Linear Algebra Review

[matt grime]

The identity map. The map that sends the vector v to 3v, or the zero map. Pick some subspace W = <w> a 1-d subspace. Define a map be v --> (v,w)w where (x,y) is the usual inner product.

 

All three are linear maps.

 

Generally, it is not advisable to start off thinking of R, but instead F where F is any field, so that we can remover any preconceived ideas about R from our mind: R is an analytic construction, vector spaces are algebraic.

 

You haven't seemed to grasp that, in taking the set theoretic approach to functions, it isn't actually that "first" one has to specify a domain etc, since they are *part of the definition of the function*.

 

Not every one requires that a a function is a set, indeed it need not be. That is "setting it in the most basic terminology". All maths must start at some point with a class of things that are not defined in terms of more basic objects. The desire to put functions on such a foundation is only one way of doing it, and is not the most informative way of doing it to most people. The whole "relation" thing is completely unnecessary. We can do all of mathematics without reference to sets if we wish, and there are arguably good reasons to use categories instead.

[Johnny5]

 

The map that sends the vector v to 3v.

 

 

Pick some subspace W = <w> a 1-d subspace. Define a map be v --> (v' date='w)w where (x,y) is the usual inner product.

 

[/quote']

 

Lets do this one.

 

So let me suppose I have a vector in the first quadrant of the XY plane.

 

 

The vector is in a three dimensional frame S.

 

Right now, its z component is zero.

 

 

Let this be the vector:

 

[math] \vec v \equiv 3 \hat i + 4 \hat j + 0 \hat k[/math]

 

 

So how can we begin talking about the function which maps this particular v, to 3v.

 

It simply stretches the vector right?

[matt grime]

Erm, do you understand the notation of functions?

 

What do you mean "begin talking about the function which maps this particular v to 3v.

 

There are an infinite number of linear maps that send (3,4,0) to (9,12,0).

 

The point was the map T(a,b,c) = (3a,3b,3c) for all vectors (a,b,c) in F^3 is a linear map.

 

All linear maps (with reference to this basis) can be realized by a matrix with matrix multpilication. Have you learned about those?

[Johnny5]

Erm' date=' do you understand the notation of functions?

 

What do you mean "begin talking about the function which maps this particular v to 3v.

 

There are an infinite number of linear maps that send (3,4,0) to (9,12,0).

 

The point was the map T(a,b,c) = (3a,3b,3c) for all vectors (a,b,c) in F^3 is a linear map.

 

All linear maps (with reference to this basis) can be realized by a matrix with matrix multpilication. Have you learned about those?[/quote']

 

One thing at a time.

 

Here is the answer to your first question:

 

[math] \text{function of x} \equiv f(x) [/math]

 

that isn't f times x, that's read "f of x". Is that the notation you are referring to, because there are others.

 

Next...

 

In your post 26, you refer to, "the map which sends the vector v, into 3v"

 

This is the map I want to discuss with you.

 

I see now, that you said there are an infinite number of linear maps that send vector v into 3v. So, pick one for starters, then explain why there are an infinite number of them.

 

Next...

 

Ok wait a minute, let me think about your point... hold on...

 

 

Let T denote the map

 

[math] (a,b,c) \rightarrow (3a,3b,3c) [/math]

 

So T is an arbitrarily chosen map? There are others which do the same thing?

 

What physical process could be associated with the stretching of a vector?

 

by the way, what is F^3.

 

I know [math] \mathcal{R}^3 [/math] is Euclidean space.

 

F is for an arbitrary field right?

 

and lastly...

 

Yes I fully understand matrix multiplication.

 

Regards

[matt grime]

The point was you gave a *specific" v=(3,4,0) and talked of "the map" that sent this to 3v=(9,12,0)

 

I gave a specific map T(w)=3w for all w.

 

There are an infinte number of maps T sending *the* specific v you chose to 3v.

 

Any map wrt to the standard basis: (a,b,c) --> (3a,3b,kc) for any k, or the map

 

(a,b,c) --> (3a+c,3b+c,0) also sends that particular v to 3v.

