An abstract introduction to Groups and Fields.

Ricky · 2006-06-22 16:35:20

Hey MathIsFun, like I promised, an introduction to group theor and fields. I'll end this by basically attaching my other post on the number systems, with a bit of information added.

A simple question most people ask when learning any sort math is, "Why do we need to know this?" Or an even more general question, "Why do we need to learn math?" So why is it that we need math? Well, the simple answer is to answer questions. And how does math answer questions? By solving equations.

Now an important thing to know is that there are different types of numbers:

Natural numbers: Integers greater than (not including) 0.
Integers: Any number without a decimal part.
Rationals: Any number that can be represented by a/b where a and b are integers, b not zero.
Reals: A bit difficult to define. One definition used are numbers which can be represented with an infinite (or less) number of decimal digits. But this definition has some problems.

So why these number systems? Well, as you'll see, they all revolve around solving problems.

First, we start off with the natural numbers. As the name implies, they arrose naturally. One of the first uses of numbers was to say how many things you have. 0 isn't included because I can't have 0 of something. If I have 0 ceiling fans, then I don't have any ceiling fans. Since I don't have any, there are no need for numbers.

So natural numbers were pretty good, but we want to be able to solve equations that arrise naturally. For example, if I have 5 apples and MathIsFun has 5, how many do I need to gain to have the same as him? In other words:

5 + x = 5, solve for x.

In a minute, I'll explain why I used the word "gain." With natural numbers, we can't solve this equation. It's impossible. Obviously, we need one number to be able to solve this equation. 0, of course. 0 is a very, very, very, special number in math. That is because:

x + 0 = 0 + x = x, for all integers x.

This is what we call an "identity". When we take the identity and add it to any element, we get that element.

But there are more equations we can't solve with the natural numbers and 0. For example, if I have 15 cups of olive oil and ganesh has 4, how many must I gain to have the same amount as him? There's that word "gain" again. In other words:
15 + x = 4.

As it should be obvious, we need negative numbers. But what really are negative numbers?

1 + -1 = 0
2 + -2 = 0
3 + -3 = 0
...

You should see a pattern forming. Negative numbers are simply numbers, which when you add them to their positive part, equal zero. The mathimatical term for this is, "inverse." More specifically, if:

a + b = 0 and b + a = 0, then a is the inverse of b and b is the inverse of a.

So all the minus sign means is "plus the inverse of..." In fact, there is no such thing as subtraction at all. You've been lied to. It's just adding the additive inverse. So for example, "5 - 3" is really "five plus the inverse of 3", or 5 + (-3).

Now we can solve all equations dealing with adding, which is "subtraction" too. We only needed two properties. We're going to introduce a third that will keep things easy.

a + b + x = c. When solving for x, we don't want the order in which we add to matter. More specifically:

a + (b + c) = (a + b) + c

Without this, things get really messy. Should the order in which I add things change how many I have in the end? In life, that doesn't happen. So in (this sort of) math, it shouldn't either.

These three properties combine form what's called a group. To restate them:

Identity exists: There exists an e in the group such that a + e = e + a = e for all elements a in the group.

Inverses exist: For all elements a in the group, there exists b in the group such that a + b = e.

Associativity: a + (b + c) = (a + b) + c for all elements a, b, c in the group.

Notice the letter e in the above. "e" just stands for the identity. So for example, for integers, e = 0.

So as we had shown above, the integers are a group. We can solve any equation with addition with them. This is no coincidence. Being a group and being able to solve equations are one and the same. So now we're done! We can solve anything! Hah... yea right...

Multiplication. Let's try some problems with integers. I have 3 grand pianos, John has 9. How many more times does he have than me?

3 * x = 9, solve for x.

The answer is of course, 3. Integers work! But lets try some more. I have 5 toe nail clippers and mathyperson has 5. How many more times does he have than me?

5 * x = 5, solve for x.

The answer? 1. Integers still work! Right now, you should be thinking to yourself, "that looks like another identity." But lets try one more. I have 4 piles of mulch and mikau has 6. How many more times does he have than me?

