Applications of Logic to Field Theory
... By the Fundamental Theorem of Algebra, the field C of complex numbers is algebraically closed. More generally, start with any field F , and let F be the set of all roots of polynomials of the form xn + an−1 xn−1 + · · · + a1 x + a0 with ai ∈ F and n > 0. (These roots exist in some extension of F ; w ...
... By the Fundamental Theorem of Algebra, the field C of complex numbers is algebraically closed. More generally, start with any field F , and let F be the set of all roots of polynomials of the form xn + an−1 xn−1 + · · · + a1 x + a0 with ai ∈ F and n > 0. (These roots exist in some extension of F ; w ...
Set 2
... p. Instead, look at the mod p reductions of the convergents pn /qn ∈ Q(T ). (First rescale so all coefficients are integral.) How do these compare with the intrinsic mod p convergents? ...
... p. Instead, look at the mod p reductions of the convergents pn /qn ∈ Q(T ). (First rescale so all coefficients are integral.) How do these compare with the intrinsic mod p convergents? ...
An algebraically closed field
... Proof. Let 2T be the family of triples (gt, hh y,) such that gx is monic, d°gi = d°g v(g-9d>0, v(h-hd>0, v(l-!gt-mhd>0, vif-gth,) = y,, S(g,)<=Q and S(hd <= Q. Define a partial order on 9~, (g,, hit y,) < (g^hj, yj), to mean y, < y,, v(gi,—gy) ^ yis v(h, — hj) ^ y;. ^" is non-empty. Any chain in 9~ ...
... Proof. Let 2T be the family of triples (gt, hh y,) such that gx is monic, d°gi = d°g v(g-9d>0, v(h-hd>0, v(l-!gt-mhd>0, vif-gth,) = y,, S(g,)<=Q and S(hd <= Q. Define a partial order on 9~, (g,, hit y,) < (g^hj, yj), to mean y, < y,, v(gi,—gy) ^ yis v(h, — hj) ^ y;. ^" is non-empty. Any chain in 9~ ...
Algebra in Coding
... 1. (a) Write down the addition and multiplication tables for GF(5) and GF(7). (b) Write down the addition and mulitplication tables for GF(4). 2. Construct GF(16) in three different ways by defining operations modulo the irreducible polynomials x4 +x+1, x4 +x3 +1, and x4 +x3 +x2 +x+1. Find isomorphi ...
... 1. (a) Write down the addition and multiplication tables for GF(5) and GF(7). (b) Write down the addition and mulitplication tables for GF(4). 2. Construct GF(16) in three different ways by defining operations modulo the irreducible polynomials x4 +x+1, x4 +x3 +1, and x4 +x3 +x2 +x+1. Find isomorphi ...
Math 614, Fall 2015 Problem Set #1: Solutions 1. (a) Since every
... Let T be the D-span of 1, x, . . . , xd−1 . To prove D = R, use induction on degree s of an element g of K[x]. If s < d, then g is in the K-span of 1, x, . . . , xd−1 . If s ≥ d, we have s = qf +ρ, where ρ is in the K-span of 1, x, . . . , xd−1 , and, by induction, since deg(q) < s, q ∈ T . Multiply ...
... Let T be the D-span of 1, x, . . . , xd−1 . To prove D = R, use induction on degree s of an element g of K[x]. If s < d, then g is in the K-span of 1, x, . . . , xd−1 . If s ≥ d, we have s = qf +ρ, where ρ is in the K-span of 1, x, . . . , xd−1 , and, by induction, since deg(q) < s, q ∈ T . Multiply ...
![(January 14, 2009) [08.1] Let R be a principal ideal domain. Let I be](http://s1.studyres.com/store/data/006005520_1-5192ab5794c8b33a00720a831d1a8a33-300x300.png)
(January 14, 2009) [08.1] Let R be a principal ideal domain. Let I be
... One approach, certainly correct in spirit, is to say that obviously k[x1 , . . . , xn ]/Rx1 + . . . + Rxj ≈ k[xj+1 , . . . , xn ] The latter ring is a domain (since k is a domain and polynomial rings over domains are domains: proof?) so the ideal was necessarily prime. But while it is true that cert ...
... One approach, certainly correct in spirit, is to say that obviously k[x1 , . . . , xn ]/Rx1 + . . . + Rxj ≈ k[xj+1 , . . . , xn ] The latter ring is a domain (since k is a domain and polynomial rings over domains are domains: proof?) so the ideal was necessarily prime. But while it is true that cert ...
Filters and Ultrafilters
... be the unit interval [0, 1] on the real line, and k could be the real numbers. Later in the course we will try to determine the maximal ideals of particular subrings of M ap(I, k). ...
... be the unit interval [0, 1] on the real line, and k could be the real numbers. Later in the course we will try to determine the maximal ideals of particular subrings of M ap(I, k). ...
A Brief Summary of the Statements of Class Field Theory
... principal ideals. The class group is Cl OK := I/P . Q We now generalize to an arbitrary modulus m = v v ev . Let Im be the subgroup of fractional ideals that do not involve the primes dividing m; i.e., Im is the free abelian group on the nonarchimedean places v satisfying ev = 0. For a ∈ K × , the n ...
... principal ideals. The class group is Cl OK := I/P . Q We now generalize to an arbitrary modulus m = v v ev . Let Im be the subgroup of fractional ideals that do not involve the primes dividing m; i.e., Im is the free abelian group on the nonarchimedean places v satisfying ev = 0. For a ∈ K × , the n ...
![Rings of constants of the form k[f]](http://s1.studyres.com/store/data/021729650_1-3a6201c0eec615e02140355abcc4b661-300x300.png)
