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?

  • $\begingroup$ How about $ab:=s_is_j+I_{i+j}$? $\endgroup$ – 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 $R\cong R'$.

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