Our course notes define that the ideal generated by a set $A$ is the intersection of all the supersets of $A$ that are themselves ideals. The notes go on to prove the following theorem.
Theorem. Let $R$ denote a (commutative, unital) ring, and suppose $A \subseteq R$. Then the ideal generated by $A$ equals $\{a_1 r_1 + \cdots + a_n r_n \,|\, n \in \mathbb{N}, a_i \in A, r_i \in R\}.$
The meaning of this theorem isn't clear to me, especially given that $A$ is not assumed finite in the premises, but the set seems to be defined in a way that assumes the existence of a bijection $a : \{1,\cdots,n\} \rightarrow A$.
What is the precise statement of the theorem that the above theorem is trying to get at?
Best Answer
It says that the ideal generated by $A$ is the collection of all (finite( "linear combinations" of elements of $A$.
There is no built in assumption that $A$ has a finite number of elements, since $n$ is variable.
Let us think about what a proof would look like. Certainly every element $a$ of $A$ can be expressed in this form, since it is equal to $a\cdot 1$. So our set of "linear combinations" contains $a$.
And certainly any ideal that contains all the elements of $A$ must contain all linear combinations.
So it remains to verify that the collection of linear combinations is an ideal. The verification is just a matter of checking off properties. Closure under multiplication by elements of $r$ is obvious. We also need to check that the linear combinations are a subring of $R$.