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**enter image description here**

The attached snapshot is from the book "Introduction to Commutative Algebra" by Atiyah & Macdonald. I have the following two questions:

  1. Can someone make me understand how the submodule $D$ is defined? since the element $x_i$ is in $M_i$ and and $u_ij (x_i)$ is in $M_j$, therefore I can't get how can we subtract these two elements.

  2. How the arbitrary element of the quotient $M=C/D$ will look like?

Prince Khan
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1 Answers1

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  1. Just before the highlighted part, $C$ is defined to be the direct sum of all $M_i$'s, and the natural inclusions $\iota_i:M_i\hookrightarrow C$ are implicitly applied here: to say it rigorously, $D$ is the submodule generated by the elements $\iota_i(x_i) - \iota_j(\mu_{ij}(x_i))$.

  2. So that, in $C/D$ we will have all elements of each $M_i$ present, and (for their equivalence classes), $x_i=\mu_{ij}(x_j)$.
    An arbitrary element of $C$ is of the form $x_{i_1}+\dots +x_{i_k}$ with $k\in\Bbb N, i_j\in I$.
    Since $I$ is assumed to be directed, and based on the above equality, there's an index $j\ge i_1,\dots, i_k$ and an element $y_j\in M_j$ so that in $C/D$, $$x_{i_1}+\dots +x_{i_k}=y_j$$

  3. Try to simplify it for the case when each $\mu_{ij}$ is an embedding (and, say, $I=\Bbb N$).

Berci
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  • When each $μ_i$$_j$ is an embedding and $I=N$, then in that case the direct limit is simply the direct sum, right? – Prince Khan Jan 23 '20 at 18:15
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    Not exactly, it's simply their union. – Berci Jan 23 '20 at 18:36
  • if you don't mind me answering a similar question [https://math.stackexchange.com/questions/3520041/direct-limit-of-arbitrary-family-of-tensor-products-of-a-algebras/3520059#3520059] – Prince Khan Jan 23 '20 at 18:42