# Ideals – Abstract Integers

When I first talked about rings, I said that a ring is an algebraic
abstraction that, in a very loose way, describes the basic nature of integers. A ring is a full abelian group with respect to addition – because the integers
are an abelian group with respect to addition. Rings add multiplication with an
identity – because integers have multiplication with identity. Ring multiplication doesn’t include an inverse – because there is no multiplicative inverse in
the integers.

But a ring isn’t just the set of integers with addition and multiplication. It’s an abstraction, and there are lots of thing that fit that abstraction beyond the basic realization of the ring of integers. So what are the elements of those
things? They can be pretty much anything – there are rings of topological spaces,
rings of letters, rings of polynomials. But can we use the abstraction of
the ring to create an abstraction of an object that resembles an integer, rather than an abstraction that resembles the set of all integers?

Obviously, the answer is yes, or I wouldn’t be asking it, right?

The answer is something called an ideal. Ideals capture some of the essence of an integer within the set of integers; there are prime ideals that
capture the essence of prime numbers within a ring.

Suppose we have a ring, (A,+,×). We can define a special subset,
R, such that (R,+) is a subgroup of (A,+), and ∀r∈R, ∀a∈A:
r×a∈R – in other words, R is a subgroup of A, and R is closed
over A with respect to multiplication when a member of R is the right operand. If that is true, then R is a right ideal of A.

We can do the same thing again, only require the subset to be closed with respect to multiplication when any member of A is the left operand; that’s called a left ideal.

An two-sided ideal I is a subset of a ring which is both a left ideal and a
right ideal of the ring. A proper ideal is an ideal that is a proper
subset of its ring. In general, when we just say ideal, we mean “two-sided proper ideal”.

The idea of an ideal of a ring is easiest to grasp by looking at
a couple of examples using the integers. This is a lot easier to grasp given an example. We know that the set Z of all integers is a ring using addition and multiplication. The set of all even integers is an ideal, usually written 2Z. Given any member of
2Z, you can multiply it by any integer, and you’ll get a result that’s a member of 2Z. We can say that 2Z is, in some sense, a representation of the number 2 within the ring of integers – it’s the set of values that can be generated from 2 using multiplication. Similarly, 3Z is the set of all integer multiples of 3, and we can say that it’s a representation of the number 3.

The easiest way to see how an ideas works as a sort of prototypical integer is by looking at prime ideals. An ideal I of a commutative ring R is prime if and only if for every a,b∈R, if a×b∈I, then either a∈I or b∈I. That’s an abstract way of saying something that works out to the definition of prime numbers in integers. A number is prime if and only if every multiple of it must be the product of two numbers, at least one of which is a multiple of the prime. So, for example, 7 is prime: you can’t get a multiple of seven except by multiplying something by seven. But 6 isn’t prime: you can multiply 4 by 9 and get 36 – 36 is a multiple of 6, but neither 4 nor 9 are multiples of 6.

That leads us to an equivalent of prime factors of the integers. We know
that in the integers, every integer can be uniquely defined as the product of a collection of prime numbers. Similarly, if you take a ring, R, and the set of prime ideals of that ring, then every ideal of R can be uniquely defined as a product of prime ideals.

## 0 thoughts on “Ideals – Abstract Integers”

1. Mark Dow

Thanks, this brings me closer to an understanding of the abstract properties of rings.
Is there a reasonably clear example of a ring that is not composed of integers (knots?), perhaps one that has prime ideals?

2. Mark C. Chu-Carroll

Mark Dow:
Sure. Polynomials. Polynomials form a ring with respect to
addition and multiplication. Prime ideals are based on non-factorable polynomials. Every polynomial is either a multiple of primes, or it’s a prime itself.

3. Torbjörn Larsson, OM

In many cases polynomials (or other power series) works as bases – and what do you know, I googled and found polynomial rings, polynomials with coefficients from a ring.
Dunno what they are good for though. The wikipage speaks in mathish. Any translators here? Or better examples?

