# Quotient ring of a graded algebra with respect to a graded ideal

An algebra $A$ over $F$ is said to be a graded algebra if as a vector space over $F$, $A$ can be written in the form

$$A=\bigoplus_{i=0}^\infty A_i$$

for subspaces $A_i$ of $A$ along with other properties.

And a graded ideal $I$ in a graded algebra $A$ is an ideal $I$ for which as a subspace of $A$,

$$I = \bigoplus_{i=0}^\infty (I \cap A_i)$$

In Steve Roman's book "Advanced Linear Algebra" it claims that if $I$ is a graded ideal in $A$, then the quotient ring $A/I$ is also graded, since

$$\frac{A}{I} = \bigoplus_{i=0}^\infty \frac{A_i+I}{I}$$ Can someone give a careful proof of this? I am familiar with the coset or equivalence relation construction behind $A/I$ as being interpreted as identifying elements of $I$ with $0$. But the numerator $A_i+I$ on the right side is an unfamiliar sight. Presumably a proof will enlighten me on how to think of it.

Quotient groups commute with direct sums. Therefore, $A/I = \bigoplus_i A_i / (I \cap A_i)$ as abelian groups. But $A_i / (I \cap A_i) \cong (A_i+I)/I$ (second isomorphism theorem). (Notice that every element in this group is the coset of some element of $A_i$, but we cannot write $A_i/I$ since $I \subseteq A_i$ does not hold.) That the multiplication rule $(A_i + I)/I \cdot (A_j + I)/I \subseteq (A_{i+j} + I)$ holds, follows directly from $A_i \cdot A_j \subseteq A_{i+j}$.
Explicitly, we call an element of $A/I$ to be homogeneous of degree $i$ if it is represented by some element of $A$ which is homogeneous of degree $i$. You could also verify that this makes $A/I$ a graded algebra this way, using elements, if you like.
• Can I ask why $A_i/(I \cap A_i)$ is an abelian subgroup of $A/I$? I dont see this, they don't share same elements, the group $A/I$ has element of form $a+I$ whereas the other group has elements of form $a + (I \cap A_i)$. Can you explain this please? – user117449 Mar 2 '15 at 1:36