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- Thread starter jimmycricket
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lavinia

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Direct sum is an algebraic structure. The Cartesian product is a set.

The underlying set of a direct sum is the Cartesian product of the underlying sets. For instance the Cartesian place is the product of the real line with itself as a set, But if you view the real line a Z-module, the the direct product is also a Z-module.

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pwsnafu

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Are you asking about the direct product ##\otimes## or the Cartesian product ##\times##?

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Fredrik

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mathwonk

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Another example, which I thought was meant, is the direct sum and product of two modules over a (say commutative) ring. In this case, one may distinguish between the defining properties and the actual set theoretic construction. It turns out the properties are opposite, or dual, to each other, but the constructions are the same. I.e. the same physical object has both properties.

defn: a (categorical) direct sum of modules A and B, is a module C plus a pair of morphisms A-->C and B-->C, that sets up a 1-1 correspondence between morphisms out of C and pairs of morphisms out of A and B. I.e. if A-->X and B-->X are any pair of morphisms out of A and B, there is a unique morphism C-->X out of C such that the two morphisms out of A and B are obtained by composition A-->C-->X, B-->C-->X.

defn: a categorical direct product is defined the same way but for morphisms going into A,B and C. I.e. one is given a pair of morphisms C-->A, C-->B, such that every pair of morphisms X-->A, X-->B, arises from a unique morphism X-->C, by composition X-->C-->A, X-->C-->B.

It turns out then that the cartesian product set AxB, together with the standard morphisms in and out, gives on the one hand the direct sum, and on the other also the direct product. So strictly speaking the two are not the same, since the sum is given by both the object AxB and the standard injective morphisms A-->AxB, B-->AxB, while the product is given by the same object AxB but the (different) standard projections AxB-->A, AxB-->B.

In the category of not necessarily commutative groups, the cartesian product construction no longer has both properties. The cartesian product again gives the direct product, but the direct sum object must be constructed in a much more involved way called a "free product". The direct product of Z and Z for instance is the cartesian product ZxZ with the usual definition of coordinatewise operations, but the (categorical) direct sum is the free group on 2 generators.

When one generalizes to the sum and product of not just two, but an infinite number of modules, the constructions are also different, i.e. this time the cartesian product set represents only the direct product, and the direct sum is obtained by restricting all but a finite number of coordinate entries in a "tuple" to be zero. Notice then the direct sum of an infinite number of copies of the integers Z has the obvious basis as a free abelian group, but it is not at all obvious (to me) how to specify a free basis of the product. My free graduate algebra notes on my website at UGA explain some of this.

http://www.math.uga.edu/%7Eroy/843-1.pdf [Broken]

http://www.math.uga.edu/%7Eroy/845-3.pdf [Broken]

defn: a (categorical) direct sum of modules A and B, is a module C plus a pair of morphisms A-->C and B-->C, that sets up a 1-1 correspondence between morphisms out of C and pairs of morphisms out of A and B. I.e. if A-->X and B-->X are any pair of morphisms out of A and B, there is a unique morphism C-->X out of C such that the two morphisms out of A and B are obtained by composition A-->C-->X, B-->C-->X.

defn: a categorical direct product is defined the same way but for morphisms going into A,B and C. I.e. one is given a pair of morphisms C-->A, C-->B, such that every pair of morphisms X-->A, X-->B, arises from a unique morphism X-->C, by composition X-->C-->A, X-->C-->B.

It turns out then that the cartesian product set AxB, together with the standard morphisms in and out, gives on the one hand the direct sum, and on the other also the direct product. So strictly speaking the two are not the same, since the sum is given by both the object AxB and the standard injective morphisms A-->AxB, B-->AxB, while the product is given by the same object AxB but the (different) standard projections AxB-->A, AxB-->B.

In the category of not necessarily commutative groups, the cartesian product construction no longer has both properties. The cartesian product again gives the direct product, but the direct sum object must be constructed in a much more involved way called a "free product". The direct product of Z and Z for instance is the cartesian product ZxZ with the usual definition of coordinatewise operations, but the (categorical) direct sum is the free group on 2 generators.

When one generalizes to the sum and product of not just two, but an infinite number of modules, the constructions are also different, i.e. this time the cartesian product set represents only the direct product, and the direct sum is obtained by restricting all but a finite number of coordinate entries in a "tuple" to be zero. Notice then the direct sum of an infinite number of copies of the integers Z has the obvious basis as a free abelian group, but it is not at all obvious (to me) how to specify a free basis of the product. My free graduate algebra notes on my website at UGA explain some of this.

http://www.math.uga.edu/%7Eroy/843-1.pdf [Broken]

http://www.math.uga.edu/%7Eroy/845-3.pdf [Broken]

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andis represented by a symbol that looks more or less like a ##+##, e.g. ##\oplus## or ##\boxplus##. The latter term is preferred in all other cases, and especially when the binary operation isnotcommutative.

I do not believe this is true. The direct product of groups is the Cartesian product where the operation is taken coordinate-wise. The direct sum is the subset of the direct product consisting of those elements having only finitely many non-identity coordinates.

In any event, the ##\otimes## symbol is typically reserved for the tensor product of modules and never for direct products.

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