Thread: Splitting over an extension field

1. Splitting over an extension field

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

2. Re: Splitting over an extension field

since $a$ is a root of $f(x)$ in $F = \mathbb{Z}_7(a)$,

we can apply the division algorithm in $F[x]$, to get:

$f(x) = (x-a)(x^2 + ax + a^2)$.

clearly f splits over F iff $x^2 + ax + a^2$ does.

since char(F) = 7 ≠ 2, we can use the quadratic formula to obtain the roots:

$x = \frac{-a \pm \sqrt{a^2 - 4a^2}}{2} = 4(-a \pm a\sqrt{4})$

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 $2a, 4a$.

(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 $(x - 2a)(x - 4a) = x^2 - 2ax - 4ax + a^2 = x^2 + 5ax - 4ax + a^2 = x^2 + ax + a^2$.

iii) $F \cong \mathbb{Z}_7[x]/(f(x))$, which means that $[F:\mathbb{Z}_7] = ?$

3. Re: Splitting over an extension field

A couple of questions, what is the notation char(F)?

Also in part iii) are you saying that the degree of F over Z7 is simply the degree of the polynomial f(x)??

thanks

4. Re: Splitting over an extension field

char(F) means the characteristic of F, that is

1) n, if 1+1+...+1 (n times) equals 0. $\mathbb{Z}_7$ 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).

5. Re: Splitting over an extension field

awesome thanks! I need to do this question not in the general case but for the polynomial f(x)=x^3-2, though I follow what you're saying (just!) Im not sure how to factor x^3-2 over Z7(a)

6. Re: Splitting over an extension field

since $F = \mathbb{Z}_7(a)$ is an extension field of $\mathbb{Z}_7$ we can consider it a vector space over $\mathbb{Z}_7$.

$x^3 - 2$ is irreducible over $\mathbb{Z}_7$. so show $\{1,a,a^2\}$ is a basis.

(hint: why must any polynomial (in $\mathbb{Z}_7[x]$) of degree < deg(f) that a satisfies imply f is reducible?)

7. Re: Splitting over an extension field

ahhh you lost me!

8. Re: Splitting over an extension field

in post #2 i gave a factorization of $x^3-2 = (x-a)(x^2 + ax + a^2)$

later in that same post, i showed how to factor (by explicitly solving) $x^2 + ax + a^2 = (x - 2a)(x - 4a)$.

now the degree of $F$ over $\mathbb{Z}_7$ can be described in a couple of different ways,

i am using the definition: $[F:\mathbb{Z}_7] = dim_{\mathbb{Z}_7}(F)$

by definition, the dimension of a vector space (over a given field), is the size of any basis.

in other words, i am regarding $c_1 + c_2a + c_3a^2,\ c_1,c_2,c_3 \in \mathbb{Z}_7$

as the vector $(c_1,c_2,c_3) \in (\mathbb{Z}_7)^3$ relative to the basis $\{1, a, a^2\}$.

it may not be obvious that this really is what $\mathbb{Z}_7(a)$ is. the usual way to show this is to form the polynomial ring:

$\mathbb{Z}_7[x]$, and consider the quotient ring $\mathbb{Z}_7[x]/(x^3-2)$, where $(x^3-2)$ is the ideal generated by $x^3-2$.

because $x^3-2$ is irreducible over $\mathbb{Z}_7$, the ideal generated by $x^3-2$ 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: $a = x + (x^3-2)$.

then $a^3 = (x^3 + (x^3-2))^3 = x^3 + (x^3-2) = x^3 + 2 - 2 +(x^3-2)$

$= 2 + x^3-2 + (x^3-2) = 2 + (x^3-2)$

it is "standard" to regard cosets of elements of $\mathbb{Z}_7$ as just "elements of $\mathbb{Z}_7$" because the mapping:

$\phi :c \mapsto c+(x^3-2)$ 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) $\mathbb{Z}_7$ and a root of $x^3-2$.

now $\mathbb{Z}_7(a)$ is actually the smallest such field. however, we can define another homomorphism

$\psi:\mathbb{Z}_7[x] \rightarrow \mathbb{Z}_7(a)$ by $\psi(f(x)) = f(a)$. so $ker(\psi) = \{h(x) \in \mathbb{Z}_7[x]: h(a) = 0\}$.

now $x^3-2$ is an element of least degree in $ker(\psi)$, so every element of this kernel is some multiple

of $x^3-2$, which is to say: $ker(\psi) = (x^3-2)$. so $im(\psi)$ is a field which contains $\mathbb{Z}_7$ and $a$,

and is contained in $\mathbb{Z}_7(a)$, so $im(\psi) = \mathbb{Z}_7(a)$.

by the First Isomorphism Theorem for rings, $\mathbb{Z}_7[x]/(x^3-2) = \mathbb{Z}_7[x]/ker(\psi) \cong \mathbb{Z}_7(a)$.

but the elements of $\mathbb{Z}_7[x]/(x^3-2)$ are of the form:

$c_1 + c_2x + c_3x^2 + (x^3-2)$, and using the isomorphism $\psi$,

we can write these as $c_1 + c_2a + c_3a^2$.

aren't those linear combinations of our basis elements?