# elements in extension

• May 5th 2008, 02:09 AM
hunkydory19
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 http://thestudentroom.co.uk/latexren...2c28b24630.png is a route of g, so it follows that http://thestudentroom.co.uk/latexren...7777bc68a8.png is also a root of g. We can apply this repeatedly to find roots http://thestudentroom.co.uk/latexren...0ed4554cd3.png and so on.

http://thestudentroom.co.uk/latexren...c9c26af62b.png
Hence http://thestudentroom.co.uk/latexren...e623e86387.png
Hence http://thestudentroom.co.uk/latexren...a4218267bb.png

Hence the roots are http://thestudentroom.co.uk/latexren...7ae20cf77b.png

Anyone know if this is OK?

• May 5th 2008, 02:34 AM
Moo
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 (Thinking)

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 :D
• May 5th 2008, 04:23 AM
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 (Worried)

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 $\displaystyle \alpha^2 + \alpha$
• May 5th 2008, 05:26 AM
Isomorphism
Quote:

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....
• May 5th 2008, 05:32 AM
Isomorphism
Quote:

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 http://thestudentroom.co.uk/latexren...2c28b24630.png is a route of g, so it follows that http://thestudentroom.co.uk/latexren...7777bc68a8.png 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?

Quote:

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 (Worried)

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? (Thinking)
• May 5th 2008, 05:50 AM
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!
• May 5th 2008, 06:05 AM
Isomorphism
Quote:

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... (Worried)

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 (Giggle)
• May 5th 2008, 07:05 AM
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 (Angry)
• May 5th 2008, 07:14 AM
ThePerfectHacker
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?
• May 5th 2008, 07:33 AM
Isomorphism
Quote:

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 (Angry)

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 (Whew)...I actually solved this problem completely for you :D
• May 5th 2008, 07:43 AM
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! :)
• May 5th 2008, 07:52 AM
Isomorphism
Quote:

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 :D

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