 

If you fully understand matrix multiplication then you fully understand the basic of linear maps. Given some basis of V an n dim vector space the linear maps V to V are exactly the same as nxn matrices.

[Johnny5]

The point was you gave a *specific" v=(3' date='4,0) and talked of "the map" that sent this to 3v=(9,12,0)

 

I gave a specific map T(w)=3w for all w.

 

There are an infinte number of maps T sending *the* specific v you chose to 3v.

 

Any map wrt to the standard basis: (a,b,c) --> (3a,3b,kc) for any k, or the map

 

(a,b,c) --> (3a+c,3b+c,0) also sends that particular v to 3v.

 

If you fully understand matrix multiplication then you fully understand the basic of linear maps. Given some basis of V an n dim vector space the linear maps V to V are exactly the same as nxn matrices.[/quote']

 

Let w denote an arbitrary three dimensional vector.

 

Let T be the following transformation:

 

 

[math] \forall v \in \mathbb{R}^3 [T(v) = 3v] [/math]

 

What this transformation does, is the following...

 

You choose a specific three dimensional vector, call it w, and throw it into transormation. The output is the vector 3w.

 

I randomly chose the following vector:

 

[math] \vec w = 3 \hat e_1 + 4 \hat e_2 + 0 \hat e_3 [/math]

 

Multiplying both sides of the equation above by 3 gives:

 

[math] 3\vec w = 9 \hat e_1 + 12 \hat e_2 + 0 \hat e_3 [/math]

 

Now, using the Pythagorean theorem, we can draw the conclusion that 3w is three times longer than w. So that the transformation stretched the vector somehow. If this happened at the same moment in time, then the transformation is abstract. If this happened over two consecutive moments in time, then the transformation is something physical.

 

 

Now, you say that there are an infinite number of maps that can do this. I don't see it.

 

you also used the phrase "standard basis" do you mean this:

 

[math] B = \mathcal{f} \hat e_1, \hat e_2, \hat e_3 \mathcal{g} [/math]

 

?

[matt grime]

No, you are confusing things here.

 

Given *the* vector w=(3,4,0) there are an infinite number of linear maps satisfying Tw=3w for this vector alone.

 

 

However there is a unique linear map satisfying Tw=3w for all w.

 

 

Plus vector spaces are by no means restricted to things over R, or C, or even Q, so the notion of "length" and something being "three times as long" is very dubious.

 

After all, over a field of char 3, the vector (1,1,1) has "length" zero.

 

 

The crap about " moments in time" is just that: utter crap.

[Johnny5]

The crap about " moments in time" is just that: utter crap.

 

 

As I am learning something old Matt' date=' I am [i']simultaneously [/i] working on something new.

 

You know what that word means dont you Matt

 

Kind regards

 

What is CHAR 3?

 

 

PS: Oh and I don't doubt that I am confusing things, but that's only temporary.

 

How in the world, can vector (1,1,1) have length zero?

 

 

Tw=3w

 

I am slowly following you.

 

 

How do you reach the conclusion that there are an infinite number of maps?

[matt grime]

As I am learning something old Matt' date=' I am [i']simultaneously [/i] working on something new.

 

You know what that word means dont you Matt

 

 

Which word in particular

 

What is CHAR 3?

 

it is to do with fields in general. A field contains 1, so it contains 1+1=2, 1+1+1=3, and so on. In a finite field this must eventually be zero, or there are an infinite number of distinct elements. The smallest p for which adding 1 p times gives 0 is called the characteristic of the field. Example: Z/pZ, the integers modulo p a prime.

 

 

How in the world, can vector (1,1,1) have length zero?

 

because in charactistic 3, or any positive characteristic, there is no notion of length, and that vector dots with itself to give 0 in char 3. this is why it is better not to presume all vector spaces are over R. In fact R is possibly the least useful field to work over since it isnt' algebraically closed.

