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Was just reading this question When is a group isomorphic to a proper subgroup of itself? and was wondering not about the conditions for being isomorphic to a proper subgroup, subring, etc., but about examples of these things happening. (Certainly one necessary condition is that our algebraic object be infinite as a set. Otherwise we cannot have a bijection between the initial set and a proper subset.)

I believe one example was given in the above link using even powers of a polynomial ring $R[x]$ and the ring $R[x]$ itself. I believe another example is in Dummit and Foote with the roots of unity or something like this. Anyway, have at it!

(Here is another related post: Rings with isomorphic proper subrings)

(Feel free to also post answers with maybe manifolds which are homeomorphic/diffeomorphic/biholomorphic to proper submanifolds or things like that if you have some favorites!)

(The only category which I know excludes this business is algebraic geometry...but only kinda if you deal with incomplete intersections...)

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    $\begingroup$ Maybe the simplest example is $\mathbb{Z} \cong n\mathbb{Z}$ as group? $\endgroup$
    – Alex Vong
    May 12, 2017 at 16:01
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    $\begingroup$ The well-known field $\mathbb{C}$ has proper subfields isomorphic as fields to $\mathbb{C}$. (Unlike the field $\mathbb{R}$.) $\endgroup$
    – ancient mathematician
    May 12, 2017 at 16:14
  • $\begingroup$ @rschwieb, if I understand you, my answer would be that the reals-of-$D$ are not the reals of $\mathbb{C}$. The degree of $D$ over the $\mathbb{R}$ of your pic will be infinite. $\endgroup$
    – ancient mathematician
    May 12, 2017 at 16:41
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    $\begingroup$ @ancientmathematician An explicit description of the construction would probably help me see the quickest, if you could be so kind. $\endgroup$
    – rschwieb
    May 12, 2017 at 16:42
  • $\begingroup$ @rschwieb I'll comment on your answer. Tell me if I am writing nonsense! $\endgroup$
    – ancient mathematician
    May 12, 2017 at 16:50

2 Answers 2

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If you have an example for fields, you have an example for rings and abelian groups at the same time:

Take the field of rational polynomials $F(x)$ where $F$ is a field. The map $x\mapsto x^2$ defines a ring homomorphism of $F(x)\to F(x)$ which is necessarily injective (since $F(x)$ is a field) but not onto (its image is $F(x^2)$. The image is an isomorphic copy of $F(x)$ strictly contained in $F(x)$.

Of course, this means there is an infinite strictly descending chain of isomorphic copies...

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    $\begingroup$ We can make more complicated examples by taking for example for $F$ the algebraic closure of the rationals; and then looking at the field of rational functions of a countable set of algebraically independent trancendentals (or indeterminates) $t_1, t_2, \dots$, that is $K=F(t_1,t_2,\dots)$. This has, for any infinite subset of the transcendentals a corresponding isomorphic subfield. And if we take the algebraic closure of $K$ we get a field isomorphic to $\mathbb{C}$ do we nt? $\endgroup$
    – ancient mathematician
    May 12, 2017 at 16:50
  • $\begingroup$ @ancientmathematician You are probably building a copy of $\mathbb C$, but I'm still not clear on how it is shown this is a proper subring of a copy of $\mathbb C$. $\endgroup$
    – rschwieb
    May 12, 2017 at 16:53
  • $\begingroup$ Well $K_1=F(t_2, t_3, \dots)\simeq K$ and is proper in $K$. Taking the algebraic closure of $K$ gives $\mathbb{C}$ in case when the $t_i$ are a transcendence basis for $\mathbb{C}$; it properly includes the algebraic closure of $K_1$ (as we don't pick up $t_1$), and this is also isomorphic to $\mathbb{C}$. $\endgroup$
    – ancient mathematician
    May 12, 2017 at 17:20
  • $\begingroup$ @ancientmathematician What $F$ are you proposing? $\mathbb Q$? $\endgroup$
    – rschwieb
    May 12, 2017 at 18:13
  • $\begingroup$ Lüroth's theorem is a generalization of this result. $\endgroup$
    – Viktor Vaughn
    May 13, 2017 at 0:39
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$\mathbb{Z} \cong n \mathbb{Z}$ by way of the group isomorphism defined by $\varphi (k) = nk$, with $k,n \in \mathbb{Z}$ of course.

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    $\begingroup$ Of course $\varphi$ is not an isomorphism of rings if $n > 1$; it is only an isomorphism of groups. $\endgroup$
    – Pete L. Clark
    May 12, 2017 at 21:07
  • $\begingroup$ Oh wow, how embarrassing. Editing now. $\endgroup$
    – Hayden
    May 12, 2017 at 21:10

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