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I am aware that the following question is a very basic one and therefore I would not be at all offended if it were to be closed. Moreover, I am not familiar at all with category theory.

Let $\mathcal{C}$ be a concrete category and $X$ be a free object of $\mathcal{C}$.

If $Y_1$ and $Y_2$ are both free subobjects of $X$, then is the intersection, $Y_1 \cap Y_2,$ free?

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2 Answers 2

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It's not in general true for the category of modules for a ring.

For example, let $R=\mathbb{C}[x]/(x^2)$, let $X=R\oplus R$ be the free module on two generators, and let $Y_1$ and $Y_2$ be the submodules of $X$ generated by $(1,0)$ and $(1,x)$ respectively. Then $Y_1$ and $Y_2$ are both free modules on one generator, but $Y_1\cap Y_2$ is one-dimensional, spanned by $(x,0)$, and is not free.

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This is true for submonoids of a free monoid, but not for free submonoids of non-free monoid. See

Tilson, B., The Intersection of Free Submonoids of a Free Monoid is Free. Semigroup Forum 4, (1972), 345-350.

Addendum: Recently I found an article:

Shubh Narayan Singh, K.V. Krishna. A sufficient condition for the Hanna Neumann property of submonoids of a free monoid. Semigroup Forum, 86(2013), pp.537–554.

It contains many useful references.

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    $\begingroup$ Indeed, I know for the while only the proof given in Lothaire's $\textit{Combinatorics on words}$. The reason why I ask this is because I consider intersection of many other free (combinatorial) algebraic structures than monoids and I would hope that a categorical argument imply their freeness. Since this is not the case (see Jeremy's answer) the only way seems to proceed case by case. $\endgroup$ Commented Jan 10, 2014 at 22:10
  • $\begingroup$ Yes, I think you are right. $\endgroup$ Commented Jan 10, 2014 at 22:23
  • $\begingroup$ And by the way, thanks for the reference! $\endgroup$ Commented Jan 10, 2014 at 22:26

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