 

 

Tw=3w

 

I am slowly following you.

 

 

How do you reach the conclusion that there are an infinite number of maps?

 

 

Now, are you asking about the map T satisfying Tw=3w *for all vectors w*, or just about a map that satisifies the relation T(3,4,0)=(9,12,0)? Because I gave an infinite number of examples (over R) that show that just specifiying where 1 vector goes when mapping R^3 to R^3 doesn't determine the map uniquely.

[Johnny5]

In fact R is possibly the least useful field to work over since it isnt' algebraically closed.

 

What does that mean Matt? "Isn't algebraically closed."

 

Real number system is "closed" under addition, and multiplication.

[matt grime]

But not algebraically. the square root of minus one doesn't exist in R.

 

A field is algebraically closed if all polynomials with coefficients in that field possess all roots in that field. So x^2+1 shows that R isn't algebraically closed since it cannot be factored over R.

[Johnny5]

I vaguely remember this now.

 

Ok... Characteristic of a field.

 

Modulo arithmetic.

 

Like a clock.

 

You break fields down into "finite" and "infinite"

 

A field is infinite if there is no limit to the number of distinct elements of it.

 

The reals, the rationals, the integers, are all examples of non-finite fields.

 

In a finite field, there are a finite number of distinct elements.

 

Consider normal arithmetic.

 

One of the field axioms, is that 1 is an element of the field.

 

And another field axiom, is closure of addition.

 

Which means that for any elements x,y in the field, x+y is also an element of the field.

 

Therefore, since 1 is an element of any field, then for any field the following statement is true:

 

1+1 is an element of the field.

 

Now, normally we associate the string "1+1" as being equivalent to 2, and we write:

 

1+1=2

 

and we say the expression above denotes a true statement, where we implicitely assume that not(1=2). This isn't really an assumption, it's a stipulation.

 

But, in higher math, the meaning of the symbol '2' doesn't have to be associated with what we normally associate it with. When using modulo arithmetic, it is quite possible that 2=1.

 

But it is also possible that there are 12 elements in the field, like a clock.

 

So that what the arabic numerals denote, can be varied.

 

Here is one fact that will not vary:

 

[math] \mathcal{A} = \mathcal{f} 0,1,2,3,4,5,6,7,8,9 \mathcal{g} [/math]

 

In the statement above, the symbol [math] \mathcal{A} [/math]denotes "the set of arabic numerals."

 

In the decimal system we have:

 

[math] 2 \equiv 1+1 [/math]

[math] 3 \equiv 2+1 [/math]

[math] 4 \equiv 3+1 [/math]

[math] 5 \equiv 4+1 [/math]

[math] 6 \equiv 5+1 [/math]

[math] 7 \equiv 6+1 [/math]

[math] 8 \equiv 7+1 [/math]

[math] 9 \equiv 8+1 [/math]

[math] 10 \equiv 9+1 [/math]

[math] 11 \equiv 10+1 [/math]

 

and so on.

 

And each successive natural number, is distinct from the elements of the natural number system which come before it.

 

We can say this using first order logic, as follows:

 

[math] \forall x \in \mathbb{N}[ \neg (x+1=x) ] [/math]

 

But, if you are doing modulo arithmetic then the expression above denotes a false statement, instead of a true statement. All that happened, is that someone changed what the arabic numerals are denoting.

[matt grime]

In a field we can prove that 1=/=2=1+1, since otherwise 0=1, which contradicts the axioms. (Ie 1=/=2 is not a "stipulation" by which we read axiom, but a theorem deduced from the axiom that 1=/=0)

[Johnny5]

In a field we can prove that 1=/=2=1+1, since otherwise 0=1, which contradicts the axioms. (Ie 1=/=2 is not a "stipulation" by which we read axiom, but a theorem deduced from the axiom that 1=/=0)

 

I think I get this, but lets just check ok?

 

In trying to answer this question, I have had to address several issues in set theory. Specifically, I am wondering how to handle a case where x=y, in a discussion which involves set theory in some way, either explicitely or implicitely.

