# I An Embarrassing Question about turning a ring into a module

#### TMO

Given a ring R, how exactly do I interpret it as a module? A lot of my homework assignments involve treating a ring as "a module over itself" and I don't know precisely what that means.

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#### Infrared

Gold Member
A module $M$ over a ring $R$ is an abelian group together with a map $R\times M\to M$ satisfying certain axioms. Take $M=R$ and let the map be given by left multiplication (so $(r,x)$ is taken to $rx$).

#### mathwonk

Homework Helper
the main thing that happens that is different from what you might expect, is to consider the meaning of a submodule of the ring. This will be a subgroup of the ring additively, but will not need to be a subring, so will not need to contain 1. It will however need to be closed not just under multiplication, but also under multiplication by any element of the ring.

thus it will be exactly what is called an ideal of the ring. in my opinion it is probably better pedadogically, to study ideals first, and then after defining modules to remark that an ideal is just a submodule when the ring is considered a module over itself.

#### lavinia

Gold Member
Given a ring R, how exactly do I interpret it as a module? A lot of my homework assignments involve treating a ring as "a module over itself" and I don't know precisely what that means.
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.

A ring can be thought of as a module over itself by letting the scalars be the elements of the ring.

As @mathwonk 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.

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"An Embarrassing Question about turning a ring into a module"

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