Kronecker’s Theorem
Suppose F is a field and p(x)  F[x]. is irreducible over F.
Then < p(x) > is a maximal ideal in F[x], so F[x]/ < p(x) > is a field.
Let  be the canonical projection of F[x] onto F[x]/ < p(x) > :
( f(x) ) = f(x) + < p(x) >, for all f(x)  F[x].
If a, b  F, and (a) = (b), then a + < p(x) > = b + < p(x) >, which implies
a – b  < p(x) >. But all elements of < p(x) > have degree  degree p  1. If a – b  0,
then degree (a – b) = 0, which is impossible. Therefore a = b. Thus  is actually 1-1 when
restricted to F. So  is actually an isomorphism between F and (F).


F
(F)
F[x]
F[x] / < p(x) >
Let Y = F[x] / < p(x) >  (F)
(F)
F
Y
F[x]
F[x] / < p(x) >
Choose set X such that X = Y , and X  ( F[x]  F[x]/ < p(x) > ) = .
Let E = X  F.
E
X
F
(F)
Y
F[x]
F[x] / < p(x) >
Let  be a bijection between X and Y. We know that  is a bijection between F and (F).
Define  =   .
Then  is a bijection between E = X  F and Y  (F) = F[x]/ < p(x) >.
Define operations  and  on E as follows:
For all x, y  E, define x  y = -1((x) + (y)) and x  y = -1((x)(y)), so that
( x  y) = (x) + (y), and ( x  y) = (x)(y).
With these definitions, E with the operations of  and , has the same structure as
F[x]/ < p(x) > with the operations of + and .
For example, ( a(bc)) = (a)((bc))= (a)((b) +  (c)) = (a)(b) + (a)(c) =
(a  b) + (a  c) = ( (a  b)(a  c) ). Since  is 1-1, we have
a(bc) = (a  b)(a  c),
showing that the distributive property holds in E.
In the same way all the field properties can be shown to hold in E.
Thus E with these operations is a field.
Since ( x  y) = (x) + (y), and ( x  y) = (x)(y),  is a ring homomorphism.
Because  is also a bijection, it is an isomorphism between E and F[x] / < p(x) >.

 :E

F [ x] /  p ( x) 
Suppose p(x) = an x n  an1 x n1    a1 x  a0 .
Let z = -1( x + < p(x) >), so that (z) = x + < p(x) >.
Then (p(z)) = ( an z n  an1 z n1    a1 z  a0 ) =
 (an ) ( z ) n   (an1 ) ( z ) n1     (a1 ) ( z )   (a0 ) =
 (an )( x   p( x) ) n   (an1 )( x   p( x) ) n1     (a1 )( x   p( x) )   (a0 ) =
(an   p( x) )( x n   p( x) )  (an1   p( x) )( x n1   p( x) )   
(a1   p( x) )( x   p( x) )  (a0   p( x) ) 
(an x n   p( x) )  (an1 x n1   p( x) )    (a1 x   p( x) )  (a0   p( x) ) 
(an x n  an1 x n1    a1 x  a0 )   p( x)   p( x)   p( x)   0   p( x)  .
Thus  maps p(z) onto the zero element of F[x] / < p(x) >, so p(z) must be the zero
element of E: p(z) = 0E.
Therefore E is an extension of F which contains z, a zero of p(x).