}^{k}}S_{v}$ now denote a $\mathbb{Z}^{k}$-graded ring. We will consider the following condition, for various elements $v\in\mathbb{Z}^{k}$:

[TABLE]

This is equivalent to saying that the left annihilator of $S_{-v}$ has trivial intersection with $S_{v}$.

The ring $S$ admits an $S_{0}$-valued “inner product” $\langle a,b\rangle=(ab)_{0}$, the degree-[math] component of the product $ab$. It is called right non-degenerate if $\langle a,S\rangle=\{0\}$ implies $a=0$, for all $a\in S$. Rings with non-degenerate inner product were investigated by Cohen and Rowen [MR696990].

Lemma 5.2**.**

1. (1)

The inner product $\langle\,\cdot\,,\,\cdot\,\rangle$ on $S$ is right non-degenerate if and only if condition (5.1) holds for every $v\in\mathbb{Z}^{n}$. 2. (2)

If $S$ is a strongly graded unital ring, the inner product $\langle\,\cdot\,,\,\cdot\,\rangle$ on $S$ is right non-degenerate.

Proof.

We prove (1) first. Suppose that $\langle\,\cdot\,,\,\cdot\,\rangle$ is right non-degenerate, and let $x\in S_{v}$ be given. Then $xS_{-v}=\langle x,S\rangle=\{0\}$ implies $x=0$, which means that condition (5.1) holds. Conversely, suppose (5.1) holds for all $v$, and let $a=\sum_{w}a_{w}\in S$ (with $a_{w}\in S_{w}$) be such that $\langle a,S\rangle=\{0\}$. Then, for any $w\in\mathbb{Z}^{k}$,

[TABLE]

so that $a_{w}=0$ and thus $a=\sum_{w}a_{w}=0$ as well. This shows that $\langle\,\cdot\,,\,\cdot\,\rangle$ is right non-degenerate.

To prove (2), let $a=\sum_{w}a_{w}\in S$ (with $a_{w}\in S_{w}$) be such that $\langle a,S\rangle=\{0\}$. For any $v\in\mathbb{Z}^{k}$ we have $S_{-v}S_{v}=S_{0}$, by definition of strong grading, hence we can find finitely many elements $y_{j}\in S_{-v}$ and $z_{j}\in S_{v}$ such that $1=\sum_{j}y_{j}z_{j}$. Now $a_{v}y_{j}=\langle a,y_{j}\rangle\in\langle a,S\rangle=\{0\}$ so that $a_{v}y_{j}=0$ for all $j$. It follows that

[TABLE]

and hence that $a=\sum_{v}a_{v}=0$. ∎

Example 5.3** (Cohen-Rowen [MR696990], Example 3).**

Let $K$ be a (possibly non-unital) ring with trivial left annihilator so that for all $x\neq 0$ there exists $y\in K$ with $xy\neq 0$, and let $S$ denote the ring of square matrices of size $n$ with entries in $K$. We equip $S$ with a $\mathbb{Z}$-grading by setting

[TABLE]

where $e_{i,j}$ is a formal matrix unit. In particular, $S_{0}$ corresponds to the main diagonal and $S_{1}$ to the first superdiagonal. Given a non-zero element $x=\sum_{j}\lambda_{j}e_{j,j+v}\in S_{v}$ there is an index $p$ such that $\lambda_{p}\neq 0$. By hypothesis on $K$ we can choose $\mu\in K$ with $\lambda_{p}\mu\neq 0$. Then $y=\mu e_{j+v,j}\in S_{-v}$ satisfies $xy=\lambda_{p}\mu e_{j,j}\neq 0$. This shows that $S$ satisfies (5.1) for all $v\in\mathbb{Z}$. As $S_{v}=\{0\}$ for $|v|\geq n$, the ring $S$ is not strongly graded.

As a matter of terminology, $S_{+}=\bigoplus_{v\in\mathbb{N}^{k}}S_{v}$ is the positive subring of $S$.

Theorem 5.4**.**

Suppose that the $\mathbb{N}^{k}$-graded ring $R=S_{+}$ is the positive subring of a $\mathbb{Z}^{k}$-graded ring $S$, and suppose that $S$ satisfies condition (5.1) for all $v\in\mathbb{N}^{k}\setminus\{0\}$. Let $A$ be a right ideal in $R$ such that $A^{\mathrm{r}}$, its right annihilator in $R$, is non-trivial. Then $A^{\mathrm{r}}\cap R_{0}\neq\{0\}$, i.e., $A^{\mathrm{r}}$ contains a non-zero homogeneous element $h$ of degree [math].

Proof.

