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. $\Box$