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For a ring $A$ and an ideal $\mathfrak{a}$ of $A$, Atiyah-Macdonald define $$A^*=\bigoplus_{n=0}^\infty \mathfrak{a}^n$$ and claim that it is a graded ring on p. 107 of their commutative algebra book. My confusion is that I cannot see how $A^*$ is a ring. Does the direct sum represent the internal direct sum or the external direct sum?

Since $\mathfrak{a}^0\supseteq\mathfrak{a}^1\supseteq\cdots$, I think the internal direct sum is not possible. If it is the external direct sum, then what would be the multiplicative identity of $A^*$? Maybe I am missing something completely obvious.

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It's the external direct sum, yes (as you remark the internal direct sum wouldn't make sense). The product is such that $\mathfrak{a}^i \cdot \mathfrak{a}^j \subset \mathfrak{a}^{i+j}$. The unit is the unit of $A = \mathfrak{a}^0$.

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  • $\begingroup$ If it is the external direct sum, its elements be of the form $(a_0, a_1, \cdots)$, $a_i\in\mathfrak{a}_i$ and $a_i=0$ for all but finitely many $i$ and the product would be componentwise. Am I correct? $\endgroup$ – PJK Oct 7 '14 at 8:33
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    $\begingroup$ The addition would be component-wise. The product, on the other hand, would be defined by $(a_0, a_1...) \cdot (b_0, b_1...) = (c_0, c_1...)$ where $c_n = \sum_{i+j=n} a_i b_j$. That's what I meant by $\mathfrak{a}^i \cdot \mathfrak{a}^j \subset \mathfrak{a}^{i+j}$. $\endgroup$ – Najib Idrissi Oct 7 '14 at 8:36
  • $\begingroup$ I didn't think of multiplication as such. It seems to be pretty clear now. Thank you. $\endgroup$ – PJK Oct 7 '14 at 8:49

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