Re: [math-fun] Rational-coefficient polynomial, applied to integers
All this talk of polynomials suggested something I hadn't noticed before: Let P(X) = Sum_{1 <= k <= n} a_n X^n for any (say) reals a_k. Then think of P(X) as the point (a_0, a_1, ..., a_n) in R^(n+1). Polynomials add just like vectors in R^(n+1). Using this we can also multiply vectors v in R^(n+1) and w in R(m+1) to get one in R^(n+m+1). So if we take the union of all R^n, 1 <= n < oo (call this R^oo), we get a (commutative, associative) ring with unit. In fact it's an integral domain (the product of nonzero elements is nonzero). Hence R^oo has a field of fractions, which I suppose is equivalent to the field of rational functions over R. It's even a topological field, since addition and multiplication are continuous functions of the coefficients. Is this not a real division algebra? I've read that Kaplansky and others proved that the only real division algebras are in 1, 2, 4, and 8 dimensions. In particular, there are *none* in infinitely many dimensions (excepting the case of vector spaces over purely inseparable fields, which I don't think applies to the reals). Whatever it is, it's a pretty cool object. —Dan
Yes, and indeed the field of rational functions in any finite or infinite number of variables. -- Gene On Friday, November 16, 2018, 12:44:26 PM PST, Dan Asimov <dasimov@earthlink.net> wrote: All this talk of polynomials suggested something I hadn't noticed before: Let P(X) = Sum_{1 <= k <= n} a_n X^n for any (say) reals a_k. Then think of P(X) as the point (a_0, a_1, ..., a_n) in R^(n+1). Polynomials add just like vectors in R^(n+1). Using this we can also multiply vectors v in R^(n+1) and w in R(m+1) to get one in R^(n+m+1). So if we take the union of all R^n, 1 <= n < oo (call this R^oo), we get a (commutative, associative) ring with unit. In fact it's an integral domain (the product of nonzero elements is nonzero). Hence R^oo has a field of fractions, which I suppose is equivalent to the field of rational functions over R. It's even a topological field, since addition and multiplication are continuous functions of the coefficients. Is this not a real division algebra? I've read that Kaplansky and others proved that the only real division algebras are in 1, 2, 4, and 8 dimensions. In particular, there are *none* in infinitely many dimensions (excepting the case of vector spaces over purely inseparable fields, which I don't think applies to the reals). Whatever it is, it's a pretty cool object. —Dan
I made an unsuccessful attempt to track down the theorem to which Dan referred, starting from https://en.wikipedia.org/wiki/Division_algebra A major nuisance is that some references involved make unstated assumptions, particularly concerning algebraic closure and finite dimensionality. I did encounter several statements which initially appear to support his claim, but on closer inspection for this reason just fail to do so. WFL On 11/16/18, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Yes, and indeed the field of rational functions in any finite or infinite number of variables.
-- Gene
On Friday, November 16, 2018, 12:44:26 PM PST, Dan Asimov <dasimov@earthlink.net> wrote:
All this talk of polynomials suggested something I hadn't noticed before: Let
P(X) = Sum_{1 <= k <= n} a_n X^n
for any (say) reals a_k.
Then think of P(X) as the point (a_0, a_1, ..., a_n) in R^(n+1).
Polynomials add just like vectors in R^(n+1). Using this we can also multiply vectors v in R^(n+1) and w in R(m+1) to get one in R^(n+m+1).
So if we take the union of all R^n, 1 <= n < oo (call this R^oo), we get a (commutative, associative) ring with unit. In fact it's an integral domain (the product of nonzero elements is nonzero). Hence R^oo has a field of fractions, which I suppose is equivalent to the field of rational functions over R. It's even a topological field, since addition and multiplication are continuous functions of the coefficients.
Is this not a real division algebra? I've read that Kaplansky and others proved that the only real division algebras are in 1, 2, 4, and 8 dimensions. In particular, there are *none* in infinitely many dimensions (excepting the case of vector spaces over purely inseparable fields, which I don't think applies to the reals).
Whatever it is, it's a pretty cool object.
—Dan
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participants (3)
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Dan Asimov -
Eugene Salamin -
Fred Lunnon