4 * x = 6, solve for x.

The answer? 3/2. But that's not an integer! We can't solve the above equation with integers. Like I said before, being a group means that we can solve equations. So what properties do we have so far?

Associative: a * (b * c) = (b * c) * a. By the definition of multiplication, this has to be true.

Identity: Yep, 1.

Inverses: No! In fact, the only integers that have multiplicative inverses are -1 and 1. So we need to start inventing more numbers.

For example, what's the multiplicative inverse of 2? 2 * x = 1. Well, the answer is of course, 1/2. Now remember what I said about subtraction? Well, the same thing is true about divison! It doesn't exist! All 1/2 isn't "one divide by two." It's just a symbol for the multiplicative inverse of 2. That's all it means.

Now we have satisfied the conditions of a group for multiplication as well as addition. This is known as a field, simply a group with respect to two different operations.

So we can solve any equations in the form of:

a + x = b
and
a * x = b

We're done! Nope. There are more things we can do to numbers.

x² = 2

The answer is the square root of 2. One very important proof in the history of mathimatics is that the square root of 2 is not rational. That is, it can't be represented by a/b where a and b are integers. So we can't solve this equations with rationals. We need to invent more numbers.

In comes the reals. The reals are very complex. You could write an entire book on how to define them. So I will not do that here. But just keep in mind that they include all nth-roots. So for example,

Are all real numbers.

Are we done? Are we ever? We got one more important equation.

x² = -1

We can't solve this with real numbers. We need more complex numbers. We need... complex numbers. That's what they're called, as well as imaginary numbers. Simply put, x = i in this equation.

So what does this all mean? If we take any polynomial:

Where n is less than 5, we can find n solutions. Perhaps even more interesting, if we keep studying Groups and Fields, we get into what's called Galois Theory which can actually show us that some degrees of polynomials can't be solved with any of the above numbers.

MathsIsFun · 2006-06-22 17:49:29

Great! Makes sense and fun to read!

Can I make it into a web page with "by Ricky" at the bottom? And maybe add some graphics (a "set" of clothes or something)?

George,Y · 2006-06-22 20:09:40

Like Euclid reduced geometry theorems into 4 or 5 assumptions as an origin, Hilbert reduced operation rules into several conditions as an origin. Great work, Ricky! Hope you have a lot of time to complete this topic!

Ricky · 2006-06-24 07:06:39

First post updated a bit. The final "third" coming tomorrow.

John E. Franklin · 2006-06-28 14:10:47

I like it!! I wish we had more operators. For example, I'd like three types of subtraction operators.
One would be like it is now so 8 - 6 = 2.
Another would include both ends so 8 -- 6 = 3. because 8 7 6 is 3 #'s.
And finally one would exclude both ends so 8 --- 6 = 1 because 7 is between 8 and 6.

Ricky · 2006-06-28 15:21:21

One thing not mentioned is that to be an operation, it just be well defined. So for example, your last one:

8 --- 6 = 7

Isn't well defined cause:

9 --- 6 = 7 or 9 --- 6 = 8

Ricky · 2006-07-01 15:18:08

Ok, I reworked the entire thing. I came to the realization that the first post was far more complex than it needed to be.

ben · 2006-07-14 06:59:15

After Ricky's nice concrete introduction to groups, let me burden you all with a degree of abstraction. First let's have a definition:

A group G is a set which has

a well-defined, associative and closed operation
an identity e
an inverse.

Ricky covered all this, so let's unpack it a little. In group theory, the group operation is always referred to as "group multiplication". But don't be fooled: group theorist don't always mean multplication in the arithmetic sense; it might mean addition (they mean something like "a multiplicity of group elements") There is a convention which I'll expailn later.

The operation, group multiplication, is said to be "closed" if the result of the operation is still an element of the group (are the primes then a group?).

The identity e is as Ricky described: 0 when the operation is addition, 1 when it is multiplication.

The inverse is again as Ricky said, respectively -x and 1/x.

