If f(x) in Z[x] is primitive, then f(x) is reducible in Z[x] if and only if f(x) is reducible in Q[x]. Unique Factorization Domain (UFD) primitive polynomial Reducibility and Primitive Polynomials Isomorphism from F[x]/ to F(a)

Reducibility and Primitive Polynomials

An integral domain R where every nonzero nonunit element can be written as a unique product of irreducible elements. Ascending Chain Condition UFD and Primes f(x) Splits Unique Factorization Domain (UFD)

Unique Factorization Domain (UFD)

Abstract Algebra Flashcards: Polynomials, Fields, Rings & Number Theory

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Algebra - Abstract Algebra

snashuhaimdinx Created by 9 mon ago

Cards in this deck(25)

The GCD of the coefficients of a polynomial.

polynomial whose content is 1.

If polynomials f and g are primitive then f(x)g(x) is primitive

If f(x) in Z[x] is primitive, then f(x) is reducible in Z[x] if and only if f(x) is reducible in Q[x].

Given f(x) in Z[x] nonconstant, and p in Z a prime then if deg(f(x)=deg(f(x)modp), and f(x)modp is irreduicble in Z/pZ[x] then f(x) is irreduicble in Q[x].

For f(x) in Z[x] degree 2 or more, if there exists a prime p such that it divides every coefficient except the leading coefficient, and p^2 does not divide the constant, the f(x) is irreduicble in Q[x]

Every irreducible element in a PID is prime.

N:Z[sqrt(d)] to Z by a+bsqrt(d) to a^2-b^2d.

x in Z[sqrt(d)] is unit if and only if N(x) is unit

An integral domain R where every nonzero nonunit element can be written as a unique product of irreducible elements.

Euclidean Domain implies Principal Ideal Domain implies Noetherian implies Unique Factorization Domain implies Integral Domain.

For all sequences of ideals such that In is contained in In+1, there exists N such that IN=IN+1=...

A ring satisfying the ascending chain condition.

is also UFD

In UFD every irreducible element is prime.

Integral domain with a map d:R\{0} to Z+: 1. d(ab) >= d(a) 2. For all a,b in R, b nonzero, there are q,r in R such that a=bq+r where r=0 or d(r)

Z[i] using norm map

Z and Z[x]

Given f(x) in F[x] (F is field) the extension E/F is a splitting field of f(x) over F if it is the smallest field that f(x) splits over.

f(x) in F[x] splits over extension E/F if there are ai in E such that f(x)=c(x-a1)...(x-an)

Given f(x) in F[x], if there are ai in E with f(x)=c(x-a1)....(x-an) then F(a1,...,an) is splitting field of f(x) over F.

Q(nthroot(a), Sn) where Sn=e^(2pi i/ n)

Given extension E1 and E2 of F, then E1 and E2 are F-isomorphic if there is an isomorphism that is identity map when restricted to F.

If p(x) in F[x] is irreducible and p(a)=0 then there exists such an isomorphism.

the splitting field of f over F and g over G are isomorphic, and restricted to F or G is the same isomorphism. (same as F-isomorphic

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Abstract Algebra Flashcards: Polynomials, Fields, Rings & Number Theory

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