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Let $R$ be a ring, and let $I_1,\ldots,I_n$ be ideals in $R$ (or submodules of some $R$-module). Consider the sequence $$ \bigoplus_{1\leq j < k\leq n} I_j\cap I_k\quad\xrightarrow{f}\quad\bigoplus_{l=1}^n I_l\quad\xrightarrow{g}\quad\sum_{k=1}^n I_k, $$ where $g$ is given by addition, and $f$ maps $x\in I_j\cap I_k$ to $x\in I_j$ and to $-x\in I_k$ (and to zero in all other components).

Clearly, $g$ is surjective and the composition $g\circ f$ vanishes.

Question: Is the above sequence exact in the middle?

(This seems to be easy for $n=2$.)

(Concerning the title: I am aware of the fact that $f$ won't be injective in general.)

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Your sum is unclear: is the sum over all ordered pairs $(j,k)$, all ordered pairs with $j\lt k$, all ordered pairs with $j\leq k$? Something else? E.g., for $n=2$, do you have just $I_1\cap I_2$, or $I_1\oplus (I_1\cap I_2)\oplus I_2$, or $I_1\oplus(I_1\cap I_2)\oplus (I_2\cap I_1)\oplus I_2$? –  Arturo Magidin Feb 8 '11 at 19:11
    
@Arturo: edited. The sum is over all pairs; however, on summands with $j=k$, the map $f$ will vanish anyway. –  Rasmus Feb 8 '11 at 21:12
    
Actually, all variants Arturo suggests give the same image. Maybe, it is nicest, to take only ordered pairs with $j<k$. Edited accordingly. –  Rasmus Feb 9 '11 at 12:26
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As discovered in the follow-up to this question, for commutative rings the property is equivalent to the ring being arithmetical. For Noetherian domains, it is equivalent to it being Dedekind. So, non-integrally closed rings such as $R=\mathbb{Z}[\sqrt{5}]$ provide simple counterexamples. For the follow-up question: math.stackexchange.com/questions/21170/… –  George Lowther Feb 11 '11 at 23:29
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2 Answers

up vote 4 down vote accepted

You can convert chandok's example to ideals in a commutative ring as well. Take R=Z[x,y,z]/(3,xx,xy,xz,yy,yz,zz), a 4-dimensional algebra over Z/3Z. It has ideals of (vector space) dimension 1 generated by x−y, y−z, and z−x. There is clearly a non-zero kernel of "g" containing the triple ( x−y, y−z, z−x ). However, the domain of "f" is 0, since all pairwise intersections of the ideals are 0.

The obvious translation of chandok's example to one over the ring Z3 fails, since the ideals generated by the xi get bigger. This ring fixes that, since it has a bunch of pointless ring elements that leave most abelian subgroups as ideals.

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interesting example -- thanks! Is there also an example for ideals in a (not necessarily commutative) algebra over the complex numbers? –  Rasmus Feb 8 '11 at 21:22
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Oh sure! R=C[x,y,z]/(x,y,z)^2 (tangent something-something at (0,0,0) in the 3-dimensional affine space). The ideals are 1-dimensional (as vector spaces) with generators x−y, y−z, and z−x. Everything works the same (for any field I think, I just chose the smallest where the things looked normal). –  Jack Schmidt Feb 8 '11 at 21:35
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If you take $I_1, \ldots I_n$ submodules of an $R$-module $M$, you can easily find a counterexample when their intersection is zero :

Pick $R = \mathbb{Z}, M = \mathbb{Z}^3, x_1 = (1,-1,0), x_2 = (0,1,-1), x_3 = (-1,0,1)$. Then you have that $x_1+x_2+x_3 = 0$, $x_i$ is in its submodule $I_i = \mathbb{Z}x_i$, and for $i \not = j, I_i \cap I_j = \{0\}$.

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