Now some notational conventions. Some authors use the centre dot to denote the group operation thus a·b, but most use juxtaposition thus ab. Don't assume this means arithmetic product, though, we might be in an additive group.

Similarly most authors use x^-1 for the inverse, to be interpreted according to the operation in question

With this and the previous posts in mind, we see that if a, b are in G, ab = c implies c is in G

We also see that ae = a , and aa^-1 = e. Note that these last equalities refer to what are called right identity and inverse, respectively. The left identity and inverse follow follow from the group axioms. (Any body want to try the simple proof?)

Now, perhaps the most important distinction we can make between different sorts of groups is this: if the operation on the elements h, g of a group G is commutative (i.e. gh = hg) the group is said to be abelian, otherwise non-abelian. I referred earlier to a notational convention: for abelian groups the operation is said to be addtion.

Now non-commutivity need not throw us into a panic - think matrix multiplication for example, but the notation can be misleading at first. ab = c need not imply that ba = c, but it is always the case that ab = c means a = cb^-1 = b^-1c, by my axioms above.

One more word on commutivity; it, or the lack of it, is a property of the group elements, not of the operation (associativity would not follow otherwise).

Now there is another, even more abstract, way to describe a group, which I think is super cool: a group G is a set of sets, these being the underlyng set written |G|, the Cartesian set |G|×|G|×|G|....(this is where our operation lives) and an axiom set (the set of rules that tell us about identity, inverse etc.)

Roughly speaking, the order of a group G is the number of elements in the underlying set |G|. There's an interesting theorem of my buddy Lagrange, but first I need to tell you this.

Consider the groups G and H. If every element of H is also an element of G clearly G = H iff every element of G is also an element of H. But if there are are elements in G not in H we say H is a subgroup of G. As H is a group in its own right it must share the identity e with G, must have an inverse and must be closed under the group operation. Evidently the operation on G and H must be the same.

OK. Lagrange says that the order of any subgroup H of G divides the order of G. This is less easy to prove than it looks, but it's a really cool result.

I'm going to close (for now) with some examples of groups and things which aren't groups:

Z, the integers, is an additive abelian group

The even integers are an additive abelian group; the odds are not a group, neither are the primes

S_n, the set of permutations on n objects is a group, non-abelian for n > 2

The set of rotations in 3-space are a non-abelian group.

Tired of typing, more another time if anyone wants (we've scarecly started, it just keeps getting better!) But I am happy to answer questions, even happier to be corrected!

numen · 2006-07-16 09:44:44

This is great reading, nice work Ricky and ben! :]

ben · 2006-07-17 02:02:23

OK, if anyone is still awake, I have a lot more to say about subgroups, but first I think I need to do this.

When thinking about group theory it is useful to return to our favourite groups, R (the reals) N (the natural numbers)and Z (the integers). But his can be misleading in lots of things I now want say. For example, a lot of stuff which seems self-evident in R, Z and N actually may not apply to other groups. So, to give a taste of a quite different sort of group, I will introduce you to an important and versatile group.

Consider the set {1, 2, 3}. Now consider the set {1, 3, 2}. It's not too hard to see what I've done, and let's write that exchange as (2, 3) i.e. exhange 2 with 3 and 3 with 2 (keep a close eye on the way brackets are used here!).

That operation is referred to as a permutation, so let's try to figure aout how many permutaions there might be on the set {1, 2, 3}.......pause for thought......That's right, for each set with n elements, there are n! permutations.

What we have here, then, is a set with 3 elements whose permutations are themselves the elements of a 6 element group called the symmetric group S_3. In fact, with a bit of practice, it's not too hard to do these sort of permutations in one's head - basically this is matrix algebra. It is easy to see that this is a non-abelian group, as are all S_n for n > 2. The connection with matrix algebra is really nice, which I'd be more than happy to expand upon (if asked).

OK, to return to subgroups. If, for some G with subgroup H we have that ghg^-1 is in H, for all g in G, all h in H, H is said to be a normal (sometimes called invariant) subgroup. Note that, as this must be true for all h in H, this condition is often written gHg^-1 = H.