 

Generally, when I use the universal quantifier, I leave open the possibility that x=y, because I can close it easily by writing....

 

[math] \forall x \forall y [ \text{if not(X=Y) then such and such} ] [/math]

 

I have no reason to change this practice. But, when using the roster method, i commonly employ the following axioms:

 

Axiom I: For any x an element of A, and any y an element of A: If A={x,y} and x=y then A={x,x} (by substitution).

 

Axiom II: For any x an element of A, If A={x,x} then A={x}.

 

Now, I started this post, trying to prove the following theorem (which I worded)...

 

[math] \text{Theorem: In any infinite field F, with 1,2 } \in \text{F, not (1=2). } [/math]

 

And in the course of trying to prove it, I had to wonder whether or not by writing

 

[math] \text{with 1,2 } \in \text{F} [/math]

 

That contained the information on whether or not 1=2. Because, if the answer is that it contains the information that not (1=2), then there can be constructed a rapid proof. But, it cannot contain the information that 1=2, since one of the axioms of a set, is that the elements of a set are distinct. So at most, it leaves open as possible that 1=2.

 

So that's where I am at now.

 

 

Suppose you see the following expression:

 

[math] 1 \in \mathbf{F} \ \& \ 2 \in \mathbf{F} [/math]

 

The serious question arises as to at what point a reasoning agent is to conclude that not(1=2).

 

What those symbols are to denote, has to be decided by the specific individual who formulated the expression using those symbols. No other conclusion is rational, so... with that in mind...

 

One of the tasks of another individual who reads the expression is, to first decide whether or not the expression denotes a statement. If the answer is yes, then that reasoning agent can go on to try to decide whether or not the statement is true. On the other hand, if the answer is no, then the individual should not bother to try and determine the truth value of a statement, since that individual would have already decided that the expression does not denote a statement.

 

But, a quite common practice by reasoning agents, is to try and look for meaning in someone else's expression. In other words, the first stage of their reasoning process, when analyzing someone else's expression, is to try and find the other individual's meaning. Once the meaning is determined, then it follows that the expression denotes a statement, and then the person can go on to try and determine the truth value of the statement.

 

So this is a complex problem.

 

 

 

 

 

 

 

Proof:

 

Suppose that 1=2, in an infinite field where 1,2 denote elements of the field.

 

Suppose that you are a reasoning agent who understands that the elements of a set must be distinct by a set theoretic axiom. Then you can use set theory to immediately draw the conclusion that since 1,2 were stipulated to be elements (plural not singular) of the field F, they must be distinct, wherefore you can immediately see that the hypothesis (1=2) contradicts a set theoretic axiom.

 

To understand the basic axiom that I am referring to, consider this:

 

In the case where {x,y} denotes a set, and x=y (meaning that whatever symbol x denotes, has also been denoted by symbol y), then {x,y}={x,x} using the substitution principle.

 

But, a set is suppose to be just a collection of distinct objects, where the order of elements is irrelevent. So, for any set {x,x} it must be the case that:

 

{x,x}={x}

 

Which is just a singleton set, a set containing only one element.

 

 

 

If {x,y} is a set then not (x=y).

 

 

On the other hand, suppose you are a reasoning agent who does not know the set theoretic axioms, but you do know the field axioms of algebra. Then you can use the following proof to prove the theorem:

 

Suppose that 1=2, in an infinite field where 1,2 denote elements of the field.

 

Now, let the definition of 2 be given by

 

[math] 2 \equiv 1+1 [/math]

 

Using the axiom of reason, known as conjunction, the following statement "should" be true (where I am using a bit of Deontic logic as well):

 

[math] 1 = 2 \& 2 = 1+1 [/math]

 

Now, using the transitive axiom of equality, the following statement "should" be true:

 

1=1+1

 

 

 

 

Then 1=1+1

 

Then, by the seldom referred to field axiom known as the axiom of addition, which is:

 

[math] \forall x,y,z \in \mathbb{F}[ x=y \Rightarrow x+a=y+a ][/math]

 

We have as a special case:

 

1+(-1)=(1+1)+(-1)

 

And we know that -1 is an element of the field, because

 

 

 

The smallest P, for which

 

[math] 1_1+1_2+1_3+... 1_p = 0 [/math]

 

Is called the "characteristic of the field."

