Unique representation in graded modules

[disregardthat]

In atiyah's book on commutative algebra page 106 it says that elements in graded modules can be written uniquely as a sum of homogeneous elements. More precisely:

If A = \oplus^{\infty}_{n=0} A_n is a graded ring, and M = \oplus^{\infty}_{n=0} M_n is a graded A-module, then an element y \in M can be written uniquely as a finite sum \Sigma y_n, where y_n \in M_n.

I can't see how that can be right. Consider any ring A. Then A = \oplus^{\infty}_{n=0} A is a graded ring, since AA \subseteq A. Let M be any non-zero A-module. Then M = \oplus^{\infty}_{n=0} M is a graded A-module, since AM \subseteq M. However, it is clear that if we pick a non-zero x \in M, then x =2x+(-x), where 2x \in M_1 =M, and -x \in M_2 = M. Maybe x should be on some "normal form" in the graded module, but how do we decide what the normal form is?

Have I got it wrong or is this an error in the book? I think the necessary and sufficient condition is that the M_n's are disjoint.

 

[disregardthat]

I also have a question regarding the notation \oplus^{\infty}_{n=0} A_n for graded rings. I thought elements of a direct sum was on component form, but in graded rings I think they are not. Shouldn't the notation really be \Sigma^{\infty}_{n=0} A_n, or at least \oplus^{\infty}_{n=0} A_n \slash \sim, where \sim relates (\delta_{ij}a)^{\infty}_{i=0} with (\delta_{ik}a)^{\infty}_{i=0} for any pair of natural k,j.

Also, just to confirm; multiplication on a graded ring is not defined componentwise, is it? Normally, (I'd think) multiplication in the ring A \oplus B is defined as (a,b)*(c,d)=(ac,bd). But in graded rings (extended to an infinite direct sum) we have (a,b)=a+b, (c,d)=c+d, and (a,b)*(c,d)=(a+b)(c+d)=ac+ad+bc+bd, which does not correspond to (ac,bd).

These are all reasons why I think the \oplus-notation is not appropriate, but even in the \Sigma-notation representations need not be unique.

 

[Landau]

@your last question:

I am not really familiar with commutative algebra, but it is my understanding that a ring A is call graded if there is a family of subgroups {A_n}_n such that
A=\bigoplus_{n\in\mathbb{N}}A_n
as Abelian groups (i.e. Z-modules), not as rings. So no, the multiplication is not 'componentwise', just the original multiplication of the ring. By the way, I haven't even heard of an infinite direct sum of rings...

 

[disregardthat]

@your last question:

I am not really familiar with commutative algebra, but it is my understanding that a ring A is call graded if there is a family of subgroups {A_n}_n such that
A=\bigoplus_{n\in\mathbb{N}}A_n
as Abelian groups (i.e. Z-modules), not as rings. So no, the multiplication is not 'componentwise', just the original multiplication of the ring. By the way, I haven't even heard of an infinite direct sum of rings...

All right I see your point. Of course it is only addition and not multiplication that is defined component-wise. Multiplication is I guess in general defined as you say. But still, in a graded ring (a,0,...) equals (0,a,...) (written a+0+... and 0+a+0+... respectively in the graded ring), but in a direct sum this is not the case (e.g. (1,0) \not = (0,1) ). This is why I think the \Sigma-notation is more proper, or that we divide out with the equivalence relation defined above.

And I still wonder how it can be so that any element can be written uniquely as a sum of homogeneous elements in the case where the modules/groups are not disjoint. If we consider the homogeneous elements of the graded ring k[x_1,x_2,...,x_n] then elements can be written uniquely as a such, but in this case the subgroups (each group consisting of the set of all homogeneous polynomials of some degree) are disjoint.

By "infinite direct sum" I just meant the representation of a graded ring as an infinite direct sum: \oplus^{\infty}_{n=0} A_n.

 

[Landau]

I think you should take a look back at the concept of direct sums (link 2) of abelian groups.

A=\bigoplus_{n\in\mathbb{N}}A_n

as Abelian groups, with the A_n's subgroups of A (here thus only considered as group), means that every ring element r can be uniquely expressed as

r=\sum_{n\in\mathbb{N}}r_n

with r_n\in A_n, and cofinitely many r_i's zero.

It is the same as saying that

A=\sum_{n\in\mathbb{N}}A_n

AND all A_n's intersect trivially. The first means that every ring element can be written as sum of r_n's, the second means that this decomposition is unique.

 

[disregardthat]

Ok, thanks for you help. I think the internal direct sum is what is meant here. I see there are different interpretations of direct sums. Generally there seems to be two different related concepts:

G = G_1 \oplus G_2 where \phi : G_1 \oplus G_2 \to G by \phi(a,b) = a+b is an isomorphism, and G_1 \oplus G_2 := G_1 \times G_2, and similarly for infinite direct sums, only in which case a finite number of components are non-zero. I assume these are the definitions of internal and external direct sums respectively, and that the internal direct sum is the one meant for graded rings and modules.