Also note that every group G has at least one normal subgroup - G itself! Where this is the case G is said to be a simple group. Note also that gHg^-1 = H implies that gH = Hg. So here's another definition: gH and Hg are called the left (resp. right) cosets of H iff g is not in H. This is an incredibly important construction, so let's see a simple example. Consider the group 3Z, those integers exactly divisible by 3. This is an abelian subgroup of Z, so has 0 as the identity e, addition as the operation. The left cosets of 3Z are 1 + 3Z, 2 + 3Z, 4 + 3Z, 5 + 3Z etc. Or are they? Elements of 3Z are ...-3, 0, 3, 6 etc. What's the difference between 1 + 6 and 4 + 3? This brings us to something really, really important.

I invite you to look at these cosets. Notice anything? What's the relation between 1, 4, 7....on the one hand and 2, 5, 8....on the other? Yup, they differ by an element of 3Z. This defines what is called an equivalence relation. To formalize: given a normal subgroup A of a group G, gh^-1 is in A defines and equivalence relation ~ on G iff the relation is

reflexive: g ~ g
symmetric: g ~ h ==> h ~ g
transitive: g ~ h and h ~ k ==> g ~ k

for all g, h, k in G.

This is an example of a universal construction, and is highly important. This post is already over long. More later, if anybody wants.

Ricky · 2006-07-17 03:39:22

I wanted to do the same, ben. That is, introduce permuatations as well as integers mod n, but I thought it would take too long to do so. I think a more thorough introduction is needed. Let me give it a shot:

A permutation is simply a mapping of one set to itself, but in a different order. If you don't know what that is, don't worry. We will use {1, 2, 3} as our set.

Now we create this "map"

1 ⇒ 1
2 ⇒ 2
3 ⇒ 3

Which is the identity map. If I have 1, I say it "goes to" 1. If I have 2, I say it "goes to" 2. If I have 3, I say it "goes to" 3. But the identity map is uninteresting, it does absolutely nothing. So let's try another:

1 ⇒ 2
2 ⇒ 1
3 ⇒ 3

Now 1 goes to 2 and 2 goes to 1. 3 stays the same. And one more:

1 ⇒ 2
2 ⇒ 3
3 ⇒ 1

But when we use this notation, it becomes rather difficult to write out. So we introduce new notation:

(1 2 3)

Start from the left and go to the right, but only do it once. 1 goes to 2. 2 goes to 3. 3 goes to 1, since there is nothing more to the right, we start back all the way on the left.

As you can see, this is the same as the last example I gave. Let's try another one:

1 ⇒ 3
3 ⇒ 2
2 ⇒ 1

First, 1 goes to 3, so we write (1 3
Now 3 goes to 2, so we write (1 3 2
Now 2 goes back to 1, so we're done (1 3 2)

If I put in 1, I get 3. If I put in 3, I get 2. If I put in 2, I get 1. But what about the earlier maps I had?

1 ⇒ 2
2 ⇒ 1
3 ⇒ 3

When number goes to itself, we just leave it out. So for this example, we write:

(1 2)

Because when we put in 3, we get out 3. How about the identity map, where every number goes to itself?

1 ⇒ 1
2 ⇒ 2
3 ⇒ 3

Here, we can just write (1). Or instead, we can just write (2). Or even (3). They all mean the same thing. It is standard to write (1) though.

Let's try a more complex example:

1 ⇒ 4
4 ⇒ 6
6 ⇒ 2
2 ⇒ 3
3 ⇒ 1
5 ⇒ 5

1 goes to 4: (1 4
4 goes to 6: (1 4 6
6 goes to 2: (1 4 6 2
2 goes to 3: (1 4 6 2 3
3 goes to 1: (1 4 6 2 3)
5 goes to 5: add nothing extra

Now it may be the case that we can't write everything with one set of numbers. For example:

1 ⇒ 4
4 ⇒ 6
6 ⇒ 1
2 ⇒ 3
3 ⇒ 2
5 ⇒ 5

1 goes to 4: (1 4
4 goes to 6: (1 4 6
6 goes to 1: (1 4 6)
2 goes to 3: Since we closed off the last parathese, we start a new one: (1 4 6)(2 3
3 goes to 2: (1 4 6)(2 3)
5 goes to 5: Add nothing new.

