Let $$S=\oplus S_i$$ be some graded ring and let $$I\subset S$$ be a graded/homogeneous ideal of $$S$$. That is to say, $$I=\oplus I_i$$, where $$I_i=S_i\cap I$$ (this is equivalent to the property that $$I$$ has a set of homogeneous generators). Put $$R=S/I$$. I want to prove $$R\cong \oplus R_i$$, where $$R_i=S_i/I_i$$, and also that $$R_iR_j\subset R_{i+j}$$. Essentially, I'm looking to prove that $$R$$ is a graded ring.

Here are my thoughts:

For the isomorphism, the map that seems obvious to me is $$s_a+\ldots+s_b\longmapsto(s_a+I_a)+\ldots+(s_b+I_b)$$ and then apply the first isomorphism theorem. My question here is, do we need to deal with the well-definedness of this map? Am I correct in saying that this map is well-defined because the sum $$s_a+\ldots+s_b$$ is unique since $$S=\oplus S_i$$ is a direct sum?

My next question is, how do we show $$R_iR_j\subset R_{i+j}$$? I have tried the following: let $$a\in R_i$$ and let $$b\in R_j$$. So $$a=s_i+I_i$$ and $$b=s_j+I_j$$ for some $$s_i\in S_i$$ and $$s_j\in S_j$$. I want to show $$ab\in R_{i+j}=S_{i+j}/I_{i+j}$$. Here is where I'm stuck, since I am confused on how $$ab$$ is defined. These are two cosets involving two potentially different ideals, so I'm not sure how to multiply them in a way that makes sense. How can we do this?

• How about $ab:=s_is_j+I_{i+j}$? – Shivering Soldier Jul 30 at 17:33

Let's first fuss about why $$R' := \bigoplus_iR_i$$ is well-defined as a graded ring (I will denote it as $$R'$$ to differentiate it for now from $$R=S/I$$). To see that $$R_iR_j\subseteq R_{i+j}$$, suppose we take $$s_i+I_i\in R_i$$ and $$s_j+I_j\in R_j$$, then just doing the "obvious" multiplication gives $$(s_i+I_i)(s_j+I_j) = s_is_j + s_iI_j + s_jI_i + I_iI_j$$ Since $$S$$ is a graded ring, $$s_is_j\in S_iS_j\subseteq S_{i+j}$$, and also $$I_iI_j\subseteq I_{i+j}$$ since $$I$$ is an ideal in the graded ring. Thus it's important to verify that $$s_iI_j\in I_{i+j}$$ (the other summand follows by symmetry). For any $$u_j\in I_j$$, we get $$s_iu_j\in I$$ from $$I$$ being an ideal, and $$s_iu_j\in S_{i+j}$$ from $$S$$ being graded, and therefore $$s_iu_j\in I\cap S_{i+j}=I_{i+j}$$, as desired. This means the "obvious" product $$(s_i+I_i)(s_j+I_j) = s_is_j + I_{i+j}$$ is well-defined in $$R'$$. Notice that we never needed the fact that $$I$$ was homogeneous for any of this to make sense, so $$R'$$ is a well-defined graded ring even if $$I$$ is not homogeneous in $$S$$.
Now regarding the isomorphism $$R\cong R'$$. Your "obvious" map is indeed well-defined if you are defining a map $$S\to R'$$ by virtue of $$S$$ being a graded ring, and by the above reasoning we will have a homomorphism of graded rings. However the kernel of this map is in question. Certainly anything that vanishes under this map is necessarily contained in the ideal $$I$$, but what of the converse? This is where homogeneity comes into play: for example, if $$S = \Bbb Q[x]$$ with the usual grading, and $$I = (x+1)$$, then $$I_j$$ is trivial for all $$j$$, so the resulting $$R'$$ will actually be $$S$$ itself, rather than $$R=S/I$$. However, if $$I$$ is homogeneous, then the homogeneous summands of any $$u\in I$$ also lie in $$I$$, and thus vanish under the map $$S\to R'$$ that you have defined. Therefore, the kernel of your map is exactly $$I$$ and by the first isomorphism theorem, we get that $$R\cong R'$$.