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It is well known that the dimension of a subspace is less than or equal to the dimension of the vector space it is contained in. The same is true e.g. for modules over a principal ring.

I am looking for situations in which what would be naturally called the dimension does not satisfy this property, that is to say a class X of objects for which what would naturally play the role of the dimension does not satisfy this property. Namely, there exists an object $x$ in $X$ and $y$ a subobject of $x$ which is such that $dim(y) > dim(x)$.

The only example I found for the moment is free groups. The "dimension" is then the rank and it is known that given a free group of rank 2, there are subgroups that are isomorphic to free groups of any rank.

Do you have any other example of such a situation? (e.g. modules over some weird ring).

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migrated from mathoverflow.net Apr 22 '15 at 15:24

This question came from our site for professional mathematicians.

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    $\begingroup$ You've almost said so in your question, but what about modules over a ring that is not principal? $\endgroup$ – James Cranch Apr 22 '15 at 13:55
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    $\begingroup$ You wrote that a free group of rank 2 has subgroups that are free of any rank; you meant "of any rank up to and including $\aleph_0$" --- not uncountable ranks. $\endgroup$ – Andreas Blass Apr 22 '15 at 15:13
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This is kind of a silly example: $\mathbb{Z} \subseteq \mathbb{Q}$, but the Krull dimension of $\mathbb{Z}$ is one while the Krull dimension of $\mathbb{Q}$ is 0.

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  • $\begingroup$ This is right. Thank you very much! $\endgroup$ – Romain Gicquaud Apr 22 '15 at 14:16
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Look at the answer to this StackExchange question: A free submodule of a free module having greater rank the submodule (and also the MO discussion referenced therein).

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