So we have (1 4 6)(2 3). When we put in numbers, we start with the right most set first, then work our way to the left most set.

(1 4 6)(2 3)

When we put in 1, we put 1 into (2 3) and get 1 back out. Then we take what we got out, 1, and we put this into the next set, (1 4 6). When we put 1 into here, we get out 4. Since there are no sets left, we're done, and 1 goes to 4.

Now when we put 2 into (2 3), we get out 3. Then we take this 3, and put it into (1 4 6), and we get out 3. So 2 goes to 3.

Try the rest for this example yourself.

Now let's try a harder example:

(1 2 3)(1 2)(1 4)

1: Put 1 into (1 4), and we get out 4. Take this 4, and put it into (1 2), and again we get out 4. Then take this 4, and put it into (1 2 3). Once again, we get out 4. We have no more sets, so 1 goes to 4.

2: Put 2 into (1 4), and we get out 2. Take this two, and put it into (1 2), and we get out 1. Take this 1, and put it into (1 2 3), and we get out 2. So 2 goes to 2.

3: Put 3 into (1 4), and then (1 2), and each time we get 3 back out. Take 3 and put it into (1 2 3), and we get out 1. So 3 goes to 1.

4: Put 4 into (1 4) and we get 1. Put 1 into (1 2) and we get out 2. Put 2 into (1 2 3) and we get out 3. So 4 goes to 3.

Now let's take what we just did, and reorder them a bit:

Start with 1: 1 goes to 4
Now start with 4: 4 goes to 3
Now start with 3: 3 goes to 1
Since we're back at 1, and we already used it, let's move on to any numbers that we missed: 2 goes to 2

And that's all four numbers. So we can rewrite this as:

(1 4 3)

Which means that

(1 4 3) = (1 2 3)(1 2)(1 4)

ben · 2006-07-17 03:58:13

Ricky, as you know I'm new here, so I may have mistaken the level of my pitch. Overpitched, you think? Your intro to symmetric groups was a lot more elementary than mine (nonetheless quite correct). I'm sure other members are grateful for that.

Nevertheless, I shall continue on my course. Remember, I am quite happy to expand, explain, illustrate and generally dance the fan-dango.

Ricky · 2006-07-17 04:43:07

Overpitched? Probably. This site is mostly geard towards highschool students and entry level college math courses. I'd be surprised if many have heard of set theory.

But maybe you can help me out on this subject. I have the following problem:

Prove that <x, y> = {xg + yh : g, h are in F[x, y]} contained in F[x, y] is not a principal ideal domain.

And I'm not entirely sure if this is valid:

Assume that (xg + yh) generates <x, y> for some polynomials g, h in F[x, y]. Then there must exist a polynomial f such that (xg + yh)f = x, since x is in <x, y>. So it must be the case that h = 0, otherwise you would have a y term. The same reasoning stands that g = 0 since we need to generate y. But if g and h are 0, then you generate nothing but 0. Contradiction.

What do you think? Valid?

George,Y · 2006-07-17 15:31:39

Ben's introduction is little bit hard to follow. If you can use Latex to prove the proof you've omitted, and give more examples, a beginner will understand your explaination more easily, Ben.

ben · 2006-07-21 06:30:42

Yes,
OK, I see I need to build from the bottom. I am happy to try that.
Ricky, George - waddya think? Where do I start? Sets? Define abstract
algebra? Any way this to Ricky

Ricky wrote:

Prove
that <x, y> = {xg + yh : g, h are in F[x, y]} contained in F[x,
y] is not a principal ideal domain.
And I'm not entirely sure if this is valid:
Assume that (xg + yh) generates <x, y> for some polynomials g, h
in F[x, y]. Then there must exist a polynomial f such that (xg + yh)f =
x, since x is in <x, y>. So it must be the case that h = 0,
otherwise you would have a y term. The same reasoning stands that g = 0
since we need to generate y. But if g and h are 0, then you generate
nothing but 0. Contradiction.
What do you think? Valid?

