Ideals generated for commutative ring

The following is a problem from the Gallian book. I'm trying to understand what exactly this ideal is and how to verify that it is in fact an ideal.

"Let $R$ be a commutative ring with unity and let $a_1, a_2,..., a_n$ belong to $R$. Then $I = \langle a_1, a_2,..., a_n \rangle = \{r_1 a_1 + r_2 a_2 + ... + r_n a_n | r_i \in R\}$ is an ideal of $R$ called the ideal generated by $a_1, a_2,...,a_n$."

My interpretation of this is that we choose some set of $a_1, a_2,..., a_n$ from R to generate an ideal with. Then we choose every combination of $n$ elements from $R$, and call each set of them $r_1, r_2,...,r_i$ and apply them to our generator set. Is this accurate? If not I could use some educating and some guidance in verifying the ideal.

EDIT:

To clarify, my questions are: a) Is my interpretation of the generator correct? b) How to verify it is an ideal and is my proof sufficient?

Adding my approach to verifying that this is an ideal:

First prove that $a - b \in I$

$(r_1 a_1 + r_2 a_2 + ... + r_n a_n) - (s_1 a_1 + s_2 a_2 + ... + s_n a_n) \in I$

$(r_1 - s_1) a_1 + (r_2 - s_2) a_2 + ... + (r_n - s_n) a_n \in I$

Next prove that $ar$ and $ra$ are in $I$ whenever $a \in I$ and $r \in R$

$r(r_1 a_1 + r_2 a_2 + ... + r_n a_n) \in I$

$r r_1 a_1 + r r_2 a_2 + ... + r r_n a_n \in I$

And thus it's verified. Is this all logical?

• i do not understand your interpretation..
– user87543
May 25, 2014 at 13:51
• It seems like you're right (only you don't need to choose just some of $a_1,\ldots,a_n$, as long as $n$ is a natural number, you can just choose all – if necessary, putting $r_i=0$ – though you certainly can only choose some, so this is not wrong, just superficial). I don't really understand: why do you even doubt? The definition you've quoted seems more or less as straightforward as it can be. To verify that it is an ideal, just check the definition. May 25, 2014 at 13:52
• I think what you posted is not a problem, it is the definition of ideal generated by elements $a_1, \ldots, a_n$ May 25, 2014 at 13:55
• @tomasz I'm in the middle of a class on abstract algebra and I am struggling and pushing to get through. I guess you could say my confidence is shot. May 25, 2014 at 13:56
• @JustinBozonier : can you explain what is your actual question...
– user87543
May 25, 2014 at 14:04

An ideal $I$ generated by a set $S\subset R$ should be the smallest ideal of $R$ containing $S$. It is universal in the sense that all ideals containing $S$ will contain $I$. Our generating set $S$ is finite with $n$ elements and we will call them $\{a_i\}_{i=1}^{n}$.
In order to be a (two-sided) ideal you must have $ra_i\in I$ and $a_ir\in I$ for any $r\in R$ and you must be closed under addition as well as additive inverses. That forces the definition of $I=<S>$ to be $$\{ r_1a_1 + ... r_n a_n +k_1a_1+...+k_na_n | r_i\in R \text{; }k_i\in \mathbb{Z}\}$$ The $k_i\in \mathbb{Z}$ are necessary because it needs to be closed under addition; i.e. the sum of an element with itself should still be in the ideal. This is often dropped from the notation. Also, "choose every" sounds imprecise.
If you choose a set of $n$ elements of $R$ and multiply each by an element of your generating set and sum, then that does indeed give an element of the ideal. However, if you are describing the whole set then you would need a different description. I think the correct way to think about it is using the universal property I described at the beginning.