Hi all,

can someone give me in general a template/algorithm for finding a splitting field ???

or more specifically, how can I do the following question ...

"Find the splitting field E for over .

Thanks.

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- Jul 19th 2012, 12:12 PMibnashrafSplitting field
Hi all,

can someone give me in general a template/algorithm for finding a splitting field ???

or more specifically, how can I do the following question ...

"Find the splitting field E for over .

Thanks. - Jul 20th 2012, 01:57 AMDevenoRe: Splitting field
well the first thing to check is if x

^{3}+ 2 is irreducible over Z_{7}. since it is a cubic, we need only check to see if it has a root in Z_{7}. working (mod 7) we have:

0^{3}+ 2 = 2

1^{3}+ 2 = 3

2^{3}+ 2 = 1 + 2 = 3

3^{3}+ 2 = 6 + 2 = 1

4^{3}+ 2 = (-3)^{3}+ 2 = -6 + 2 = -4 = 3

5^{3}+ 2 = (-2)^{3}+ 2 = -1 + 2 = 1

6^{3}+ 2 = (-1)^{3}+ 2 = -1 + 2 = 1

this tells us that if we adjoin any root of x^{3}+ 2 to Z_{7}, say u, then Z_{7}(u) is isomorphic to the field Z_{7}[x]/<x^{3}+ 2> (where u corresponds to the coset of x in the quotient ring).

this alone is no guarantee that x^{3}+ 2 splits in Z_{7}(u). however, if we find a second root in Z_{7}(u), say v, we must have the splitting field, since we can then divide x^{3}+ 2 by (x - u)(x - v) to get the third factor.

now considering Z_{7}(u) as a vector space over Z_{7}, we see that {1,u,u^{2}} is a basis (because of the irreducibility of x^{3}+ 2) so we know that Z_{7}(u) has 343 elements. we could (if we were masochistic) try the 335 elements that might be roots, but there is a better way.

as we saw above, 2^{3}= 1 (mod 7), so (2u)^{3}= 2^{3}u^{3}= u^{3}. this means that 2u is also a root of x^{3}+ 2. similarly, 4^{3}= (4^{2})(4) = (2)(4) = 1 (mod 7), so 4u is also a root of x^{3}+ 2, hence x^{3}+ 2 splits in Z_{7}(u) as:

x^{3}+ 2 = (x - u)(x - 2u)(x - 4u).

checking, we have that:

(x - u)(x - 2u) = x^{2}- 3ux + 2u^{2}= x^{2}+ 4ux + 2u^{2}, and:

(x^{2}+ 4ux + 2u^{2})(x - 4u) = (x^{2}+ 4ux + 2u^{2})(x + 3u)

= x^{3}+ 4ux^{2}+ 2u^{2}x + 3ux^{2}+ 5u^{2}x + 6u^{3}

= x^{3}+ 6u^{3}.

but since u^{3}= -2 = 5, 6u^{3}= (6)(5) = 2.

(this happy state of affairs occurs because Z_{7}happens to contain all 3 cube roots of 1). - Jul 20th 2012, 06:11 AMibnashrafRe: Splitting field
Thank you so much Deveno !

At least now i think i have a much better idea of how to find a splitting field over the integers mod p.

My question now is, how do you find a splitting field over the rationals ???

for example, how to find a splitting field F for over Q ?

I know that the first step (as shown by deveno before) is to show that f(x) is irreducible over Q

using Eisenstein's Criterior and if that fails we resort to Rational Roots theorem.

But how do i proceed after showing it is irreducible over Q ??? - Jul 20th 2012, 03:18 PMDevenoRe: Splitting field
let's look at the question in full generality. suppose F is ANY field, and suppose f(x) in F[x] is irreducible over F.

consider the ideal <f(x)> in F[x] generated by f. i claim this ideal is maximal in F[x].

for suppose J is any ideal in F[x] containing f(x) (and thus <f(x)>), with some element g(x) not in <f(x)>. by the irreducibility of f(x), gcd(f(x),g(x)) = 1.

this means there are polynomials h(x),k(x) in F[x], with:

h(x)f(x) + k(x)g(x) = 1. since J is an ideal, h(x)f(x) and k(x)g(x) are in J, thus h(x)f(x) + k(x)g(x) is in J, hence 1 is in J.

this mean J = F[x], since 1 is the multiplicative identity of F[x]. thus, <f(x)> is maximal.

since <f(x)> is maximal, F[x]/<f(x)> (the quotient ring) has only 2 ideals: 0 + <f(x)>, and F[x]/<f(x)> itself.

now F[x] is an integral domain (it is actually euclidean, but we don't need that, here), therefore so is F[x]/<f(x)>. since F[x]/<f(x)> has no non-trivial proper ideals, it is a field.

we can regard this field as an extension of F via the monomorphism:

a → a + <f(x)> (cosets of constant polynomials).

now here is the cool thing: in F[x]/<f(x)>, f(x) has a root: namely, the coset of x: x + <f(x)>:

f(x + <f(x)>) = f(x) + <f(x)> = 0 + <f(x)>, the additive identity of F[x]/<f(x)>, because f(x) is IN <f(x)>.

in fact, we have an ISOMORPHISM, of F(u), where u is a root of f(x), with F[x]/<f(x)>, given by:

g(u) ↔ g(x) + <f(x)> (note this isomorphism extends our earlier embedding of F into F[x]/<f(x)>).

the easiest way to see that this is an isomorphism, is to note that:

h(x) → h(u) is an onto ring homomorphism of F[x] → F(u) whose kernel is <f(x)>.

**********

so, the upshot of all of this is that given an irreducible polynomial f(x) over F, we can enlarge F to E = F(u), so that it contains a root, u. however, this extension might not be a splitting field for f(x) (we might only get some roots, instead of all of them). we might have to repeat the process, after "dividing out" the factors in E[x], to get an irreducible polynomial in E[x], but...it will be of lesser degree, so this cannot continue forever.