I hope you have solved this because I'm really rusty on this, so I'm
sort of feeling my way. But I suspect
you need to take advantage of the fact that every PID is also a UFD.
But I'm a bit confused by your notation. The way I learned it, F[x,y]
means a polynomial ring in 2 variables, x and y, with coefficients
taken from the ring F. But you have <x,y> generated by {xg + yh},
g and h in F[x, y]. I don't quite see what's going on here. Please tell.

Anyway, this is as close as I can get. Let F be a field and F[x, y] be the ring of polynomials over. Let <x,y> be generated by (xg + yh).
The ideal I is generated by the polynomials f(xh + yg) = x and
k(xh + yg) = y. Since no element of F[x, y] divides both f and g except for elements of the field F, and elements of the field are not contained in the ideal, the ideal cannot be principal.

And I tell you now, I am not in the slightest convinced by that. What did you you finally get?

Ricky · 2006-07-21 07:09:19

Ben, I just wrote a short introduction to sets:

http://www.mathsisfun.com/sets/sets-introduction.html

And I'm right now going through the editing process for that of groups. But feel free to write your own here on the forum, they are most certainly welcome.

But you have <x,y> generated by {xg + yh},
g and h in F[x, y]. I don't quite see what's going on here. Please tell.

All elements in <x, y> are in the form of {xg + yh} for some two polynomials g and h in F[x, y]

Here is what I got from another forum, The Art of Problem Solving (aops):

The ideal <x,y> consists of the polynomials without constant term. Especially, it's not F[x,y]. So, if <x,y> = <g> for some g in F[x,y], then g is no unit. On the other hand, it's a greatest common divisor of x,y. But then g = ax and g = by for some a,b in F^*, a contradiction.

I understand everything except for why it contradicts that g is not a unit.

ben · 2006-07-21 10:49:49

Ricky wrote:

The ideal <x,y> consists of the polynomials without constant term. Especially, it's not F[x,y]. So, if <x,y> = <g> for some g in F[x,y], then g is no unit. On the other hand, it's a greatest common divisor of x,y. But then g = ax and g = by for some a,b in F^*, a contradiction.

Umm. Maybe we should agree what we mean by an ideal in general terms? Like in any ring?

But you're being really naughty here. What for the love of all that's holy is F*? Where do a and b come from? What do you by a unit? You really must define your terms. I'm not just being nit-picky here, I truly have no idea what you're talking about

All_Is_Number · 2006-07-22 03:59:02

Ricky wrote:

These three properties combine form what's called a group. To restate them:
Identity exists: There exists an e in the group such that a + e = e + a = e for all elements a in the group.
Inverses exist: For all elements a in the group, there exists b in the group such that a + b = e.
Associativity: a + (b + c) = (a + b) + c for all elements a, b, c in the group.
Notice the letter e in the above. "e" just stands for the identity. So for example, for integers, e = 0.

A very enlightening post. However...

Shouldn't the highlighted equation read: "a + e = e + a = a" ? The way you have it written, it would seem that a = 0, and e could equal any integer (and more).

If I'm incorrect, which is always a possibility, could you please elaborate further?

--All Is Number

ben · 2006-07-22 04:58:28

All_Is_Number wrote:

Shouldn't the highlighted equation read: "a + e = e + a = a" ?
--All Is Number

Yes you are right. Why didn't I spot that? However, we may not assume that, if e is a right (resp. left) identity then it is also a left (resp. right) identity.

However, the proof is trivial. Let ab = c and ae = a. Then, because we insist on associativity of the group operation, we have that (ae)b = a(eb) = c, which implies that eb = b, so e is also a left identity.

The proof that inverses are identically left and right is a bit more tedious, but equally trivial. If anyone wants a crack, remember that (a^-1)^-1 = a.

