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I am interested in the following question raised by André Nicolas in this question.

If $(R,+,\cdot, 0,1)$ is the ring of real sequences (with pointwise addition and multiplication) and $(S,+,\cdot, 0,1)$ is the subring of convergent sequences, then are $R$ and $S$ elementarily equivalent?

Eric Wofsey
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abc
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  • Well, the inclusion $S \hookrightarrow R$ isn't an elementary embedding: $(\frac{1}{n})$ has no inverse in $S$, but it does have an inverse in $R$. – Daniel Schepler Jan 24 '18 at 18:57
  • True. But that only proves that $S$ is not an elementary substructure of $R$, which is a stronger property than just being elementarily equivalent. – abc Jan 24 '18 at 18:59
  • Hmm... $S$ is "almost" a local ring with maximal ideal "sequences converging to 0" - except that sequences converging to a nonzero number could have a finite number of zero terms which would still make them nonunits. – Daniel Schepler Jan 24 '18 at 19:03

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They are not elementarily equivalent. The ring $R$ has the (first-order) property that every non-unit is a zero-divisor. Indeed, if $x\in R$ is not a unit, then one of its coordinates is $0$. Letting $y$ be a sequence which is nonzero on that coordinate but $0$ on all other coordinates, $y\neq 0$ and $xy=0$, so $x$ is a zero-divisor.

On the other hand, $S$ does not have this property. Indeed, if $x\in S$ is any sequence of nonzero numbers which converges to $0$, then $x$ is not a unit, but it is also not a zero-divisor.

A perhaps more difficult question would be whether $S$ is elementarily equivalent to the ring of bounded sequences.

Eric Wofsey
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    $S$ is not elementarily to the ring of bounded sequences. In $S$ the sum of two non-invertible non-zero-divisors can not be invertible, but in the ring of bounded sequences it can. – Jeremy Rickard Jan 25 '18 at 09:54