direct product of rings
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From: Kusanagi on 3 Dec 2009 01:11 > > > On Dec 3, 12:13 am, Kusanagi > > <Kusan...(a)hotmail.com> > > > wrote: > > > > Why the direct product of two fields, > considered > > as > > > a ring is never itself a field? > > > > > > > > Thanks. > > > > > > What happens when you multiply (1,0) by (0,1)? > Is > > > either of them equal > > > to the zero of the product? > > > > > > -- > > > Arturo Magidin > > > > I see. > > > > How the tensor product of two fields, considered > as > > a ring can get away with the above problem and > become > > a field? > > > > Thanks. > > The tensor product over which ring or field? > > To form a tensor product of two commutative rings > say, you have to consider them as algebras over a > base ring R. > > In general the reason why the tensor product of two > fields over some base can be a field is that the > tensor product involves a lot of identification of > pairs (you can contruct it using a certain > equivalence > relation on the product). > > H For example, (0,1) (x) (1, 0) = (0,0), where (x) is a tensor product and R^2 (R:a set of real numbers) is a base ring. This cannot be a field,becasue multiplication of nonzero elements cannot be zero. Now going back to the original question, how the tensor product of two fields, considered as a ring can get away with the above problem (a problem like multiplication of (0,1) (x) (1,0)=(0,0)) and become a field? Do you have a specific example how to avoid those problems? Thanks. Kusanagi
From: Arturo Magidin on 3 Dec 2009 13:01 On Dec 3, 10:11 am, Kusanagi <Kusan...(a)hotmail.com> wrote: > > > > On Dec 3, 12:13 am, Kusanagi > > > <Kusan...(a)hotmail.com> > > > > wrote: > > > > > Why the direct product of two fields, > > considered > > > as > > > > a ring is never itself a field? > > > > > > Thanks. > > > > > What happens when you multiply (1,0) by (0,1)? > > Is > > > > either of them equal > > > > to the zero of the product? > > > > > -- > > > > Arturo Magidin > > > > I see. > > > > How the tensor product of two fields, considered > > as > > > a ring can get away with the above problem and > > become > > > a field? > > > > Thanks. > > > The tensor product over which ring or field? > > > To form a tensor product of two commutative rings > > say, you have to consider them as algebras over a > > base ring R. > > > In general the reason why the tensor product of two > > fields over some base can be a field is that the > > tensor product involves a lot of identification of > > pairs (you can contruct it using a certain > > equivalence > > relation on the product). > > > H > > For example, (0,1) (x) (1, 0) = (0,0), where (x) is a tensor product and R^2 (R:a set of real numbers) is a base ring. This cannot be a field,becasue multiplication of nonzero elements cannot be zero. But R^2 is not a field either, so what is the problem? On the other hand, if you do R(x)R over R, then there is no problem, because the tensor 1(x)0 is actually equal to the zero element of the tensor product. > > Now going back to the original question, how the tensor product of two fields, considered as a ring can get away with the above problem (a problem like multiplication of (0,1) (x) (1,0)=(0,0)) and become a field? Do you have a specific example how to avoid those problems? Do you have a specific pair of elements that you think should multiply to zero? Look: take Q(sqrt(2)) and Q(sqrt(3)), and then do their tensor over Q. The elements of this field F = Q(sqrt(2))(x)Q(sqrt(3)) are of the form a_1(x)b_1 + ... + a_n(x)b_n where a_i is in Q(sqrt(2)), b_i is in Q(sqrt(3), and n is a natural number. However, there is a lot of identification, as was mentioned. You have that (a1+a2)(x)b = a1(x)b + a2(x)b a(x)(b1+b2) = a(x)b1 + a(x)b2 (ra)(x)b = r(a(x)b) = a(x)rb for all r in Q Multiplication is defined by letting (a(x)b)*(c(x)d) = ac(x)bd, and extending linearly. Now, why is a product of two "pure" nonzero tensors never equal to 0? A pure tensor is equal to 0 if and only if at least one of its components is equal to 0. So if (a(x)b)*(c(x)d) = ac(x)bd is 0, then either ac=0 or bd=0, from which you get that either a, b, c, or d is 0, from which you get that ether a(x)b or c(x)d is zero. Then you can show that the product of a pure tensor times a general element is 0 if and only if one of the two factors is zero; and then finally you can show that the product of two general elements is zero if and only if at least one of them is zero. This can be done in generality; for this particular case, every element of Q(sqrt(2)) can be written as a1+b1*sqrt(2), and every element of Q(sqrt(3)) as c1+d1*sqrt(3), so each pure tensor can be written as (a1+b1*sqrt(2))(x)(c1+d1*sqrt(3)) = a1(x)c1 + a1(x)d1*sqrt(3) + b1*sqrt (2)(x)c1 + b1*sqrt(2)(x)d1*sqrt(3) = (a1*c1)(1(x)1) + (a1*d1)(1(x)sqrt (3))+(b1*c1)(sqrt(2)(x)1) + (b1*d1)(sqrt(2)(x)sqrt(3)). So this is a vector space over Q of dimension 4, with basis 1(x)1, sqrt(2)(x)1, 1(x) sqrt(3), (sqrt(2)(x)sqrt(3); it is in fact an algebra (you can multiply these vectors), and from there it is not hard to see you won't get a product equal to 0 if you start with things that are not equal to zero. In fact, you will get a field that is isomorphic to Q (sqrt(2),sqrt(3)). But you really need to work these things out and play with them yourself. Your first question in this thread was trivial, and suggests you did not bother to think too hard about it before asking. Doing that will not gain you any insight, it will only give you answers that don't stick. -- Arturo Magidin
From: Arturo Magidin on 3 Dec 2009 13:11 On Dec 3, 12:46 am, Kusanagi <Kusan...(a)hotmail.com> wrote: > > On Dec 3, 12:13 am, Kusanagi <Kusan...(a)hotmail.com> > > wrote: > > > Why the direct product of two fields, considered as > > a ring is never itself a field? > > > > Thanks. > > > What happens when you multiply (1,0) by (0,1)? Is > > either of them equal > > to the zero of the product? > I see. > > How the tensor product of two fields, considered as > a ring can get away with the above problem and become a field? Rings don't "become" fields; they either *are* fields, or they are not. The reason the tensor product "can get away with the above problem" is that the "above problem" doesn't occur in the tensor product. The tensor product is a quotient of the free modulo on the *set* FxK, and its ring structure has almost nothing to do with the ring structure of the direct product. Your question is like asking why is it that a dog without legs cannot walk, but a healthy horse can get away with the above problem and run... -- Arturo Magidin
From: Bill Dubuque on 3 Dec 2009 14:34
Kusanagi <Kusanagi(a)hotmail.com> writes: > > Now going back to the original question, how the tensor product of > two fields, considered as a ring can get away with the above problem > (a problem like multiplication of (0,1) (x) (1,0)=(0,0)) and become > a field? Do you have a specific example how to avoid those problems? HINT tensors are bilinear so are 0 if either argument is 0 (recall that, generally, f linear => f(0) = 0). Therefore while (0,1) (1,0) = (0,0), (0,1) != (0,0) in a direct product it becomes 0 * 0 = 0 as a tensor product. --Bill Dubuque |