# Internal Direct Sum

## Main Question or Discussion Point

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 $$I + J = R$$ ?

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Hurkyl
Staff Emeritus
Gold Member
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....

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!

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
Staff Emeritus
Gold Member
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​

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:

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 $$G$$ ?