A conceptual question in ring theory?

Question

What is the main(conceptual) difference between an ideal of a ring and a submodule over a ring?

Answer 1

An ideal is a submodule of its ring.

The difference is that the idea of module conveys that the coefficients and the elements of the module can be different kinds of objects.

Answer 2

The difference is the ideal has to be a subset of the ring, inheriting the ring multiplication; while a module over the ring can be any set with any map $R × M → M$ satisfying the module conditions. Think of an R-module as a generalization of the concept of a vector space over a given field: in fact, the latter is a special case of the former.

Answer 3

FROM WIKIPEDIA

For an arbitrary ring $(R,+,\cdot)$, let $(R,+)$ be its additive group. A subset $I$ is called a '''two-sided ideal''' (or simply an '''ideal''') of $R$ if it is an additive subgroup of ''R'' that "absorbs multiplication by elements of ''R''". Formally we mean that $I$ is an ideal if it satisfies the following conditions:

$(I,+)$ is a subgroup of $(R,+)$

$\forall x \in I, \forall r \in R :\quad x \cdot r, r \cdot x \in I $

Suppose that ''R'' is a ring and 1_R is its multiplicative identity. A '''left ''R''-module''' ''M'' consists of an abelian group (''M'', +) and an operation : ''R'' × ''M'' → ''M'' such that for all ''r'', ''s'' in ''R'' and ''x'', ''y'' in ''M'', we have: $ r \cdot ( x + y ) = r \cdot x + r \cdot y $ $ ( r + s ) \cdot x = r \cdot x + s \cdot x $ $ ( r s ) \cdot x = r \cdot ( s \cdot x ) $ $ 1_R \cdot x = x .$

Suppose $M$ is a left $R$-module and $N$ is a subgroup of $M$. Then $N$ is a submodule (or $R$-submodule, to be more explicit) if, for any $n$ in $N$ and any $r$ in $R$, the product $r ⋅ n$ is in $N$ (or $n ⋅ r$ for a right module).

So what we understand from here is that an ideal is just a subset of $R$ but a submodule consists of the ring $R$ and an abelian group $M$ and an operation defined between $R$ and $M$ whereas $I$ is just a subset of $R$ that satisfies some criterion.