## Archive for the ‘Graded Algebras & Modules’ Category

Posted: April 22, 2011 in Graded Algebras & Modules, Noncommutative Ring Theory Notes
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For basic definitions and examples see here. We continue our discussion with a few more examples.

Example 1. Let $R=M_m(A)$ be the algebra of $m \times m$ matrices with entries in an algebra $A.$ Let $e_{ij}$ be the $m \times m$ matrix with $1$ in the $i$-th row and $j$-th column and $0$ anywhere else. Let $R_n = \sum_{i \geq 1} Ae_{i, i+n}$ for all integers $|n| < m.$ For $|n| \geq m$ we define $R_n=0.$ So, for example, if $R=M_2(A),$ then $R_{-1}=Ae_{21}, \ R_0=Ae_{11}+Ae_{22}$ and $R_1=Ae_{12}$ and $R_n = 0$ for all $|n| \geq 2.$ The reason that $R = \bigoplus_{|n| < m} R_n$ holds is that for every $1 \leq i,j \leq m$ we may let $n=j-i.$ Then obviously $|n| < m$ and $e_{ij}=e_{i,i+n} \in R_n.$  Besides, every $e_{ij}$ appears in $R_n$ for the unique $n = j-i.$ Finally $R_nR_k = \sum_{i,j \geq 1} Ae_{i,i+n}e_{j,j+k}$ and $e_{i,i+n}e_{j,j+k}$ is non-zero if and only if $i+n=j.$ Thus

$R_nR_k \subseteq \sum_{i \geq 1} Ae_{i,i+n+k} = R_{n+k}.$

Example 2. Suppose $G_1, G_2$ are two groups and $f: G_1 \longrightarrow G_2$ is an onto group homomorphism. If $R$ is $G_1$-graded, then $R$ is also $G_2$-graded if we define $R_h = \bigoplus_{ \{g \in G_1: \ f(g)=h \} } R_g$ for all $h \in G_2.$ Clearly

$\bigoplus_{h \in G_2} R_h = \bigoplus_{g \in G_1} R_g = R.$

It is also easy to see that $R_h R_u \subseteq R_{hu}$ for all $h,u \in G_2.$

Example 3. This example is an application of Example 2. Let $G_1 = \mathbb{Z}$ and $G_2= \mathbb{Z}/2\mathbb{Z}.$ Let $f: G_1 \longrightarrow G_2$ be the natural group homomorphism. Let $R=M_m(A)$ with the grading introduced in Example 1. So, by Example 2, we have a $G_2$-grading for $R=R_0 \oplus R_1,$ where

$R_0 = \bigoplus_{ \{n \in G_1: \ f(n)=0 \}} R_n = \bigoplus_{2 \mid n} R_n = \bigoplus Ae_{i,i+2n} = \bigoplus_{2 \mid i+j} Ae_{ij}$

and, similarly, $R_1$ is the direct sum of all $Ae_{ij}$ such that $i+j$ is odd.

We now move on to define an important concept, i.e. a graded ideal of a graded algebra.

Definition. Let  $R=\bigoplus_{g \in G}R_g$ be a $G$-graded algebra. A graded or homogenous ideal of $R$ is an ideal $I$ such that $I= \bigoplus_{g \in G} (I \cap R_g).$ Graded left or right ideals and graded subalgebras of $R$ are defined analogously.

Theorem. Let  $R=\bigoplus_{g \in G}R_g$ be a $G$-graded algebra and let $I$ be an ideal of $R.$ Then $I$ is graded if and only if, as an ideal, $I$ is generated by a subset of $\bigcup_{g \in G} R_g.$ In other words, $I$ is graded if and only if $I$ can be generated by a set of homogeneous elements of $R.$

Proof. If $I$ is generated by $S \subseteq \bigcup_{g \in G} R_g,$ then $I = \sum_{s \in S} Rs.$ Now, if $s \in S,$ then $s \in R_u,$ for some $u \in G.$ Let $r \in I.$ Then $r = r_{g_1} + \ldots + r_{g_n},$ where $r_{g_i} \in R_{g_i}.$ Clearly $r_{g_i}s \in I$ because $s \in I$ and $I$ is an ideal of $R.$ Also, $r_{g_i}s \in R_{g_iu}$ because $s \in R_u$ and $r_{g_i} \in R_{g_i}.$ Thus $r_{g_i}s \in I \cap R_{g_iu}$ and hence $r \in \bigoplus_{i=1}^n (I \cap R_{g_iu}) \subseteq \bigoplus_{g \in G} (I \cap R_g).$ So we have proved that $I \subseteq \bigoplus_{g \in G}(I \cap R_g)$ and, since the other side of the inclusion is trivial, we get that $I$ is graded. Conversely, suppose that $I = \bigoplus_{g \in G}(I \cap R_g)$ and let $S = \bigcup_{g \in G} (I \cap R_g).$ We want to prove that $I=\sum_{s \in S} Rs.$  Obviously $\sum_{s \in S} Rs \subseteq I$ because $S \subseteq I$ and $I$ is an ideal.  Also every element of $I$ is a finite sum of elements of $S$ because $I = \bigoplus_{g \in G}(I \cap R_g).$ Thus $I \subseteq \sum_{s \in S} Rs$ and we are done. $\Box$

## Graded algebras; basic definitions and examples

Posted: December 18, 2010 in Graded Algebras & Modules, Noncommutative Ring Theory Notes
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Throughout $G$ is a group, $C$ is a commutative ring with 1 and $R$ is a $C$algebra.