All_Is_Number · 2006-07-22 05:05:14

ben wrote:

All_Is_Number wrote:
Shouldn't the highlighted equation read: "a + e = e + a = a" ?
--All Is Number
Yes you are right. Why didn't I spot that? However, we may not assume that, if e is a right (resp. left) identity then it is also a left (resp. right) identity.
However, the proof is trivial. Let ab = c and ae = a. Then, because we insist on associativity of the group operation, we have that (ae)b = a(eb) = c, which implies that eb = b, so e is also a left identity.
The proof that inverses are identically left and right is a bit more tedious, but equally trivial. If anyone wants a crack, remember that (a^-1)^-1 = a.

You lost me after "Why didn't I spot that."

I need more math classes!

ben · 2006-07-22 05:27:36

All_Is_Number wrote:

I need more math classes!

Well, you're in the right place then! But what I did was even easier than falling off a log. Look.

Group theory is an element of Abtract Algebra (so called), where we are not allowed to take our intuition about real numbers for granted, so everything has to be proved from the ground up, so to speak.

Here's a secret. It ain't that hard.

All_Is_Number · 2006-07-22 05:49:48

ben wrote:

Here's a secret. It ain't that hard.

The difference between a hard question and an easy question is knowing the answer!

It's the left v right that seems confusing, BTW. I've never seen them used in such context.

Last edited by All_Is_Number (2006-07-22 05:52:42)

Ricky · 2006-07-22 09:11:42

Umm. Maybe we should agree what we mean by an ideal in general terms? Like in any ring?

And ideal I in a ring R is a subset of that ring such that for any element i in I, and r in R, ir is in the ideal.

A principle ideal is an ideal that is generated by one element. So there exists an i in I such that {ir : r in R} = I.

But you're being really naughty here. What for the love of all that's holy is F*? Where do a and b come from? What do you by a unit? You really must define your terms. I'm not just being nit-picky here, I truly have no idea what you're talking about

I understand you aren't being nit-picky. But when I write a proof out, I don't define e, G (rarely), or something like the set of integers. These things are implied because they are simply notation and common definitions. So are F* and unit.

F*: F is a field and F* simply means F \ {0}, or rather the entire field F without the additive identity.

unit: An element (normally in a ring) without a multiplicative inverse

Now let me go through the proof slowly, as it took me some time to understand as well.

<x, y> = {xg + yh : g, h in F[x, y]}. So it can be seen upon inspection that <x, y> doesn't contain any constants. This means that <x, y> ≠ F[x, y], because F[x, y] does contain constants.

So it has to be the case that 1 is not in the ideal, because if it were, then we have {1f : f in F[x, y]} and so that would be the same as {f : f in F[x,y]} or rather just F[x,y]. We know that isn't the case, so 1 isn't in the ideal.

Now let's assume that g generates <x, y> for some g in F[x, y]. Then it must be the case that g doesn't have an inverse. If g did have an inverse, then g-¹ would be in F[x, y] and so gg-¹ would be in the ideal since g is in the ideal and g-¹ is in F[x, y]. But gg-¹ = 1, and we already know that 1 is not in the ideal. So it has to be the case that g does not have an inverse, or rather, g is not a unit.

As for the gcd part, I'm still trying to understand. I can see that g | x and g | y because g must generate both x and y. But other than that, I'm a bit lost.

ben · 2006-07-23 08:59:41

All_Is_Number wrote:

It's the left v right that seems confusing, BTW. I've never seen them used in such context.

Yes, it can be confusing. Just remember this. Obviously, 1 + 2 = 2 + 1, it's called commutativity. But there are plenty, plenty situations where xy does not equal yx. So every time you see xy = z you may not simply assume that yx =z. This would imply that xy = yx, and you gotta prove it! That's all I meant by "left and right" identity, if ae = a, I have to prove that ea = a.

Go on! Try it with the inverses, it's easy, just inch forward.

Last edited by ben (2006-07-23 09:12:13)

Ricky · 2006-07-23 12:15:14

Matrix multiplication is an easy example where xy ≠ yx, although sometimes it does.

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