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Find link is a tool written by Edward Betts .
searching for Algebraic integer 8 found (43 total)

alternate case: algebraic integer

Ring class field
(219 words)
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completely in L if and only if p splits completely in K. L=K(a) for a an algebraic integer with minimal polynomial over Q of degree h(-4n), the class number

Proofs of quadratic reciprocity
(3,994 words)
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{\textstyle \tau ^{2}=2} . Because τ {\displaystyle \tau } is an algebraic integer , if p is an odd prime it makes sense to talk about it modulo p. (Formally

Lindemann–Weierstrass theorem
(4,699 words)
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f i ( j ) ( α k ) {\displaystyle f_{i}^{(j)}(\alpha _{k})} is an algebraic integer which is divisible by p! for j ≥ p {\displaystyle j\geq p} and vanishes

Lehmer's conjecture
(1,675 words)
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algebraic number so m ( P ) {\displaystyle m(P)} is the logarithm of an algebraic integer . It also shows that m ( P ) ≥ 0 {\displaystyle m(P)\geq 0} and that

Baker's theorem
(3,372 words)
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point are given by algebraic integers times known constants. If an algebraic integer has all its conjugates bounded by a known constant, then it cannot

Mersenne prime
(9,421 words)
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former is not a prime. This can be remedied by allowing b to be an algebraic integer instead of an integer: In the ring of integers (on real numbers),

Cayley–Hamilton theorem
(10,729 words)
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_{1},\ldots ,\alpha _{k}]} of Q {\displaystyle \mathbb {Q} } and an algebraic integer α ∈ Q [ α 1 , … , α k ] {\displaystyle \alpha \in \mathbb {Q} [\alpha

Principalization (algebra)
(8,949 words)
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integer if necessary we may assume that A {\displaystyle A} is an algebraic integer . The non-unit A {\displaystyle A} is generator of an ambiguous principal