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I'm trying to work through Exercise III.27 of Lang's Algebra:

Let $A$ be a filtered algebra, $A=\bigcup_{j\geq 0}A_{j}$. For $j\geq 0$, define $R_{j}=A_{j}/A_{j-1}$, with $A_{-1}=\{0\}$. Let $R=\bigoplus_{j\geq 0}R_{j}$. Define a natural product on $R$ making $R$ into a graded algebra.

My Attempt: I believe that the product is defined by $$(x+A_{n-1})(y+A_{m-1})=xy+A_{n+m-1}$$ for all $x\in A_{n}$ and all $y\in A_{m}$. Then $R$ is clearly a direct sum of subspaces and $R_{j}R_{k}\subset R_{j+k}$ for all $j$ and $k$, since $$ \frac{A_{j}}{A_{j-1}}\cdot\frac{A_{k}}{A_{k-1}}\subset\frac{A_{j+k}}{A_{j+k-1}}$$ (in the above containment, we use the fact that $xy\in A_{j+k}$ if $x\in A_{j}$ and $y\in A_{k}$ ; Also, $A_{j-1} \cdot A_{k-1} \subset A_{j+k-1}$). Hence, $R$ is a graded algebra.

My Questions: Does this argument look okay? In particular, how can I show that the product is well-defined?

Thanks in advance for any suggestions.

Adithya S
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user608571
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1 Answers1

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You definition is correct. However, it's a bit nicer to define the product (or, rather, justify that the product you give is well-defined) by using some universal properties of quotients. Consider the composition $$A_n \times A_m \to A_{n+m} \twoheadrightarrow R_{n+m},$$ where the first map is the multiplication in $A$ and the second is the natural projection. Since $A_{n-1} \times A_m$ and $A_n \times A_{m-1}$ are in the kernel of the composition, you get an induced map $R_n \times R_m \to R_{n+m}$, which is precisely the product you wrote down.

Alistair Savage
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