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

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.

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.

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.

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.

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)

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.

Which word in particular 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. 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. 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.

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.

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.

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?

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.

Erm, is the earth's surface a) euclidean, b) riemannian (hyperbolic) or c) spherical, or d) none of the above necessarily, though eulcidean geometry provdes a reasonably good approximation/model for local questions, and spherical for global questions? What about the observation that thinking of modern vector spaces (kernels, linear maps, spectral theory), set theory (a function defined as a subset of AxB, domains, codomains), and propositonal logic in terms of Newton, who died many years before any of those systems were created, formalized, studied and presented in the manner of the text books you are reading is probably not going to be a great help? A vector is an element of a vector space, V. A linear map is a function from V to V (or some other vector space) satisfying certain properties. A function is a way of associating to each element in the domain a unique element in the codomain. The propositional form for this uniqueness is, if you'll forgive the lack of formality, something like: for all x in the domain, there is a y in the codomain such that f(x)=y [or whatever form you prefer your functions to take], and such that for all z in the codomain f(x)=z implies z=y.

I just posted a rpely pointing out that doign evctor spaces accoridng to someone how's been dead for 280 years isn't a good idea. only it didn't appear. plus your triangle walking experiment doesn't accout for the fact that the earth is curved... learn to differentiate between a model and the reality it is modelling

Hmm, if you want to do maths as if you're a physicist who's been dead for the best part of 300 years good luck to you. Vectors are not, in modern maths, "things that have length and direction", they are elements of a vector space, things that satisfy certain axioms. A model of which is useful for describing what are commonly called "vectors" such as force adn displacement.

It shuold be possible to use an array within an equation environment. Or just define your own numbering. Any decent intro to latex will tell you how, though asking people here to regurgitate it would be probably a waste of time given the Tex users groups that exist and all the online resources.

Vector addition isn't *defined* geometrically, and it is part of the axioms that the vector space is commutative under addition, so if iti s a theorem it is a trivial one. Moreover, it is an axiom for a metric space that d(x,y)=0 if and only if x=y. And it is clear from the euclidean metric on R^n that this is true. It has nothing to do with lengths of string, or twisting them. If d(x,y) = 0 for some x not equal to y, or d(x,x) isn't zero it isn't a metric, that is all. As it happens, the universe is more properly modelled with hyperbolic geometry. But note the emphasisi here onthe world modelled. You apparently seem to think the universe IS euclidean space. If you don't think it is a suitable model that is different. Nor do I see why you think there are an infinite number of paths from a point to itself no one of which is any shorter than any other, and why this isn't true of two other points. The number of paths isn't important anyway, ony the inf of all the lengths. There are an infinite number of paths all of the same shortest lengths between antipodal points on the sphere. What exactly do you mean by path anyway? Do you in any book see anything that specifies uniqueness of paths? Oh, and you don't need to write out the (usual) definitions for me, I already know them. When I say "what do you think a path is", for example, I am pointing out that you don't appear to be using standard fixed ideas.

ANd now you'er doing metric spaces? ALmost no vector spaces have a metric. You can define, rigorously, relations and then fucntions using set theory if you want to, but why bother? It doesn't aid you in understanding what functions do, or proving things about functions. Define it in words.

Hamilton upon noting that the complex numbers are a 2-d vector space over the reals (this isn't actualyl how he'd've stated it, but it is how we say it now) wanted to find a 3-dimensional version. HE struggled, then infamously, and probably apocryphally in a flash of insight saw that you need to make it 4-d, and the quaternions were born and he carved in to the underside of a bridge in dublin. there are 8-d things to called octonians, and even sedonions. thoguh you have to drop commutativity for the quaternions and associativity for the others. We can show there is no 3-d version.

Yes, it is perfectly acceptable and very common practice to write the compositoin wrt to * as juxtaposition. In fact I can think of no one who uses * except in basic undergrad texts. The axioms of a ring are easily googlable, wolfram will have them.

Ok, the reason why dividing by zero isn't allowed algebraically is becuase 0*x=0*y for all x and y. this holds in a division ring too, as well as a field. a division rnig is one where for all non-zero elements x there is a y such that xy=1, it is a field if it is commutative (wrt *) i just wondered what made you think of quaternions - the simplest example of a division ring that isn't a field.

No, of course not. Division by zero is not defined in any division ring for obvious reasons. Though why the quarternions?

Well, the reals by one of the standard DEFINITIONS is the completion of the rationals in the euclidean norm. So this "step" follows trivially, in the mathematical sense. Ie two two sequences are declared to be equivalent if their difference is a sequence that tends to zero. There are othere ways of giving a model for the real numbers, but they are all models of a totally ordered complete field, and any such is unique up to (unique) isomorphism, hence they are called the real numbers. Remember, decimals are just representations of real numbers, they are not in any sense "actually real numbers". And, Dave, yes, most people are misusing the notions of infinity, however, if you state they are doing something that is just plain wrong, one of them will come back with spaces where infinity is a member. None of these is the real numbers though.

Actualy, Dave, 1/0=infinity in both the extended reals, which aren't very interesting, and the one point compactification of the complex plane, aka the riemann sphere, which is very widely used in mathematics, especially complex analysis: it's automorphisms (in the proper sense) are the mobius maps, which are essentially SU(2). It is easy to show that any riemann surface that is locally euclidean is a quotient space of this sphere by some finite group action, and this tells us that hyperbolic geometry is the proper geometry of mathematics.