[matt grime]

Are there bits missing from your post? for instance:

 

"And we know that -1 is an element of the field, because"

 

has no conclusion. -1 is defined to be the additive inverse of 1, and as a field is a group under addition, this exists. Is that what you're getting at.

 

I thought you wanted to only do "one bit at a time". For a thread about linear algebra this has wandered far and near from that topic.

[Johnny5]

Are there bits missing from your post? for instance:

 

"And we know that -1 is an element of the field' date=' because"

 

has no conclusion. -1 is defined to be the additive inverse of 1, and as a field is a group under addition, this exists. Is that what you're getting at.

 

I thought you wanted to only do "one bit at a time". For a thread about linear algebra this has wandered far and near from that topic.[/quote']

 

I am re-working everything Matt, hold on until its finished, then correct it if necessary.

[matt grime]

Why on earth does the field have to be infinite? Actually it is true in a finite field that 1=/=2. 2 is always defined to be 1+1. This may also equal zero, which is iff the field has char 2. So? I really struggle to follow what it is you're working towards. You seem to be vastly over complicating a very simple idea, and I say that as professional mathematician.

[Johnny5]

Why on earth does the field have to be infinite? Actually it is true in a finite field that 1=/=2. 2 is always defined to be 1+1. This may also equal zero, which is iff the field has char 2. So? I really struggle to follow what it is you're working towards. You seem to be vastly over complicating a very simple idea, and I say that as professional mathematician.

 

I am trying to recall modulo arithmetic Matt, and then express the logic behind it correctly is all.

 

Can you sort of give me a push in the right direction please? What I mean by that is this, how would you start off a lecture on the subject? There will be a branch from what is well known, to modulo arithmetic. I want to see it.

 

Thanks

 

2=1+1 regardless... Yeah but then you can reach a contradiction this way...

 

 

Suppose you are working with the following field:

 

[math] \mathbb{F} [/math]

 

So the field axioms are all true statements.

 

Now, one of the axioms of a field, is that the symbol 1 denotes an element of any field conceivable. Therefore, regardless of whether the field chosen was the reals, the rationals, the complex numbers, whatever... the following expression is guaranteed to denote a true statement:

 

[math] 1 \in \mathbb{F} [/math]

 

Now, the following expression denotes a true statement, regardless of the field chosen. The field could be chosen at random, and the following expression would still denote a true statement:

 

[math] 1+1 \in \mathbb{F} [/math]

 

And this is so, because a field is closed under addition. Now, the closure axiom of a field can be stated logically, as follows:

 

 

[math] \text{Axiom Of Closure of a randomly chosen field} \ \ \mathbb{F} [/math]

 

[math] \forall x \in \mathbb{F} \forall y \in \mathbb{F} [x+y \in \mathbb{F}][/math]

 

The possibility that x=y has been left open. Consider the case where x=y, the axiom of closure reduces to:

 

[math] \forall y \in \mathbb{F} [y+y \in \mathbb{F}][/math]

 

 

Now, one of the field axioms is the following:

 

[math] \text{Existence of a multiplicative identity 1} [/math]

 

[math] \exists 1 \forall x \in \mathbb{R} [1 \cdot x = x] [/math]

 

 

Uniqueness is left as a theorem.

 

But the point is, that the previous axiom ensures that any field has at least one point in it, being denoted by the symbol 1.

 

And so now, using the special case of the closure axiom stated earlier, together with existential instantiation, the following expression denotes a true statement, regardless of what field was randomly chosen:

 

[math] 1+1 \in \mathbb{F} [/math]

 

And repeating the process, it follows that the following expression denotes a true statement, for any field F:

 

[math] 1+1+1 \in \mathbb{F} [/math]

 

And so on.