4. Torbjörn Larsson, OM

Ah, crossposting.
And, um, of course power series can work as bases. What I was wondering about was the goodies the wikipedia page threw in as extra spice. Never mind.

5. Anonymous

R is a subgroup of A, and R is closed over A with respect to multiplication when a member of R is the right operand.

A right ideal R is closed under right multiplication by elements from A.
From some reason your algebra mistakes stick out in my mind, like misstating the group axioms or saying a dense ordering is the defining property of the continuum. They were pointed out, but not fixed.

6. Davis

Dunno what they are good for though. The wikipage speaks in mathish. Any translators here? Or better examples?

Polynomial rings are my favorite example — they’re fundamental to algebraic geometry. Ideals have a very geometric meaning in that realm; let’s look at the polynomial ring in 2 variables with coefficients in the complex numbers, C[x,y].
Suppose you have an ideal generated by one polynomial, f(x,y). Geometrically, we can think of this as defining a subset of C2, call it V, given by the equation f(x,y)=0 (we work in the complex realm to guarantee solutions). Now the ideal generated by f(x,y) is the set of all polynomials P(x,y) that are multiples of f: P(x,y)=p(x,y)f(x,y). But that means P=0 whenever f=0, so the ideal is simply the set of polynomials which vanish on V (I’m assuming f is not a power of another polynomial here).
In this scenario, prime ideals have a special meaning — prime ideals correspond to irreducible subsets V (if f(x,y) factors, each factor defines a component of V). And you can extend to ideals generated by more than one polynomial to construct intersections of these geometric objects.
In algebraic geometry we abstract this idea, and end up identifying a geometric object with the ideal of polynomials which vanish on that object, allowing us to treat algebra and geometry as one and the same.

7. Mark C. Chu-Carroll

Davis:
Thanks for that. One of the problems I’m having writing these articles lately is that I’m getting far outside my comfort zone. I know that the good examples of rings and ideals are in algebraic geometry – but I’ve never studied algebraic geometry, and I don’t have the time to learn it. So I’m struggling to find examples.

8. Torbjörn Larsson, OM

In algebraic geometry we abstract this idea, and end up identifying a geometric object with the ideal of polynomials which vanish on that object,

Just to make sure, like the 1-dim sphere as x^2+y^2 -1 = 0? So from bases to basic objects?

9. Austin

My favorite example of ideals involves the ring of continuous functions on ℜ, C(ℜ), with standard addition and multiplication of functions. It is not difficult to verify that Ia={f∈C(ℜ) | f(a)=0 } is an ideal for any a∈ℜ. This ideal is, in fact prime and maximal, which is most easily shown by recognizing that C(ℜ)/Ia is isomorphic to the reals, which is a field.

10. Davis

Just to make sure, like the 1-dim sphere as x^2+y^2 -1 = 0? So from bases to basic objects?

Exactly like that, yes. That equation corresponds to an ideal generated by the polynomial x^2+y^2-1.
Incidentally, you can form the basis for a topology on C2 by declaring these sorts of objects to be the closed sets. The empty set corresponds to the non-proper ideal generated by the constant polynomial f(x,y)=1 (if you examine the definition of an ideal, you’ll note any ideal containing 1 must contain the entire ring); the entire space corresponds to the ideal generated by the constant polynomial g(x,y)=0 (0 is the only thing in this ideal). This is the Zariski topology, which is mighty weird — if you play a little, you’ll discover that the open sets are all huge!

11. Davis

Thanks for that. One of the problems I’m having writing these articles lately is that I’m getting far outside my comfort zone.

No problem. I love to talk algebraic geometry to anyone who’ll listen. 🙂

12. Torbjörn Larsson, OM

Thanks Davis (and Mark)! I think I finally got the idea of ideals abstracted away from the basic case, and why they matter in the general case.

13. Howard

Come back, Mr. Algebraic Geometer!
You said that the factors of f(x,y) correspond to the components of its variety. I’ve been wondering for awhile if one cannot use the standard topology (on R^n or C^n for instance) instead of the Zariski topology for some of the classification concerning irreducible components.

