# Extension fields and Kronecker's theorem. confusion.

• Jun 17th 2011, 05:15 AM
abhishekkgp
Extension fields and Kronecker's theorem. confusion.
definition: A field $\displaystyle E$ is an extension field of a field $\displaystyle F$ if $\displaystyle F \leq E$. (correct?)

Kronecker's theorem: Let $\displaystyle F$ be a field and let $\displaystyle f(x)$ be a non-constant polynomial in $\displaystyle F[x]$. Then there exists an extension field $\displaystyle E$ of $\displaystyle F$ and an $\displaystyle \alpha \in E$ such that $\displaystyle f(\alpha)=0$.

PROOF: (this is where i have a confusion) $\displaystyle f(x)$ has a factorization in $\displaystyle F[x]$ into polynomials that are irreducible over $\displaystyle F[x]$. Let $\displaystyle p(x)$ 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:
$\displaystyle E=F[x]/(p(x))$ is an extension field of $\displaystyle F$.
now how is that??... the elements of $\displaystyle F$ even look different that the elements of $\displaystyle E$. i mean to say that no element of $\displaystyle F$ is in $\displaystyle E$.
I understand that $\displaystyle E$ has a sub-field which is isomorphic to $\displaystyle F$ but according to the definition of an extension field how do we regard $\displaystyle E$ as an extension of $\displaystyle F$??
(phewf!)
• Jun 17th 2011, 05:43 AM
Deveno
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.
• Jun 17th 2011, 07:12 AM
FernandoRevilla
Re: Extension fields and Kronecker's theorem. confusion.
Equivalently, if $\displaystyle p(x)\in \mathbb{K}[x]$ is irreducible and degree $\displaystyle n$ then, $\displaystyle \Sigma=\mathbb{F}^n=\{(c_0,\ldots,c_{n-1}):c_i\in\mathbb{F}\}$ is a field with the operations

Sum $\displaystyle (c_0,\ldots ,c_{n-1})+(c'_0,\ldots,c'_{n-1})=(c_0+c'_0,\ldots,c_{n-1}+c'_{n-1})$

Product $\displaystyle (c_0,\ldots,c_{n-1})\cdot (c'_0,\ldots,c'_{n-1})=(c''_0,\ldots,c''_{n-1})$ where $\displaystyle c''_0+c''_1x+\ldots c''_{n-1}x^{n-1}$ is the remainder of the division

$\displaystyle (c_0+c_1x+\ldots+c_{n-1}x^{n-1})(c'_0+c'_1x+\ldots+c'_{n-1}x^{n-1})$ : $\displaystyle p(x)$

Then, $\displaystyle \mathbb{F}_1=\{(c_0,0,\ldots,0):c_0\in \mathbb{F}\}$ is a subfield of $\displaystyle \Sigma$ isomorphic to $\displaystyle \mathbb{F}$ by means of $\displaystyle (c_0,0,\ldots,0)\to c_0$ , we can identify those elements (so , $\displaystyle \mathbb{F}\subset \Sigma)$ and $\displaystyle \alpha=(0,1,0,\ldots,0)$ is a root of $\displaystyle p(x)$ in $\displaystyle \Sigma$ .
• Jun 17th 2011, 08:32 AM
Deveno
Re: Extension fields and Kronecker's theorem. confusion.
surely you meant $\displaystyle p(x) \in \mathbb{F}[x]$.

while that does show that $\displaystyle \mathbb{F}[x]/(p(x))$ 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 $\displaystyle \mathbb{F}^n$ as the corresponding

polynomial) for every non-zero g(x) represented by $\displaystyle (c_0,\dots,c_{n-1})$, we have gcd(g(x),p(x)) = 1.

it also appeals to an (unspecified) isomorphism: $\displaystyle (c_0,c_1,c_2,\dots,c_{n-1}) \rightarrow c_0 + c_1x + c_2x^2 +\dots +c_{n-1}x^{n-1} + (p(x))$

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 $\displaystyle \mathbb{F}[x]/(p(x))$, or the one embedded in $\displaystyle \mathbb{F}^n$).
• Jun 17th 2011, 11:06 AM
FernandoRevilla
Re: Extension fields and Kronecker's theorem. confusion.
Quote:

Originally Posted by Deveno
surely you meant $\displaystyle p(x) \in \mathbb{F}[x]$.

Of course, just a typo. The rest is all correct.
• Jun 17th 2011, 11:10 AM
Deveno
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.
• Jun 19th 2011, 09:51 PM
abhishekkgp
Re: Extension fields and Kronecker's theorem. confusion.
A field $\displaystyle E$ is an extension field of $\displaystyle F$ if there is a subfield $\displaystyle K$ of $\displaystyle E$ such that $\displaystyle K$ is isomorphic to $\displaystyle F$.

• Jun 19th 2011, 11:05 PM
FernandoRevilla
Re: Extension fields and Kronecker's theorem. confusion.
Quote:

Originally Posted by abhishekkgp
A field $\displaystyle E$ is an extension field of $\displaystyle F$ if there is a subfield $\displaystyle K$ of $\displaystyle E$ such that $\displaystyle K$ is isomorphic to $\displaystyle F$.

That is a good definition. Reflects exactly what we want to mean.
• Jun 20th 2011, 01:22 PM
Deveno
Re: Extension fields and Kronecker's theorem. confusion.
Quote:

Originally Posted by abhishekkgp
A field $\displaystyle E$ is an extension field of $\displaystyle F$ if there is a subfield $\displaystyle K$ of $\displaystyle E$ such that $\displaystyle K$ is isomorphic to $\displaystyle F$.