 

Now, is the time to cover the issue of whether or not the field is finite, or infinite, and introduce the definition of:

 

[math] \text{Char} \ \mathbb{F} [/math]

 

... the characteristic of a field F.

 

So now, to modulo n arithmetic...

 

Let it be stipulated(that means that both you and I agree) that the randomly chosen field has three and only three elements in it.

 

It only makes sense that we should (deontic logic word) choose different symbols to denote each of the three elements.

 

Let the following be stipulated to be true:

 

[math] \mathbb{F} = \mathcal{f} 1,2,3 \mathcal{g} [/math]

 

So the elements of F are distinct by design. And saying that the elements of the set are distinct, is just a quick way to say this:

 

[math] \text{not(1=2) and not(1=3) and not(2=3)} [/math]

[matt grime]

And another field axiom is that its mutliplicative identity (1) is always different from its additive identity (0).

 

 

I would, if I were to have to start with something "nice" say, modulo arithmetic is something you use all the time. Suppose that today is wednesday, and that in 17 days time event X happens. What day would that be? We work "modulo 7" all the time. We work modulo 12, or 24 everyday too. All we are doing is saying that after 7 (days) or 12/24 hours, we start counting from 0 again.

 

However, I wouldn't choose to do so since I am a mathematician and therefore prefer talking maths.

 

Let p be a prime, define an equivalanence relation on Z via x~y iff there is a k in Z such that x=y+kp, or equivalently p divides x-y. Then, using Euclid's algorithm it is easy to show that the set of equivalence classes [0], [1],...,[p-1] is a field: define [x]+[y]=[x+y], and [x][y]=[xy]. This is independent of the choice of representative of equivlance class:

 

proof. If [a]= and [c]=[d] then there are integers s and t such that a=b+sp and c=d+tp, hence a+b=c+d+p(s+t) so that [a+b]=[c+d]. The proof for multiplication is analogous.

 

 

Thus we shall by standard abuse of notation refer to the equivlaence classes as 0,1..,p-1. This set is a field. Exercise (very easy!) show that given x=/=0 there is a y such that xy=1. Hint: I didn't mention Euclid's algorithm for nothing. This is the only non-trivial thing to show when demonstrating that 0,1,...,p-1 is a field with addition and mutliplication as defined above.

 

 

Consider the polynomial p(x)= x^2+x+1 over the field with two elements 0,1 with addition etc modulo 2. Then p(0)=p(1)=1, and thus we see that this is not algebraically closed. Can you produce a proof to show that F_p = {0,1,..,p-1} with addition and mult mod p is not algebraically closed? Hint: it is a finite set.

[Johnny5]

Can you produce a proof to show that F_p = {0,1,..,p-1} with addition and mult mod p is not algebraically closed? Hint: it is a finite set.

 

I'll get right on on it.

 

 

[math] \text{Non-triviality field axiom} [/math]

 

[math] \text{not(0=1)} [/math]

[matt grime]

I notice that you choose to denote the field with three elements to contain 1,2,3. This is most unusual. It is standard practice to use 0,1,2 as the field with three elements. Actually, there are infinitely many fields with three elements, so you shouldn't use "the¨, however, it is trivial to show that any finite field with three elements is canonically isomorphic to another field with three elements. This isn't necessarily true for other finite fields with size different from p.

[Johnny5]

I notice that you choose to denote the field with three elements to contain 1,2,3. This is most unusual. It is standard practice to use 0,1,2 as the field with three elements. Actually, there are infinitely many fields with three elements, so you shouldn't use "the¨, however, it is trivial to show that any finite field with three elements is canonically isomorphic to another field with three elements. This isn't necessarily true for other finite fields with size different from p.

 

My bad. One of the field axioms is that there is an element 0, of any field distinct from 1.

