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I'm struggling with the idea of ideals (both the definitions and the common notation). I'm in a basic collegiate algebra course, just looking for a bit of help. As simply defined as possible, if you could. thanks!

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Did you do any group theory? Ideals are the analogue of normal subgroups. They're the kind of subring you need to have to be able to mod out. –  Mike Miller Feb 25 at 18:52
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Note: if rings are defined to be unital, then ideals are usually not subrings. –  Dustan Levenstein Feb 25 at 18:54
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@eMJayeS We want to be able to make a sort of quotient object by setting a sub-something "equal to zero" (think the case of the integers mod $n$ - it's equal to $\Bbb Z/n\Bbb Z$, and you can then think of it as sending that whole subgroup to zero - normality just means you can do this in a consistent manner). This is a decent way to think about it, but it's far from rigorous. –  Mike Miller Feb 25 at 18:57
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What is the source you are learning from? –  Bill Dubuque Feb 25 at 19:04
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@eMJayeS Knowing which book you are using will help us to give you more focused answer. Also it would help to know more specifically which things you are struggling with, e.g. provide some specific examples. The more details you can provide the greater possibility that you will receive pertinent replies. –  Bill Dubuque Feb 25 at 21:09

7 Answers 7

You can think of ideals as subsets that behave similarly to zero. For example, if you will add $0$ to itself, it is still $0$, or if you multiply $0$ with any other element, you still get $0$. So ideals are like "zeros with several elements". It is a subset $I$ of a ring $R$ that is closed under addition, and also if $a\in I, b\in R$, then $ab\in I$. In particular, one single element $0$ forms an ideal in any ring $R$.

I hope this helps a bit.

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I really like that conceptualization, especially when thinking that this train of thought might even help when trying to understand quotient rings later on. (The ideal really becomes the zero in that case) –  HalloDu Feb 25 at 20:38

An ideal is a generalization of a number that makes factoring work better if you allow square roots of some numbers.

What numbers does $n$ divide?

When studying how numbers factor, we can take a specific number, $n$, and look at all the numbers it divides, $(n) = \{ nk : k \in R \}$ where $R$ is the universe of numbers we are examining.

If you've worked with factoring problems much, you probably know these two basic rules: (1) if $n$ divides $a$ then $n$ divides $ab$ too, and (2) if $n$ divides $a$ and $b$, then $n$ divides $a+b$. The first is true, since if $a=nk$ then $ab= n(kb)$ is still a multiple of $n$. The second is true, since if $a=ni$ and $b=nj$ then $a+b=n(i+j)$, so a sum of multiples of $n$ is still a multiple of $n$.

Ideal numbers

An ideal number is a (nonempty) set of numbers that satisfies those two properties, even if we don't know what $n$ is. We can pretend there is an $n$ if we want, but later we begin to think of the ideal numbers like $(n)$ as being even better than just plain old numbers like $n$.

It turns out we can multiply ideal numbers pretty easily, we just multiply their members: if $mn$ divides $a$, then $a=mnk$ and we can write $a=m(nk)$ as a product, one from $(m)$ and one from $(n)$. Conversely, if $a=mi$ and $b=nj$ then $ab=(mi)(nj) = (mn)(ij)$ is in $(ij)$.

So we define $IJ$ to be the smallest ideal that contains all the products $ij$ for $i \in I$ and $j \in J$.

We can also find a nice analogue GCDs of numbers using ideals. If $d$ divides both $m$ and $n$, then it also divides $m+n$ and $m-n$ and $m+3n$ and $7m-4n$ and all sorts of other multiples. In other words, $d$ divides every single element of $(m) + (n) = \{ mi + nj : i,j \in R\}$. In the integers we get that $(m)+(n) = (\gcd(m,n))$, but for general rings, we just think of $I+J$ as the analogue of the GCD of $I$ and $J$.

Factoring with numbers is weird

Now it turns out there are nice rings like $R=\mathbb{Z}[\sqrt{-5}]$ where factoring numbers is very weird, $6=2\cdot 3$ cannot be factored further, but it can be factored differently, $6=(1+\sqrt{-5})\cdot(1-\sqrt{-5})$. One use of ideals is to clearly state that these are different: $(2) \neq (1\pm\sqrt{-5})$ so the way in which things are multiples of 6 is not very well described by only one of $6=2\cdot 3$ or $6=(1+\sqrt{-5})\cdot(1-\sqrt{-5})$.

Factor by any means necessary

A second use is simpler and more satisfying to me: why CAN'T we keep factoring those two until we get a common factorization? In early grade school we couldn't subtract 3 from 2, but then we invented -1 (which is like saying... but you still need to subtract 1). In later grade school, we couldn't divide 7 into 2, but then we invented $7/2$ (which is totally cheating, what is 7 divided by 2? It is 7/2 of course!). So, why not invent a new way of factoring? We just need to find the common divisor of $2$ and $1+\sqrt{-5}$. Can't do it for numbers (try it! You get $1$ as the answer, which is just useless.) But for ideal numbers it is TOOOOO easy! The GCD of the ideal numbers $(2)$ and $(1+\sqrt{-5})$ is just $$(2) + (1+\sqrt{-5}) = \{ 2i + (1+\sqrt{-5})j : i,j \in R = \mathbb{Z}[\sqrt{-5}] \}$$

We usually abbreviate that as $(2,1+\sqrt{-5})$, just leaving off the $\gcd$. Now we can refine the factorization: $$(6) = (2,1+\sqrt{-5})(2,1-\sqrt{-5})(3,1+\sqrt{-5})(3,1-\sqrt{-5})$$

It turns out this is the only unrefinable way to factor the ideal number $6$ into ideal numbers, none of which is equal to the silly ideal number $(1)=R$ (which is like saying $2=1\cdot 2$ is factoring the prime number $2$).

