Hi, I've done part i) of this question by showing that f(x) has no zeros over Z[7] but I'm now stuck on part ii) completely lost!
Do I need to factorize x^3-2 to obtain it's extension field? and how do I show it splits?
Thanks
Hi, I've done part i) of this question by showing that f(x) has no zeros over Z[7] but I'm now stuck on part ii) completely lost!
Do I need to factorize x^3-2 to obtain it's extension field? and how do I show it splits?
Thanks
since is a root of in ,
we can apply the division algorithm in , to get:
.
clearly f splits over F iff does.
since char(F) = 7 ≠ 2, we can use the quadratic formula to obtain the roots:
so x will be in F iff 4 is a square in F. and indeed, in F, both 2 and 5 (that is, -2), are square roots of 4, giving us the two roots .
(here, the "±" is just to indicate that we have (possibly) two square roots, there is no clear reason to call one "positive).
to verify, note .
iii) , which means that
char(F) means the characteristic of F, that is
1) n, if 1+1+...+1 (n times) equals 0. has characteristic 7.
2) 0, if there is no such n (like in the rationals, or the real numbers).
for iii) ONLY if f is irreducible. you can show this by establishing the linear independence of {1,a,a^2....,a^(deg(f)-1)} over the base field, where a is a root of f(x).
in post #2 i gave a factorization of
later in that same post, i showed how to factor (by explicitly solving) .
now the degree of over can be described in a couple of different ways,
i am using the definition:
by definition, the dimension of a vector space (over a given field), is the size of any basis.
in other words, i am regarding
as the vector relative to the basis .
it may not be obvious that this really is what is. the usual way to show this is to form the polynomial ring:
, and consider the quotient ring , where is the ideal generated by .
because is irreducible over , the ideal generated by is maximal, so the quotient ring has to be a field.
in this new quotient ring, we can DEFINE a to be the coset of x: .
then
it is "standard" to regard cosets of elements of as just "elements of " because the mapping:
is a ring homomorphism with 0 kernel.
all of this is just to show that we can create a field that contains (an isomorphic copy of) and a root of .
now is actually the smallest such field. however, we can define another homomorphism
by . so .
now is an element of least degree in , so every element of this kernel is some multiple
of , which is to say: . so is a field which contains and ,
and is contained in , so .
by the First Isomorphism Theorem for rings, .
but the elements of are of the form:
, and using the isomorphism ,
we can write these as .
aren't those linear combinations of our basis elements?