 

[math] \text{Existence of an Additive Identity} [/math]

 

[math] \exists 0 \forall x \in \mathbb{R} [0+x=x] [/math]

 

Hence... if the randomly chosen field F was {1,2,3} then one of the symbols 1,2,3 must denote that which 0 denotes, in the field axioms. So I should have written:

 

[math] \mathbb{F} = \mathcal{f} 0,1,2 \mathcal{g} [/math]

[Johnny5]

And another field axiom is that its mutliplicative identity (1) is always different from its additive identity (0).

 

 

I would' date=' if I were to have to start with something "nice" say, modulo arithmetic is something you use all the time. Suppose that today is wednesday, and that in 17 days time event X happens. What day would that be? We work "modulo 7" all the time. We work modulo 12, or 24 everyday too. All we are doing is saying that after 7 (days) or 12/24 hours, we start counting from 0 again.

 

However, I wouldn't choose to do so since I am a mathematician and therefore prefer talking maths.

 

Let p be a prime, define an equivalanence relation on Z via x~y iff there is a k in Z such that x=y+kp, or equivalently p divides x-y. Then, using Euclid's algorithm it is easy to show that the set of equivalence classes [0'], [1],...,[p-1] is a field: define [x]+[y]=[x+y], and [x][y]=[xy]. This is independent of the choice of representative of equivlance class:

 

proof. If [a]= and [c]=[d] then there are integers s and t such that a=b+sp and c=d+tp, hence a+b=c+d+p(s+t) so that [a+b]=[c+d]. The proof for multiplication is analogous.

 

 

Thus we shall by standard abuse of notation refer to the equivlaence classes as 0,1..,p-1. This set is a field. Exercise (very easy!) show that given x=/=0 there is a y such that xy=1. Hint: I didn't mention Euclid's algorithm for nothing. This is the only non-trivial thing to show when demonstrating that 0,1,...,p-1 is a field with addition and mutliplication as defined above.

 

 

Consider the polynomial p(x)= x^2+x+1 over the field with two elements 0,1 with addition etc modulo 2. Then p(0)=p(1)=1, and thus we see that this is not algebraically closed. Can you produce a proof to show that F_p = {0,1,..,p-1} with addition and mult mod p is not algebraically closed? Hint: it is a finite set.

 

 

You start out, let p be a prime. A prime number is a number which is only divisible by one and itself. The first few primes are in the following set:

 

{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53...}

 

Now, focus on the following statement:

 

there is a k in Z such that x=y+kp

 

In the expression above, let Z denote the integers? Shouldn't it be the natural numbers Matt?

 

The standard symbol for the integers is:

 

[math] \mathbb{Z} [/math]

 

The standard symbol for the natural numbers is:

 

[math] \mathbb{N} [/math]

 

I'm going to assume you meant the natural numbers.

 

 

 

[math] \exists k \in \mathbb{N} [ x=y+kp ][/math]

 

Now, there is a question as to what x,y denote. They are just arbitrary natural numbers.

 

So...

 

[math] \forall x \in \mathbb{N} \forall y \in \mathbb{N} \exists k \in \mathbb{N} [ x=y+kp ][/math]

 

Where p denotes a prime.

 

But the x,y are used to define an equivalence relation on the set.

[matt grime]

actually i failed to make it clear that you CAN say "the field with three elements" since given any two fields with three elements we x,y,z and a,b,c with x and a the addtive identity and y,b the mult ids, then the map a-->x, b-->y must be uniquely completed to c-->z, so the iso is unique, hence we really can say "the" field with 3 elts

[Johnny5]

actually i failed to make it clear that you CAN say "the field with three elements" since given any two fields with three elements we x,y,z and a,b,c with x and a the addtive identity and y,b the mult ids, then the map a-->x, b-->y must be uniquely completed to c-->z, so the iso is unique, hence we really can say "the" field with 3 elts

 

hmmm...

 

What is an isomorphism again?

 

Also, what is Euclid's algorithm again.

 

Also, is 1 a prime number?

 

Thanks Matt