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Mathematics
Linear and Abstract Algebra
An Embarrassing Question about turning a ring into a module
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[QUOTE="lavinia, post: 6036249, member: 243745"] Formally, a ring can be thought of as a generalization of a number system such as the the integers. It has a commutative law of addition and a multiplication that distributes over addition. A module generalizes the idea of a vector space. It has a commutative law of addition and a multiplication by "scalars" which are elements of a ring (not necessarily a field as in a vector space) and which distributes over addition. However, the elements of the ring may not be elements of the module. For instance any abelian group is a module over the ring of integers. [U]A ring can be thought of as a module over itself by letting the scalars be the elements of the ring.[/U] As [USER=13785]@mathwonk[/USER] points out, modules naturally arise as "ideals" in rings. An ideal is a subgroup of a ring in which the scalars are the ring itself. For example if the ring is the integers, then the even integers are an ideal. Multiplication of an even integer by an arbitrary integer is still even, so the even integers are a module over all of the integers. If the ring is not commutative multiplication by scalars can be different if one multiplies on the right or on the left. One gets the idea of a "left ideal" and a "right ideal" or a "two sided ideal" if multiplication on the left and right both work as scalar multiplication. [/QUOTE]
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An Embarrassing Question about turning a ring into a module
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