14. Davis

I’ve been wondering for awhile if one cannot use the standard topology (on R^n or C^n for instance) instead of the Zariski topology for some of the classification concerning irreducible components.

What exactly did you have in mind? The Zariski topology is strictly coarser than the standard topology (Zariski-open implies standard-open), so you can probably do most things with the standard topology as well. The idea is for it to encode only the “algebraic” open and closed sets, though, so moving to the standard topology means drifting away from algebra.
If you want something finer than Zariski but still algebraic, it’s worth looking into the étale topology (which is much more difficult to grasp).

15. revere

Interesting. Thanks. If I’m not mistaken, however, Rings don’t have to have multiplicative identities. That would be a Ring with unity. But if you are far from your comfort zone, I can’t even see my comfort zone from here. So you might want to check if what I said is right!

16. JuanCarlos

Long time since the last time I read this blog, and when I read this post I just remembered a question I made myself when I studied something on these abstract objects for the first time… where are the names of algebraic structures come from?, I mean, why something like (A,+,×) is called “a ring”?
I hope I’m not being disappointing 😉

17. Jonathan Vos Post

“… The word ring is short for the German word ‘Zahlring’ (number ring). The French word for a ring is anneau, and the modern German word is Ring, both meaning (not so surprisingly) ‘ring.’ Fraenkel (1914) gave the first abstract definition of the ring, although this work did not have much impact. The term was introduced by Hilbert… ”
Weisstein, Eric W. “Ring.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/Ring.html

18. Jonathan Vos Post

How many different rings are there with n elements? [extra credit if you can say exactly what I mean by “different”]. Here’s a table if we restrict ourselves to rings with 1.
http://www.research.att.com/~njas/sequences/A037291
A037291 Number of rings with 1 containing n elements.
n a(n)
1 1
2 1
3 1
4 4
5 1
6 1
7 1
8 11
9 4
10 1
11 1
12 4
13 1
14 1
15 1
16 50
17 1
18 4
19 1
20 4
21 1
22 1
23 1
24 11
25 4
26 1
27 12
28 4
29 1
30 1
31 1
32 208
33 1
34 1
35 1
36 16
37 1
38 1
39 1
40 11
41 1
42 1
43 1
44 4
45 4
46 1
47 1
48 50
49 4
50 4
51 1
52 4
53 1
54 11
55 1
56 11
57 1
58 1
59 1
60 4
61 1
62 1
63 4
CROSSREFS
Cf. A027623 [number of rings with n elements], A037221 [Number of near-rings (or nearrings) definable on cyclic group of order n].
KEYWORD
nonn,nice,hard
AUTHOR
Christian G. Bower (bowerc(AT)usa.net), Jun 15 1998.
EXTENSIONS
a(16) and a(32)-A(63) from Christof Noebauer (christof.noebauer(AT)algebra.uni-linz.ac.at), Sep 29, 2000
COMMENT
Here a ring means (R,+,*):
(R,+) is abelian group,
* is associative, a*(b+c) = a*b+a*c, (a+b)*c = a*c+b*c.
Need not contain “1”,
* need not be commutative.

19. scooter

Is infinity an ideal of Z, the integers? I ask because I am wondering if Z/infinity is a ring. If not, why not please.

20. Simon

> Is infinity an ideal of Z, the integers? I ask because I am wondering
> if Z/infinity is a ring
No. Infinity isn’t an integer (or even a real number), so there’s no such thing as ‘the ideal generated by infinity’.
So there’s no such object as Z/infinity. But even if there was, it’s probably just be the same as Z — after all, Z/(n) is isomorphic to Z modulo n; and “Z modulo infinity” doesn’t look that different from Z to me.
(BTW: For anyone wondering, no, that last sentence wasn’t actually *expressely* written to send pure mathematicians (aka ‘pedants’) into paroxysms of anger — that’s just a nice side-effect 😉 )