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Another place where you find ideals quite naturally is in polynomial rings of various kinds. Working with ideals helps to generalise ideas about dimension. And it is easy to give simple examples of ideals which are generated by more than one element.

So for example, if there are three variables and integer coefficients we have $\mathbb Z[x,y,z]$ with elements like $5x^2y^4z-yz^3+xy+29$.

The ideal generated by $x$ and $y$ consists of all the polynomials which become zero on setting $x=0$ and $y=0$. This is an ideal, because multiplying anything by zero gives zero, and adding zero to zero gives zero. So $5x^2y^4z-yz^3+xy+29$ wouldn't be in the ideal, but $5x^2y^4z-yz^3+xy$ would.

Intuitively, factoring out by this ideal leaves the constant terms and polynomials in $z$.

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One characterisation that is implicit in some of the other answers, especially Sasha's, is the following:

Ideals are exactly those subsets which can arise as the kernel of a ring homomorphism.

That kernels are ideals is immediate (and can hence motivate the definition of an ideal), and the quotient construction (modding out) yields an epimorphism whose kernel is a given ideal.

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Why add the word "proper"? –  Tobias Kildetoft Feb 26 at 13:47
    
@TobiasKildetoft, because I am not sure if the zero map counts as a ring homomorphism. If it does, then proper must certainly be omitted. I do not know what it means that I have to admit this, but I am never sure how $1$s are handled in rings. –  Carsten Schultz Feb 26 at 14:43
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The $0$-map is a ring homomorphism precisely when the codomain is the $0$-ring (when we are working with unital rings). –  Tobias Kildetoft Feb 26 at 18:23
    
Tak, @TobiasKildetoft. –  Carsten Schultz Feb 26 at 18:29

For most algebraic structures, we are interested in transformations that preserve the structure and, if the algebraic structure contains a neutral element (usually zero), we are interested in the preimage of zero under such a transformation because this is a measure of how much of the structure is "forgotten" by the transformation. Usually, this permits to form quotient structures, too.

So, for rings, the transformation should preserve addition, multiplication and neutral elements. If we now check what restriction this gives for the preimage of zero, it turns out that we find exactly the definition of an ideal: It is stable under addition, and it is stable under multiplication by any ring element.

For vector spaces, the corresponding objects turn out to be simply vector spaces as well (the kernel of the linear homomorphism). For groups, the corresponding objects are normal subgroups which is more restrictive than just taking subgroups.

So, you can see that ring ideals take something of a middle ground: They do not have all the laws of a ring (no multiplicative identity required), but they do have some additional laws (stable under multiplication by external elements).

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The simplest case to start with is ideals in a commutative ring, and the most familiar commutative ring is $\mathbb{Z}$, the ring of integers. Let $I$ be the set of all multiples of $2$. We know from basic arithmetic that the sum of two multiples of $2$ is again a multiple of $2$. So, the set $I$ is closed under the addition of $\mathbb{Z}$. Now, if we take a multiple of $2$ and multiply it by any integer $n$, then we get another multiple of $2$. This is a consequence of the associativity of multiplication in $\mathbb{Z}$: $n(m2)=(mn)2$. So, the set $I$ has the property that anything in $I$, multiplied by any element of $\mathbb{Z}$, is again in $I$. Some people refer to this property as absorption of products.

The definition of ideal just generalizes these two properties to a general ring $R$. An ideal $I$ in a ring $R$ is a nonempty subset of $R$ that is closed under the addition of $R$ (i.e., $a+b \in I$ whenever $a,b \in I$) and satifies $ra, ar \in I$ whenever $r \in R$, and $a \in I$. The requirement that both $ar$ and $ra$ belong to $I$ is a concession to the fact that the general ring may not be commutative. What we have defined is actually referred to as a two-sided ideal, because we require "closure" under multiplication (that is, absorption of products) regardless of the side on which the general ring element $r$ appears. If we require only $ra$ to be in $I$, we have what is called a left ideal (a right ideal if we only require $ar$ in $I$).

I hope this helps some.

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Say we have a ring, $R = \{r_1, r_2, r_3, ...\}$

An ideal of $R$ is the subset generated by $r_n x$ where $x \in R$. The notation we use to denote this ideal is $(r_n)$. That is: In order to form the ideal generated by an element of $R$, we take that element and multiply it by every other element in $R$—the result of each multiplication forms the set that we call an ideal.

So $(r_n) = \{r_nr_1,~ r_nr_2,~ r_nr_3,~ ... \}$ is the ideal generated by $r_n$ in $R$.

Since elements of $(r_n)$ just consist of multiples of $r_n$, then $x \equiv r_n \equiv 0$ mod($r_n$) for all $x \in (r_n)$. In other words, elements in $(r_n)$ are multiples of $r_n$ and are thus divisible by $r_n$ (so they are equal to zero modular $r_n$).

Here may be an intuitive example: Consider the numbers on the face of a clock. Let's take any number and multiply it by the other numbers:

We'll consider $2$.

The ideal generated by $2$ is $(2) = \{2, 4, 6, 8, 10, 12\}$. That is, $~2*1$, $~2*2$, $~2*3$, etc.

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