Definition 1. $R$ is called a $G$graded $C$-algebra if for every $g \in G$ there exists a $C$-module $R_g \subseteq R$ such that

1) $R_gR_h \subseteq R_{gh},$ for all $g,h \in G,$

2) $R = \bigoplus_{g \in G} R_g,$ as $C$-modules.

A homogeneous element of $R$ is any element of $R_g, \ g \in G.$ If $0 \neq r_g \in R_g,$ then $r_g$ is called homogeneous of degree $g.$ If $r \in R,$ then $r$ is written uniquely as $r = \sum_{g \in G} r_g,$ where $r_g \in R_g$ and all but finitely many of $r_g$ are zero. Each $r_g$ is called a homogeneous component of $r.$

Definition 2. We say that $R$ is positively graded if $R$ is $\mathbb{Z}$-graded and $R_n = 0$ for all $n < 0.$

The concept of graded algebras is just a very natural generalization of polynomial algebras.

Example 1. Let $A$ be a $C$-algebra and consider the polynomial algebra $R=A[x].$ Then $R$ is positively graded because clearly $R=\bigoplus_{n=0}^{\infty}R_n,$ where $R_n=Ax^n, \ n \geq 0.$ Note that $R_nR_m=Ax^{n+m}=R_{n+m}.$ This grading for $R$ is called the standard grading of $R.$

Example 2. Let $R=A[x,y],$ the polynomial algebra in the indeterminates $x$ and $y.$ Let $R_n, \ n \geq 0,$ be the set of all polynomials of total degree $n.$ For example, $R_0=A, \ R_1=Ax+Ay, \ R_2=Ax^2+Axy+Ay^2,$ etc. Then $R=\bigoplus_{n=0}^{\infty}R_n.$ Note that $R_nR_m=R_{n+m}.$ So $R$ is positively graded and we call this the standard grading of $R.$ In genral, the standard grading of the polynomial algebra $R=A[x_1, \ldots , x_m]$ in the indeterminates $x_1, \ldots , x_m$ is defined by $R = \bigoplus_{n=0}^{\infty}R_n,$ where $R_n$ is the set of polynomials of total degree $n.$

Example 3. The standard grading of the Laurent polynomial algebra $R=A[x,x^{-1}]$ is $R=\bigoplus_{n \in \mathbb{Z}} Ax^n.$

Remark. Let $R=\bigoplus_{n=0}^{\infty} R_n$ be a (positively) graded algebra and let $R_{+}=\bigoplus_{n=1}^{\infty} R_n.$ Then

1) $R_0$ is a subring of $R$ and $R_n$ is an $R_0$-module for all $n \geq 0.$

2) $1_R \in R_0.$

3) $R_{+}$ is a two-sided ideal of $R.$

Proof. 1) By the property 1) in Definition 1, we have $R_0R_n \subseteq R_n$ and $R_nR_0 \subseteq R_n.$ That means $R_n$ is both left and right $R_0$-module. Also, $R_0R_0 \subseteq R_0$ and so $R_0$ is a subring of $R.$ To prove 2), let $1 = \sum_{n=0}^{\infty} r_n,$ where $r_n \in R_n$ and only a finitely many of $r_n$ are non-zero. We need to show that $r_n = 0$ for all $n > 0.$ To prove this, let $m \geq 0.$ Then $r_m = \sum_{n \geq 0} r_mr_n.$ But $r_mr_n \in R_{m+n}$ and if $n > 0,$ then $m + n \neq m.$ Thus $r_m r_n = 0,$ for all $m \geq 0$ and $n > 0.$ Thus if $t > 0,$ then $r_t = \sum_{n \geq 0}r_nr_t = 0.$ Part 3) of the remark is trivial. $\Box$

Definition 3. The ideal $R_{+}$ in the above remark is called the augmentation ideal of $R.$

Example 4. Let $R$ be the polynomial algebra in Example 1. Then $R_{+}=\bigoplus_{n=1}^{\infty} Ax^n = \langle x \rangle,$ the ideal generated by $x.$ Similarly, the augmentation ideal of the polynomial algebra $R=A[x_1, \ldots , x_m]$ (see Example 2) is $\langle x_1, \ldots , x_m \rangle.$