Internal Direct Sum of Commutative Rings: Is I + R/I = R?

[markiv]

For a commutative ring $$R$$ and an ideal $$I$$, is it true that $$I \oplus R/I \cong R$$ ? 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 $$R/I$$ is isomorphic to some ideal in $$R$$, call this $$J$$. It's clear that $$I \cap J = 0$$, but does $$I + J = R$$ ?

 

[Hurkyl]

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 $G \cong G/H \times H$ for all abelian groups G and subgroups H -- e.g. let G be the cyclic group of 4 elements...

 

[markiv]

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!

 

[markiv]

What is true? It is certainly not true that $G \cong G/H \times H$ 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 $$H$$ is Sylow in $$G$$, then $$G \cong G/H \oplus H$$, right? Still assuming $$G$$ is FG Abelian, of course.

 

[Hurkyl]

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​

 

[markiv]

Well this is what I was thinking. If G is a FG Abelian group, then it can be decomposed as $$G \cong \mathbb{Z}^r \oplus \mathbb{Z}_{p_1^{\alpha_1}} \oplus \cdots \oplus \mathbb{Z}_{p_n^{\alpha_n}}$$ where $$n = p_1^{\alpha_1}\cdots p_n^{\alpha_n}$$, and where $$p_i$$ are (not necessarily distinct) primes. So if $$H$$ is q-Sylow, then $$H$$ has to be of the form $$H \cong \mathbb{Z}_{q^\beta_1} \oplus \cdots \oplus \mathbb{Z}_{q^\beta_m}$$. All summands in the decomposition of $$G$$ that are q-groups must be included in the decomposition of $$H$$, otherwise $$H$$ would not be q-Sylow. But that means, if I mod out by $$H$$, then I am killing entire summands in $$G$$. I think that means then that $$G \cong G/H \oplus H$$. 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 $$G$$. Like in the cyclic group of order 4, the whole group itself is the only summand. And so $$\mathbb{Z}_2$$ is not Sylow in $$\mathbb{Z}_4$$, so I wouldn't be killing off a whole summand if I mod out by that. $$\mathbb{Z}_2$$ would still remain. At least I think this makes sense :tongue:

 

[markiv]

Should be $$H \cong \mathbb{Z}_{q^\beta_1} \oplus \cdots \oplus \mathbb{Z}_{q^\beta_m}$$

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 $$G \cong G/H \oplus H$$ if $$H$$ is a Sylow subgroup of nilpotent group $$G$$ ?