Often while reading in Jacobson's Basic Algebra textbook, I find him referring to elements of a set in a particular, weird way.

Instead of saying for example "let $x_1,\ldots,x_n$ be elements of $S$", he says let $i \mapsto x_i$ be a map of $\{1,\ldots,n\}$ into $S$.

Why does he do this? Is it just for rigour purposes?

Added: here is a concrete one.

For any ring $R$ and any positive integer $r$ there exists a ring $R[x_1,\ldots,x_r]$ with the following "universal" property. If $S$ is any ring and $\eta$ is a homomorphism of $R$ into $S$ and $i\mapsto u_i$ is a map of $\{1,\ldots,r\}$ into $S$, then there exists a unique extension of $\eta$ to a homomorphism of $R[x_1,\ldots,x_r]$ into $S$ sending $x_i$ to $u_i$, $1\le i \le r$.

Another one is on page $69$ (first volume) when talking about free groups.

  • 2
    $\begingroup$ Without context, it is impossible to answer. He might use this special form for some arguments later. $\endgroup$ – Mundron Schmidt Jul 10 '17 at 10:43
  • $\begingroup$ @MundronSchmidt I added context. $\endgroup$ – Cauchy Jul 10 '17 at 10:49
  • $\begingroup$ The additional information is that in that way the elements of $S$ are "ordered". $\endgroup$ – Mauro ALLEGRANZA Jul 10 '17 at 10:59
  • $\begingroup$ @MauroALLEGRANZA in what sense? $\endgroup$ – Cauchy Jul 10 '17 at 11:01
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    $\begingroup$ @MauroALLEGRANZA I really don't see the significance of this. To me this is exactly the same as "let $x_1,\ldots,x_r$ be elements of $S$". $\endgroup$ – Cauchy Jul 10 '17 at 11:09

The issue is the uniqueness of the extension, as claimed in the book. If you didn't have a map $i\mapsto u_i$ to begin with, you couldn't make such a claim. More importantly, notice that the map $i\mapsto u_i$ is not required to be injective. The uniqueness claim remains true.

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    $\begingroup$ I still don't see what difference the map thing makes. Why is uniqueness affected if we just say "let $x_1,\ldots, x_r\in S$"? Unless what you are referring to is just the remark Dirk already gave. $\endgroup$ – Cauchy Jul 10 '17 at 11:47
  • $\begingroup$ If you change the order, the extended map changes as well. Again, you don't need to have $r$ distinct members of $S$ either. $\endgroup$ – Mikhail Katz Jul 10 '17 at 11:48
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    $\begingroup$ I really don't think uniqueness has to do with anything here. I think it is just what Dirk mentioned in the comments (and what you subsequently mentioned in your answer). $\endgroup$ – Cauchy Jul 10 '17 at 11:51
  • $\begingroup$ @MikhailKatz I know the direction you're headed in is the right one... but the answer could be improved by being a bit more explicit! Probably even one example would suffice to show how order and repetition are both handled by the map scheme Jacobson gave. $\endgroup$ – rschwieb Jul 10 '17 at 13:09
  • $\begingroup$ Hi @rschwieb, thanks. I think the OP seems in command of the material. $\endgroup$ – Mikhail Katz Jul 10 '17 at 13:13

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