1. ## elements in extension

g = X^3 + X + 1

Let E be the extension of F2 with a root alpha of g. Write down a complete list of elements of E without repetitions.

I've tried by best at this question, but I still don't feel that I fully understand what I'm doing, (which will probably be proved with it being wrong...)

We know is a route of g, so it follows that is also a root of g. We can apply this repeatedly to find roots and so on.

Hence
Hence

Hence the roots are

Anyone know if this is OK?

2. Hello,

Hm, I don't think it is what you've been asked.

If there is a root alpha of g, then any $\displaystyle \alpha^3$ can be transformed in $\displaystyle \alpha+1$.

This means that any element of E won't contain any $\displaystyle \alpha^3$ (or highest powers).

Actually, the elements of E will be all polynomials of degree < 3, which is the degree of g(x).
So it's 0, 1, x, x+1, x², x²+x, x²+1, x²+x+1.

The transition x <-> $\displaystyle \alpha$ can be done because $\displaystyle \alpha$ is an equivalent class of x (sort of).

I don't know if it is clear

As I told you in another thread, this is like working in R[X]/(g(x)), which is actually the set of all possible remainders of the division of any polynomial by g(x)...

Perhaps Isomorphism will be able to explain it better

3. Thank you Moo Sorry to make you repeat yourself I'm just really really bad at understanding this stuff...I'm much more of a calculus person

Ok the the next question on this exam paper is:

Find the other roots of g and the roots of h in E.

This is the question to what my answer before was to right?! so with my above answer for g and answer below for h is this ok?

The roots of h are: $\displaystyle \alpha, \alpha^2, \alpha + 1, \alpha^2 + 1$ and $\displaystyle \alpha^2 + \alpha$

4. Originally Posted by hunkydory19
g = X^3 + X + 1
Let E be the extension of F2 with a root alpha of g. Write down a complete list of elements of E without repetitions.

$\displaystyle \alpha^4 = \alpha \cdot \alpha^3 = \alpha^2 + \alpha$

$\displaystyle \alpha^5 = \alpha \cdot \alpha^4 = \alpha^3 + \alpha^2 = (1+\alpha)+\alpha^2$

$\displaystyle \alpha^6 = \alpha \cdot \alpha^5 = \alpha+\alpha^2+\alpha^3 = \alpha+\alpha^2+\alpha + 1 = 1 + \alpha^2$

And so on....

5. Originally Posted by hunkydory19
I've tried by best at this question, but I still don't feel that I fully understand what I'm doing, (which will probably be proved with it being wrong...)

We know is a route of g, so it follows that is also a root of g.
You are right up to here.

Now if $\displaystyle \alpha$ is a root, then $\displaystyle \alpha^2,\alpha^4,\alpha^8$.... are all roots of g. Why should $\displaystyle \alpha^3$ be a root?

Originally Posted by hunkydory19
Thank you Moo Sorry to make you repeat yourself I'm just really really bad at understanding this stuff...I'm much more of a calculus person

Ok the the next question on this exam paper is:

Find the other roots of g and the roots of h in E.

This is the question to what my answer before was to right?! so with my above answer for g and answer below for h is this ok?

The roots of h are: $\displaystyle \alpha, \alpha^2, \alpha + 1, \alpha^2 + 1$ and $\displaystyle \alpha^2 + \alpha$

What is h?

6. Thanks Iso, right I've had another go...

So sorry for missing out h by the way, so stupid!

h = $\displaystyle X^3 + X^2 + 1$

The roots of g are: $\displaystyle \alpha, \alpha^2, \alpha + 1, \alpha^2 + \alpha$

The roots of h are: $\displaystyle \alpha, \alpha^2, \alpha + 1, \alpha^2 + \alpha, \alpha^2 + \alpha + 1$

This any better? Please say yes!

7. Originally Posted by hunkydory19
Thanks Iso, right I've had another go...

So sorry for missing out h by the way, so stupid!

h = $\displaystyle X^3 + X^2 + 1$

The roots of g are: $\displaystyle \alpha, \alpha^2, \alpha + 1, \alpha^2 + \alpha$

The roots of h are: $\displaystyle \alpha, \alpha^2, \alpha + 1, \alpha^2 + \alpha, \alpha^2 + \alpha + 1$

This any better? Please say yes!
Unfortunately no...

The roots of g are: $\displaystyle \alpha, \alpha^2, \alpha^2 + \alpha$
Why did you add $\displaystyle \alpha^3$?

If the elements are from $\displaystyle \mathbb{F}_q$ with characteristic 'p', then the roots are $\displaystyle \{\alpha, \alpha^p, \alpha^{p^2},\alpha^{p^3}....\}$.

