Internal Direct Sum

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  • #1
markiv
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For a commutative ring [tex]R[/tex] and an ideal [tex]I[/tex], is it true that [tex]I \oplus R/I \cong R[/tex] ? I know in some cases this is true, and I know it's true for finitely-generated Abelian groups, but is it true for any commutative ring?

In other words, we know that [tex]R/I[/tex] is isomorphic to some ideal in [tex]R[/tex], call this [tex]J[/tex]. It's clear that [tex]I \cap J = 0[/tex], but does [tex]I + J = R[/tex] ?
 

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  • #2
Hurkyl
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You're missing some relevant details. What kind of isomorphism are you talking about? Ring isomorphism? Group isomorphism? Set isomorphism? When you say "ring", does it have to contain a multiplicative identity? When you say "ideal", is it even allowed to contain a multiplicative identity? (if so, there is only one such ideal)


I know it's true for finitely-generated Abelian groups
What is true? It is certainly not true that [itex]G \cong G/H \times H[/itex] for all abelian groups G and subgroups H -- e.g. let G be the cyclic group of 4 elements....
 
  • #3
markiv
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Heh well that last part answers my questions right away. I was talking about rings with identity and ring isomorphisms. And I was talking about just any ideal. But if it's not true for FG Abelian groups, then it can't be true for commutative rings either!

Thanks!
 
  • #4
markiv
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What is true? It is certainly not true that [itex]G \cong G/H \times H[/itex] for all abelian groups G and subgroups H -- e.g. let G be the cyclic group of 4 elements....

I'm just trying to understand why I thought this, but if [tex]H[/tex] is Sylow in [tex]G[/tex], then [tex]G \cong G/H \oplus H[/tex], right? Still assuming [tex]G[/tex] is FG Abelian, of course.
 
  • #5
Hurkyl
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I'm just trying to understand why I thought this
I never really remember the Sylow stuff. I have three other guesses as to why you might have thought it, though:
  1. It's true for vector spaces.
  2. It's "often" true for abelian groups (e.g. if the quotient is torsion free)
  3. There is an isomorphism of sets between G and H (+) G/H

(For the third point, I'm pretty sure I'm using the axiom of choice)


Or, maybe you're partially remembering a general theorem:
If there is a group homomorphism f : G/H --> G with the property that f(x) = x (mod H), then G is isomorphic to H (+) G/H​
 
  • #6
markiv
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Well this is what I was thinking. If G is a FG Abelian group, then it can be decomposed as [tex]G \cong \mathbb{Z}^r \oplus \mathbb{Z}_{p_1^{\alpha_1}} \oplus \cdots \oplus \mathbb{Z}_{p_n^{\alpha_n}}[/tex] where [tex]n = p_1^{\alpha_1}\cdots p_n^{\alpha_n}[/tex], and where [tex]p_i[/tex] are (not necessarily distinct) primes. So if [tex]H[/tex] is q-Sylow, then [tex]H[/tex] has to be of the form [tex]H \cong \mathbb{Z}_{q^\beta_1} \oplus \cdots \oplus \mathbb{Z}_{q^\beta_m}[/tex]. All summands in the decomposition of [tex]G[/tex] that are q-groups must be included in the decomposition of [tex]H[/tex], otherwise [tex]H[/tex] would not be q-Sylow. But that means, if I mod out by [tex]H[/tex], then I am killing entire summands in [tex]G[/tex]. I think that means then that [tex]G \cong G/H \oplus H[/tex]. On the other hand, if I didn't factor out a Sylow subgroup, then I'd be leaving parts here and there in some of the summands of [tex]G[/tex]. Like in the cyclic group of order 4, the whole group itself is the only summand. And so [tex]\mathbb{Z}_2[/tex] is not Sylow in [tex]\mathbb{Z}_4[/tex], so I wouldn't be killing off a whole summand if I mod out by that. [tex]\mathbb{Z}_2[/tex] would still remain. At least I think this makes sense :tongue:
 
  • #7
markiv
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Should be [tex]H \cong \mathbb{Z}_{q^\beta_1} \oplus \cdots \oplus \mathbb{Z}_{q^\beta_m}[/tex]

Wait actually... FG Abelian groups are nilpotent, and a group is nilpotent if and only if it is a direct sums of its Sylow subgroups. So [tex]G \cong G/H \oplus H[/tex] if [tex]H[/tex] is a Sylow subgroup of nilpotent group [tex]G[/tex] ?
 

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