Modern Algebra
Ring Theory (XXXII)
Polynomial Ring
Irreducible Polynomial

If f(x) is in F[x], where F is the field of integers mod p, p a prime, and f(x) is irreducible over F of degree n,
prove that F[x]/(f(x)) is a field with p^n elements.

See the attached file for the solution of the problem.

Modern Algebra
Ring Theory (XXXII)
Polynomial Ring
Irreducible Polynomial

By:- Thokchom Sarojkumar Sinha

If is in , where is the field of integers mod , a prime , and is irreducible over of ...

Solution Summary

It id the explanation of the following topic:

If f(x) is in F[x], where F is the field of integers mod p, p a prime, and f(x) is irreducible over F of degree n, then F[x]/(f(x)) is a field with p^n elements.
The solution is given in detail.

... You start with Z_p = Z/pZ, which is a field with p elements (integers modulo p) and consider the splitting field of the polynomial f(x) = x^{p^n} - x over Z_p. ...

... u is algebraic over K with minimial polynomial p(x) then consider how multiplicative inverses are found in K[x]/<p(x)>. ... Given : F is an extension field of K ...

... Theorem 10.7 Let K be an extension field of F and u an algebraic element over F with minimal polynomial p(x) of degree n.Then (1) F(u) F[x]/(p(x)) (2) { ,u ...

1. If E, F are fields and F is a subring of E, show each q in Aut(E/F... So, f is a homomorphism ... Then, x=ap+bq for some integer p and q. But then x=pnd+qmd=(pn+qm)d ...

... Irreducible Polynomial A polynomial p ( x ) in F [x ] is called an irreducible polynomial over the field F if whenever p ( x ) = a ( x )b( x ) where a ...

... On the sphere's surface the vector field is: é1 ù1 = ê2 ú= 2 F rr ... Ay ¶ Ax ¶ Az ¶f ¶f ¶f Ñ ×(f A ) = Ax + f + Ay + f + Az + f ¶x ¶x ¶y ¶y ...

... Then, m(x)=a(x)b(x), and m(g)=a(g)b(g)=0. Since a(g) and b(g) are elements of the field E, and so E has no zero divisors ... 2) The[n] suppose q(x) in F[x] with q ...

... a_2)(x^2) + ... where a_i belongs to the field F), {p(x)/q(x...x) which is contained in a maximal ideal N≠M. But ... by the Corollary, both x and 1-x are elements...

...X ) is isomorphic to Z / 2 Z , the field with 2 ... The same proof applies to any ideal (p,X), of course. ... It is also true, more generally, that (p,f(X)) is maximal ...