Since the field we are considering is $\displaystyle \mathbb{F}_{2^3}$ and it has a characteristic 2, the roots of g are $\displaystyle \{\alpha, \alpha^2, \alpha^{2^2},\alpha^{2^3}....\}$. But observe that since the field has 8 elements, $\displaystyle \alpha^8 = \alpha$(Why?). So the roots start repeating and thus we truly have only three roots.
They are $\displaystyle \{\alpha, \alpha^2, \alpha^{4}\}$

For h, find one root in $\displaystyle \mathbb{F}_{2^3}$ using the previous exercise of constructing roots. Lets call it $\displaystyle \beta$. Then the remaining roots are $\displaystyle \beta^2,\beta^4$

Understood?

P.S: A small hint to see if you are going wrong: A cubic can have only 3 roots and you are getting four roots

8. So for h, $\displaystyle \beta$ is a root so it follows that $\displaystyle \beta^2$ is also a root.

Using $\displaystyle \beta^3 = \beta^2 + 1$

$\displaystyle \beta^4 = \beta.\beta^3 = \beta(\beta^2 + 1) = \beta^3 + \beta = \beta^2 +\beta + 1$

$\displaystyle \beta^8 = (\beta^2 + \beta + 1)^2 = \beta^4 + 1 = \beta(\beta^2 + 1) + 1 = \beta^3 + \beta + 1 = \beta^2 + \beta$

Why am I getting this fourth root? I actually think I will die doing this

9. Theorem: If $\displaystyle p(x)$ is irreducible $\displaystyle n$-degree polynomial over $\displaystyle F$ and $\displaystyle \alpha$ is a zero in an extension field then $\displaystyle F(\alpha) = \{ a_0+a_1\alpha+...+a_{n-1}\alpha^{n-1} \}$.

This means given $\displaystyle F=\mathbb{F}_2$ and $\displaystyle \alpha$ solving the irreducible polynomial $\displaystyle x^3+x+1$ it means $\displaystyle \mathbb{F}_2(\alpha) = \{ a+b\alpha+c\alpha^2+d\alpha^3\}$ where $\displaystyle a,b,c,d\in \mathbb{F}_2$.

So can you now list all the elements?

10. Originally Posted by hunkydory19
So for h, $\displaystyle \beta$ is a root so it follows that $\displaystyle \beta^2$ is also a root.

Using $\displaystyle \beta^3 = \beta^2 + 1$

$\displaystyle \beta^4 = \beta.\beta^3 = \beta(\beta^2 + 1) = \beta^3 + \beta = \beta^2 +\beta + 1$

$\displaystyle \beta^8 = (\beta^2 + \beta + 1)^2 = \beta^4 + 1 = \beta(\beta^2 + 1) + 1 = \beta^3 + \beta + 1 = \beta^2 + \beta$

Why am I getting this fourth root? I actually think I will die doing this
Ok lets go over this carefully

$\displaystyle g(x) = x^3 + x + 1$ and $\displaystyle g(\alpha) = 0$

The elements of $\displaystyle \mathbb{F}_{2^3}$ are:
$\displaystyle 0, \alpha,\alpha^2,\alpha^3(= \alpha + 1),\alpha^4(= \alpha^2 + \alpha),\alpha^5(= \alpha^2 + \alpha + 1),\alpha^6(= \alpha^2 + 1),\alpha^7(= 1)$
So,
$\displaystyle \mathbb{F}_{2^3} = \{0,1,\alpha,\alpha^2,\alpha + 1,\alpha^2 + \alpha ,\alpha^2 + \alpha + 1,\alpha^2 + 1\}$

Now consider $\displaystyle h(x) = x^3 + x^2 + 1$.
Dont try a new $\displaystyle \beta$, but try one of the above elements. We clearly see that $\displaystyle h(\alpha^3) = (\alpha^3)^3 + (\alpha^3)^2 + 1 = \alpha^2 + (\alpha^2 + 1) + 1 = 0$

So $\displaystyle \alpha^3$ is a root of h(x). But this means $\displaystyle (\alpha^3)^2$ and $\displaystyle (\alpha^3)^4 = \alpha^{12} = \alpha^5$ are also roots of h(x).

Thus the $\displaystyle \alpha^3, \alpha^6 ,\alpha^5$ are the roots of $\displaystyle h(x)$

P.S:There ...I actually solved this problem completely for you

11. Well ring a ding dong its finally hit home and I get it now! 200 years later!

I didn't know you could actually stick elements in to see if they gave 0...

Thank you so much for being SO patient with me, I know I wouldn't have been if it was the other way round!

12. Originally Posted by hunkydory19
Well ring a ding dong its finally hit home and I get it now! 200 years later!

I didn't know you could actually stick elements in to see if they gave 0...

Thank you so much for being SO patient with me, I know I wouldn't have been if it was the other way round!
There is a trick to it

Since $\displaystyle \alpha, \alpha^2, \alpha^4$ were already roots of g(x), I figured h should have the others and tried $\displaystyle \alpha^3$