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 $\displaystyle a$ is a root of $\displaystyle f(x)$ in $\displaystyle F = \mathbb{Z}_7(a)$,
we can apply the division algorithm in $\displaystyle F[x]$, to get:
$\displaystyle f(x) = (x-a)(x^2 + ax + a^2)$.
clearly f splits over F iff $\displaystyle x^2 + ax + a^2$ does.
since char(F) = 7 ≠ 2, we can use the quadratic formula to obtain the roots:
$\displaystyle x = \frac{-a \pm \sqrt{a^2 - 4a^2}}{2} = 4(-a \pm a\sqrt{4})$
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 $\displaystyle 2a, 4a$.
(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 $\displaystyle (x - 2a)(x - 4a) = x^2 - 2ax - 4ax + a^2 = x^2 + 5ax - 4ax + a^2 = x^2 + ax + a^2$.
iii) $\displaystyle F \cong \mathbb{Z}_7[x]/(f(x))$, which means that $\displaystyle [F:\mathbb{Z}_7] = ?$
char(F) means the characteristic of F, that is
1) n, if 1+1+...+1 (n times) equals 0. $\displaystyle \mathbb{Z}_7$ 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).
since $\displaystyle F = \mathbb{Z}_7(a)$ is an extension field of $\displaystyle \mathbb{Z}_7$ we can consider it a vector space over $\displaystyle \mathbb{Z}_7$.
$\displaystyle x^3 - 2$ is irreducible over $\displaystyle \mathbb{Z}_7$. so show $\displaystyle \{1,a,a^2\}$ is a basis.
(hint: why must any polynomial (in $\displaystyle \mathbb{Z}_7[x]$) of degree < deg(f) that a satisfies imply f is reducible?)
in post #2 i gave a factorization of $\displaystyle x^3-2 = (x-a)(x^2 + ax + a^2)$
later in that same post, i showed how to factor (by explicitly solving) $\displaystyle x^2 + ax + a^2 = (x - 2a)(x - 4a)$.
now the degree of $\displaystyle F$ over $\displaystyle \mathbb{Z}_7$ can be described in a couple of different ways,
i am using the definition: $\displaystyle [F:\mathbb{Z}_7] = dim_{\mathbb{Z}_7}(F)$
by definition, the dimension of a vector space (over a given field), is the size of any basis.
in other words, i am regarding $\displaystyle c_1 + c_2a + c_3a^2,\ c_1,c_2,c_3 \in \mathbb{Z}_7$
as the vector $\displaystyle (c_1,c_2,c_3) \in (\mathbb{Z}_7)^3$ relative to the basis $\displaystyle \{1, a, a^2\}$.
it may not be obvious that this really is what $\displaystyle \mathbb{Z}_7(a)$ is. the usual way to show this is to form the polynomial ring:
$\displaystyle \mathbb{Z}_7[x]$, and consider the quotient ring $\displaystyle \mathbb{Z}_7[x]/(x^3-2)$, where $\displaystyle (x^3-2)$ is the ideal generated by $\displaystyle x^3-2$.
because $\displaystyle x^3-2$ is irreducible over $\displaystyle \mathbb{Z}_7$, the ideal generated by $\displaystyle x^3-2$ 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: $\displaystyle a = x + (x^3-2)$.
then$\displaystyle a^3 = (x^3 + (x^3-2))^3 = x^3 + (x^3-2) = x^3 + 2 - 2 +(x^3-2)$
$\displaystyle = 2 + x^3-2 + (x^3-2) = 2 + (x^3-2)$
it is "standard" to regard cosets of elements of $\displaystyle \mathbb{Z}_7$ as just "elements of $\displaystyle \mathbb{Z}_7$" because the mapping:
$\displaystyle \phi :c \mapsto c+(x^3-2)$ 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) $\displaystyle \mathbb{Z}_7$ and a root of $\displaystyle x^3-2$.
now $\displaystyle \mathbb{Z}_7(a)$ is actually the smallest such field. however, we can define another homomorphism
$\displaystyle \psi:\mathbb{Z}_7[x] \rightarrow \mathbb{Z}_7(a)$ by $\displaystyle \psi(f(x)) = f(a)$. so $\displaystyle ker(\psi) = \{h(x) \in \mathbb{Z}_7[x]: h(a) = 0\}$.
now $\displaystyle x^3-2$ is an element of least degree in $\displaystyle ker(\psi)$, so every element of this kernel is some multiple
of $\displaystyle x^3-2$, which is to say: $\displaystyle ker(\psi) = (x^3-2)$. so $\displaystyle im(\psi)$ is a field which contains $\displaystyle \mathbb{Z}_7$ and $\displaystyle a$,
and is contained in $\displaystyle \mathbb{Z}_7(a)$, so $\displaystyle im(\psi) = \mathbb{Z}_7(a)$.
by the First Isomorphism Theorem for rings, $\displaystyle \mathbb{Z}_7[x]/(x^3-2) = \mathbb{Z}_7[x]/ker(\psi) \cong \mathbb{Z}_7(a)$.
but the elements of $\displaystyle \mathbb{Z}_7[x]/(x^3-2)$ are of the form:
$\displaystyle c_1 + c_2x + c_3x^2 + (x^3-2)$, and using the isomorphism $\displaystyle \psi$,
we can write these as $\displaystyle c_1 + c_2a + c_3a^2$.
aren't those linear combinations of our basis elements?