Extension fields and Kronecker's theorem. confusion.

definition: A field is an **extension field of a field** if . (correct?)

Kronecker's theorem: Let be a field and let be a non-constant polynomial in . Then there exists an extension field of and an such that .

PROOF: (this is where i have a confusion) has a factorization in into polynomials that are irreducible over . Let be an irreducible polynomial in such a factorization. (till here everything is fine).

i am now skipping some details and coming to the point which troubles me:

is an extension field of .

now how is that??... the elements of even *look* different that the elements of . i mean to say that *no* element of is in .

I understand that has a sub-field which is isomorphic to but according to the definition of an extension field how do we regard as an extension of ??

somebody please comment.

(phewf!)

Re: Extension fields and Kronecker's theorem. confusion.

well, you are correct. if b is an element of F, then b isn't "really" an element of F[x]/(p(x)), the actual element of F[x]/(p(x)) is b + (p(x)).

however, the field E obtained in such a way is isomorphic to F(α), where F(α) is created from F by "adjoining α, a root of p(x)".

here is a simple example: x^2 - 2 is irreducible over Q (it has no rational roots). now, we can construct Q[x]/(x^2-2), whose elements are all of the form:

(a + (x^2-2)) + (bx + (x^2-2)) = a+bx + (x^2-2). so we form the following isomorphism between Q(√2) and Q[x]/(x^2-2):

a+b√2 <--> a+bx + (x^2-2).

note that this map is totally determined by the images of 1, and √2. so we would expect that 1 + (x^2-2) is the multiplicative identity of Q[x]/(x^2-2),

and that x + (x^2-2) is an element that when squared, is the image under the isomorphism of 2: to make the algebra simpler,

i will write I for the principal ideal generated by x^2-2.

(x+I)^2 = (x+I)(x+I) = x^2 + I = x^2 + 2 - 2 + I = 2 + (x^2-2) + I = 2 + I (because x^2-2 is IN I).

so the coset x+I really does act like "a square root of 2".

in general, algebraic objects are only considered "up tp isomorphism". isomorphism is an equivalence relation on the set (more properly class, because of "free objects")

of all groups/rings/fields, etc. and since we are only interested in how things objects behave, equality is too strict a requirement.

for example, the complex field is often regarded as R^2 with a certain multiplication. well, as a complex vector space,

C is of dimension one, whereas as a real vector space, R^2 is of dimension 2. so clearly C and R^2 aren't "the same thing".

in fact, you have been doing this kind of "identification" all your life. consider the field of quotients of the integers:

a fraction is often regarded as a/b, and integers are "identified with" the fractions k/1 (for an integer k).

not only is k/1 a PAIR of numbers, this pair isn't even unique: k/1 = 2k/2 = 3k/3......etc.

so integers are not the same sort of thing as rational numbers, but we agree to identify the integers with the subring of Q they are isomorphic to.

Re: Extension fields and Kronecker's theorem. confusion.

Equivalently, if is irreducible and degree then, is a field with the operations

*Sum*

*Product* where is the remainder of the division

:

Then, is a subfield of isomorphic to by means of , we can identify those elements (so , and is a root of in .

Re: Extension fields and Kronecker's theorem. confusion.

surely you meant .

while that does show that is a ring (which being a quotient of a ring by an ideal it has to be anyway),

it does not actualy show it is a field (that is, that the product you have defined admits inverses of non-zero elements), although

you could do this by appealing the the fact that p(x) is irreducible, so (regarding an element of as the corresponding

polynomial) for every non-zero g(x) represented by , we have gcd(g(x),p(x)) = 1.

it also appeals to an (unspecified) isomorphism:

while this is fairly standard, from a logical point of view, there is no reason to prefer one isomorph of F over another

(that is, the one embedded in , or the one embedded in ).

Re: Extension fields and Kronecker's theorem. confusion.

Quote:

Originally Posted by

**Deveno** surely you meant

.

Of course, just a typo. The rest is all correct.

Re: Extension fields and Kronecker's theorem. confusion.

indeed it is, and it provides another way of looking at an extension field, which might make more sense to the original poster.

Re: Extension fields and Kronecker's theorem. confusion.

thank you fernando and denevo (that rhymes..). how about this definition of an extension field..

A field is an extension field of if there is a subfield of such that is isomorphic to .

please comment.

Re: Extension fields and Kronecker's theorem. confusion.

Quote:

Originally Posted by

**abhishekkgp** how about this definition of an extension field..

A field

is an extension field of

if there is a subfield

of

such that

is isomorphic to

.

That is a good definition. Reflects exactly what we want to mean.

Re: Extension fields and Kronecker's theorem. confusion.

Quote:

Originally Posted by

**abhishekkgp** thank you fernando and denevo (that rhymes..). how about this definition of an extension field..

A field

is an extension field of

if there is a subfield

of

such that

is isomorphic to

.

please comment.

now you're thinking like an algebraist! :)

Re: Extension fields and Kronecker's theorem. confusion.

Quote:

Originally Posted by

**Deveno** now you're thinking like an algebraist! :)

haha.. thanks!