Reading on wiki:
"A related concept is that of the direct product, which is sometimes the same as the direct sum, but at other times can be entirely different."
http://en.wikipedia.org/wiki/Direct_sum

So according to this I guess it is so that the subgroups of which graded rings and graded modules are composed of intersect trivially, which indeed implies that any element can be written uniquely as a sum of homogeneous elements.

Thus in this case as you say \oplus is equivalent to \Sigma (+trivial intersections) (the latter being defined in my book, the former not).

 

[Landau]

In general, the "direct product" and the "direct sum" are the "product" and "coproduct" from category theory, respectively; they have a different defining universal property.

However, usually in algebra (groups, modules), the two concepts are the same in the finite case (i.e. when the collection you are taking the sum or product of is finite). In the infinite case the concepts differ.

Sometimes a distinction between "internal" and "external" direct sum is made. The internal direct sum is only defined for a collection of subgroups (submodules) of a given group (module), while the external direct sum is defined for an arbitrary collection of groups (modules). However, the two are isomorphic, so no-one really cares about the distinction:

If M is the external direct sum of some collection of modules (M_i), then M is the internal direct sum of some collection of submodules (A_i) with A_i isomorphic to M_i. Conversely, if M is the internal direct sum of some collection of submodules (A_i) , then M is isomorphic (not equal) to the external direct sum of the collection (A_i).

You can read about all this in most decent algebra texts; Abstract Algebra by Grillet is another good choice.

So yes, in this case (A graded ring) the internal direct sum is meant (because the A_n are subgroups, and A is equal to the direct sum). It is, indeed, equivalent to say that A_n is the usual sum of the A_n's, and that they intersect trivially. Which is again equivalent to say that every ring element has a sum decomposition into r_n's, and this is unique.

 

[disregardthat]

Thanks for the explanations, this all makes sense.

Could you explain what the universal properties of the external direct sum and external direct product are? For tensor products I guess the universal property is that M \otimes N is a module T with a bilinear map g from M \times N to T such that any bilinear map from M \times N to P factors uniquely through T and any T' with such a bilinear map with this property is isomorphic to T and the isomorphism composed with g is this bilinear map. Is it something similar for sums and products?

 

[Landau]

Thanks for the explanations, this all makes sense.

You're welcome!

Could you explain what the universal properties of the external direct sum and external direct product are?

Certainly. Direct sum: see coproduct. Direct product: see product. They are dual to each other.

In case of R-modules, you should read 'R-module' for object and 'R-linear homomorphism' for 'morphism' in these definition. In case of 'groups' the objects are 'groups' and the morphism are 'group homomorphisms'.

For tensor products I guess the universal property is that M \otimes N is a module T with a bilinear map g from M \times N to T such that any bilinear map from M \times N to P factors uniquely through T and any T' with such a bilinear map with this property is isomorphic to T and the isomorphism composed with g is this bilinear map. Is it something similar for sums and products?

Yes, if the ring R is commutative this is correct.

 

[disregardthat]

Hi, in the same book page 107 they introduce the graded ring A^* = \oplus_{n \in \mathbb{N} } \mathfrak{a}^{n} where \mathfrak{a} \subseteq A is an ideal. Similarly for modules they set M^* = \oplus_{n \in \mathbb{N} } M_n, where \mathfrak{a}^mM_n \subseteq M_{m+n} As far as I understood it the components would need to have trivial intersections, but in this case they do not. Why are they using internal direct sum notation? Might they mean external direct sum?

 

[Landau]

There is no difference between the notation for internal and external direct sum, as far as I know. (Like I said, no-one really cares about the distinction because it is usually clear from the context, and they are isomorphic anyway.) So I don't really understand your question?

 

[disregardthat]

There is no difference between the notation for internal and external direct sum, as far as I know. (Like I said, no-one really cares about the distinction because it is usually clear from the context, and they are isomorphic anyway.) So I don't really understand your question?

I am confused because they first say that any element may be written as a unique sum of homogeneous elements in a graded ring. We concluded the intersections between the components had to be trivial in order for this to be true. Now the intersections are not trivial, so how can elements be written as a unique sum? I guess it need to be interpreted as an external direct sum? In that case my question becomes whether (b,0,0,...) and (0,b,0,...) are equal in the graded ring defined above? (b is an element of the ideal) It cannot be an internal direct sum since the powers of the ideals does not intersect trivially, and for an external sum the elements above must be distinct, however I am not sure if this is intended.

 

[Landau]

I guess it needs to be interpreted as an external direct sum?

Correct.

 

[disregardthat]

Correct.

I edited my post above with another question. Are (0,b,0,...) and (b,0,...) considered equal in A^*? I guess not though.
.