non-zero elements in E

• May 5th 2008, 05:53 AM
hunkydory19
non-zero elements in E
$\displaystyle g = X^4 + X + 1$ over F2 and let E be the extension of F2 with a root alpha of g.

Show that every non-zero element of E is a power of $\displaystyle \alpha$.
I can do this question which is worth 8 marks...but cannot do:
Show that if $\displaystyle \beta$ is an element of E - F2 then every non-zero element of E is a power of $\displaystyle \beta$.

Which is worth 4 marks...I'm guessing this answer is much easier since it is only 4 marks, so there is a way I can adapt my answer to the former question for this one? Or is it a totally different method?

• May 5th 2008, 06:06 AM
topsquark
Quote:

Originally Posted by hunkydory19
$\displaystyle g = X^4 + X + 1$ over F2 and let E be the extension of F2 with a root alpha of g.

Show that every non-zero element of E is a power of $\displaystyle \alpha$.
I can do this question which is worth 8 marks...but cannot do:
Show that if $\displaystyle \beta$ is an element of E - F2 then every non-zero element of E is a power of $\displaystyle \beta$.

Which is worth 4 marks...I'm guessing this answer is much easier since it is only 4 marks, so there is a way I can adapt my answer to the former question for this one? Or is it a totally different method?

This is a question about your question. Simple curiosity. What is the definition of the field (I presume) F2?

-Dan
• May 5th 2008, 06:13 AM
Isomorphism
Quote:

Originally Posted by hunkydory19
Show that if $\displaystyle \beta$ is an element of E - F2 then every non-zero element of E is a power of $\displaystyle \beta$.

Which is worth 4 marks...I'm guessing this answer is much easier since it is only 4 marks, so there is a way I can adapt my answer to the former question for this one? Or is it a totally different method?

I wonder whether I have understood this correctly.

$\displaystyle E - F_2$ has all the elements of E except 0 and 1. Consider $\displaystyle \beta = \alpha^3 \in E- F_2$, We can never write $\displaystyle \alpha$ as a power of this $\displaystyle \beta$. So the statement is not true.

Either I have misunderstood the question or you have not copied it correctly(Worried)
• May 5th 2008, 07:09 AM
ThePerfectHacker
Quote:

Originally Posted by hunkydory19
$\displaystyle g = X^4 + X + 1$ over F2 and let E be the extension of F2 with a root alpha of g.

Show that every non-zero element of E is a power of $\displaystyle \alpha$.

If $\displaystyle E$ is an extension field containing $\displaystyle \alpha$ over $\displaystyle \mathbb{F}_2$ then $\displaystyle E$ contains $\displaystyle \mathbb{F}_2$.
It means if $\displaystyle x\in \mathbb{F}_2(\alpha)$ then $\displaystyle x=a+b\alpha+c\alpha^2+d\alpha^3$ where $\displaystyle a,b,c\in \mathbb{F}_2$.
So maybe you meant to say $\displaystyle E=\mathbb{F}_2$.
• May 5th 2008, 07:17 AM
Isomorphism
I would like to give a detailed explanation on why the question is wrong.

As a side note, I would like to add that if $\displaystyle g(x) = x^3 + x + 1$, the what you said would have been true. Since then $\displaystyle \alpha^7 = 1$ and 7 is prime.

Here(that is to say $\displaystyle g(x) = x^4 + x + 1$), we have 16 elements and $\displaystyle \alpha^{15} = 1$. Observe that this means $\displaystyle (\alpha^3)^5 = 1$.

Now your questions says if $\displaystyle \beta \in E - \mathbb{F}_2$, then $\displaystyle \forall x \in E - \mathbb{F}_2, \exists i \in \mathbb{Z}: x = \beta^i$

I claim this is wrong by a counterexample:

Consider $\displaystyle \beta = \alpha^3$, now:

$\displaystyle \beta = \alpha^3 , \beta^2 = \alpha^6, \beta^3 = \alpha^9,\beta^4 = \alpha^{12},\beta^5 = \alpha^{15} = 1$

So clearly in this entire cycle $\displaystyle \alpha$ never appears. And these are exactly all the powers of $\displaystyle \beta$.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~
Finally, I think that the question really asked you to disprove their statement.Or g(x) was different.

Hope you understand why it is wrong (Wink)
• May 5th 2008, 07:33 AM
hunkydory19
I've attached the question to prove it's definitely minus! It comes up every year with either $\displaystyle g = X^3 + X + 1$ or $\displaystyle g = X^4 + X + 1$.

Think I'll show my lecturer your explanation Iso and see what he says!