By Theorem 4.1 there exist $v\in\mathbb{N}^{k}$ and a non-zero element $g\in R_{v}=S_{v}$ such that $g\in A^{\mathrm{r}}$. If $v=0$ we are done. Otherwise, the hypothesis on $S$ guarantees that there exists an element $y\in S_{-v}$ with $gy\neq 0$. Set $h=gy$; by construction $h\in S_{0}=R_{0}$, and $h\in A^{\mathrm{r}}$ as $A^{\mathrm{r}}$ is a (right) ideal. ∎

Corollary 5.5**.**

In the situation of Theorem 5.4, the ideal $A^{\mathrm{r}}$ contains a graded ideal of $R$ intersecting $R_{0}$ non-trivially. ∎

Theorem 5.4 applies to the $\mathbb{N}^{k}$-graded ring $R=K[X_{1},\,X_{2},\,\cdots,\,X_{k}]$ of polynomials with coefficients in a ring $K$ with trivial left annihilator (that is, $xK=\{0\}$ implies $x=0$). Indeed, we have $R=S_{+}$ where $S$ is the Laurent polynomial ring

[TABLE]

equipped with the usual $\mathbb{Z}^{k}$-grading, giving the indeterminate $X_{j}$ degree $e_{j}$, the $j$th unit vector. For unital $K$ we recover the classical result of McCoy [McC57, Theorem].

To go any further, we need to put stronger conditions on our rings:

Theorem 5.6**.**

Let $S$ be a strongly $\mathbb{Z}^{k}$-graded unital ring, and let $R=S_{+}$ be its positive subring. Let $f_{1},\,f_{2},\,\cdots,\,f_{\ell}\in S$, and suppose that there exists a non-zero element $g\in S$ such that

[TABLE]

Then there exists a non-zero element $h\in R_{0}$, homogeneous of degree [math], such that $f_{j}rh=0$ for all $j$ and all $r\in R$.

Proof.

Without loss of generality we may assume that $g\in R$. Indeed, we may choose a vector $w\in\mathbb{N}^{k}$, with sufficiently large positive entries, such that $w+\mathrm{supp}(g)=\{w+x\,|\,x\in\mathrm{supp}(g)\}\subset\mathbb{N}^{k}$. As $S$ is strongly graded there are finitely many elements $x_{i}\in S_{w}$ and $y_{i}\in S_{-w}$ such that $\sum_{i}x_{i}y_{i}=1$. As $g=g\cdot 1=\sum_{i}(gx_{i})\cdot y_{i}$ is non-zero there is an index $p$ with $gx_{p}\neq 0$. By construction $gx_{p}\in R$ and $f_{j}r(gx_{p})=(f_{j}rg)x_{p}=0$ for all $j$, so we can replace $g$ with $gx_{p}\in R$ in (5.7).

Similarly, we may choose a $v\in\mathbb{N}^{k}$ such that $v+\mathrm{supp}(f_{j})\subset\mathbb{N}^{k}$ for all $j$. As $S$ is strongly graded there are finitely many elements $x_{i}\in S_{-v}$ and $y_{i}\in S_{v}$ such that $\sum_{i}x_{i}y_{i}=1$. Then $y_{i}f_{j}\in R$ by choice of $v$, and for all indices $i$ and $j$, and all $r\in R$, we have

[TABLE]

by our hypotheses on $g$ and the $f_{j}$. This means that the right ideal A=\big{\langle}\{y_{i}f_{j}\}\big{\rangle} of $R$ generated by the elements $y_{i}f_{j}$ has a non-trivial right annihilator in $R$, viz., the element $g$. By Theorem 5.4 there exists a non-zero element $h\in R_{0}$ annihilating $A$ from the right. In particular, $y_{i}f_{j}rh=0$ for all indices $i$ and $j$, and all $r\in R$. But then we also have the equality

[TABLE]

for all $j$ and all $r\in R$, proving the Theorem. ∎

Corollary 5.8**.**

Let $S$ be a strongly $\mathbb{Z}^{k}$-graded unital ring. Suppose that $S$ is semi-commutative so that $ab=0$ implies $asb=0$ for all $s\in S$. Let $f_{1},\,f_{2},\,\cdots,\,f_{\ell}\in S$, and suppose that there exists a non-zero element $g\in S$ such that $f_{j}g=0$ for all $j$. Then there exists a non-zero element $h\in S_{0}$, homogeneous of degree [math], such that $f_{j}h=0$ for all $j$.

Proof.

By definition of semi-commutativity the condition $f_{j}g=0$ implies $f_{j}sg=0$ for all $s\in S$. In particular, condition (5.7) is satisfied, hence Theorem 5.6 applies. ∎