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


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