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In general, it is not true that intersections of free submodules of a free module are free, see here. What if our ring is particularly nice?

Over $k[x_1, x_2]$, the intersection of free submodules $F_1, F_2 \subseteq F$ of a free module can be written as the pull-back $F_1 \times_F F_2$, which is free by the syzygy theorem.

What about more general polynomial rings? Are intersections of free submodules of a free modules free for $k[x_1,\dotsc,x_n]$-modules, or other Noetherian rings?

Bubaya
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1 Answers1

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The answer is no. Here is an example.

Let the ring be $k[x,y,z]$. Let $F$ be the free module of rank $4$ on $e_1,e_2,e_3,e_4$. Let $F_1$ be the free submodule generated by $e_1,e_2,e_3$ and $F_2$ be the free submodule generated by $e_1+xe_4,e_2+ye_4,e_3+ze_4$. I will let you check that $F_1\cap F_2$ is not free.

Mohan
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  • I see. $F_1 \cap F_2$ has a mininimal generating system $g_1, g_2, g_3$, where $g_1 = ye_1 - xe_2$, $g_2=ze_2 - ye_3$, $g_3 = ze1 - xe_3$. Then $zg_1 + xg_2 - yg_3 = 0$, so this cannot be free. – Bubaya Jun 02 '22 at 08:37