1. ## algebra, field

Sorry for all the algebra questions everybody, I have my exam on Wednesday (which I'm going to fail) and seem to be struggling with everything as it's so abstract...

Let $g = X^3 +2X - 1$
Let F be the smallest subfield of R containing a root $\alpha$ of g. What are the elements of F?

Would this just be the identity element 1?

What is the dimension of F considered as a vector space over the rationals?

This I just do not understand at all....can anyone please explain?

2. Hallo,

(which I'm going to fail)
Never say this before you write the exam

Would this just be the identity element 1?
1 is not a root, but -1 is. I don't know if it helps...
I don't get the thing with the subspace F... sorry

3. Ok cheers for the relpy Moo, think I'll just pray that those questions don't come up!

I've got another question which seems to come up every year, seems a little easier but I still don't see what to do?!

Write down the inverse of $\alpha^2 - 2\alpha$ in the form $\lambda \alpha^2 + \mu\alpha + v$ where $\lambda, \mu, v$ are all rational numbers.

Can anyone please explain the method for this?

Thanks again.

4. Originally Posted by hunkydory19
Sorry for all the algebra questions everybody, I have my exam on Wednesday (which I'm going to fail) and seem to be struggling with everything as it's so abstract...

Let $g = X^3 +2X - 1$
Let F be the smallest subfield of R containing a root $\alpha$ of g. What are the elements of F?

Would this just be the identity element 1?
Note that any field should contain 0 and 1. Here it should also contain $\alpha$ according to the posed question. Now wonder what happens $\alpha^2, \alpha^3,...$. Since closure would force them into the field. The real thing you should observe here is that :

$g(\alpha) = 0 \Rightarrow \alpha^3 + 2\alpha - 1 = 0 \Rightarrow \alpha^3 = - 2\alpha + 1$-----------------------------------(*)
This means every power of $\alpha$ in this field can be brought down to the form $\lambda \alpha^2 + \mu\alpha + v$

For example:
Write down the inverse of in the form where are all rational numbers.

Can anyone please explain the method for this?
$
(\alpha^2 - 2\alpha)^{-1} = \lambda \alpha^2 + \mu\alpha + v \Rightarrow (\lambda \alpha^2 + \mu\alpha + v)\cdot(\alpha^2 - 2\alpha) = 1$

$(\lambda \alpha^2 + \mu\alpha + v)\cdot(\alpha^2 - 2\alpha) = \lambda \alpha^4 - (2\lambda- \mu)\alpha^3 - (2\mu+v)\alpha^2 - 2v\alpha = 1$

Now use the property of $\alpha$ we just proved in (*)

$\lambda (-2\alpha^2 + \alpha) - (2\lambda- \mu)(-2\alpha+1) - (2\mu+v)\alpha^2 - 2v\alpha - 1 = 0$

$(-2\lambda-2\mu-v)\alpha^2 + (5\lambda- 2\mu -v)\alpha -2\lambda +\mu - 1= 0$

So solve this system of equations:

$-2\lambda-2\mu-v = 0$

$5\lambda- 2\mu -v = 0$

$-2\lambda +\mu-1 = 0$

Unless I have been careless in my arithmetic you should get the right answer
Originally Posted by hunkydory19
What is the dimension of F considered as a vector space over the rationals?
Clearly it is 3. Since $\{1,\alpha,\alpha^2 \}$ is the basis.

5. Yo Iso,

Why should F contain 0 and 1 ? Because they're null elements ?

6. Originally Posted by Moo
Yo Iso,

Why should F contain 0 and 1 ? Because they're null elements ?
Every Field should have additive and multiplicative identities since a field is both an additive and a multiplicative group.

7. Oh, I misunderstood "field" then -__-

8. Thank you so much Iso, you have literally just saved my life!

9. OK I've just worked through this example, and understood everything...EXCEPT where the $-2\alpha + 1$ comes from? Obviously I see that it comes from rearranging, but what I mean is do you always have to know what g is before you can find this alpha equation?

Also there were a few signs wrong in your working, but got the answers to be $\lambda = -\frac{6}{11}, \mu = -\frac{1}{11} v = -\frac{14}{11}$

Is there a way to check these answers correct after doing this method?

Thanks again

10. Originally Posted by hunkydory19
OK I've just worked through this example, and understood everything...EXCEPT where the $-2\alpha + 1$ comes from? Obviously I see that it comes from rearranging, but what I mean is do you always have to know what g is before you can find this alpha equation?
Yes! The field generated is due to a particular polynomial. Its called the generating polynomial of the field.

Originally Posted by hunkydory19
Also there were a few signs wrong in your working, but got the answers to be [tex] \lambda = -\frac{6}{11}, \mu = -\frac{1}{11} and v = -\frac{14}{11} [tex]
Sorry for that... I did not work it out on paper

Originally Posted by hunkydory19
Is there a way to check these answers correct after doing this method?

Thanks again
Yes multiply out the obtained answer with $\alpha^2 - 2\alpha$ and check if you get 1

11. Iso do not dare apologise to me, haha! Would be failing this exam if it wasn't for you! Was just pointing it out incase you got to different answers so thought it would save time. Thank you SO much for all your help!

12. I've just tried your tip for checking if $(\alpha^2 - 2\alpha)(-\frac{6}{11}\alpha^2 -\frac{1}{11}\alpha -\frac{14}{11})$ multiply out to give 1 and they don't, nowhere near since none of the alphas cancel! What am I doing wrong?

13. Originally Posted by hunkydory19
I've just tried your tip for checking if $(\alpha^2 - 2\alpha)(-\frac{6}{11}\alpha^2 -\frac{1}{11}\alpha -\frac{14}{11})$ multiply out to give 1 and they don't, nowhere near since none of the alphas cancel! What am I doing wrong?
Well I get 1 perfectly...

Let $b =-\frac1{11}$

$(\alpha^2 - 2\alpha)(6b\alpha^2 + b\alpha + 14b)$

$= 6b\alpha^4+(b - 12b)\alpha^3 + (14b - 2b)\alpha^2 - 28b\alpha$

$= 6b(-2\alpha^2 + \alpha) -11b(-2\alpha + 1) + 12b\alpha^2 - 28b\alpha$

$= -12b \alpha^2 + 6b\alpha + 22b\alpha -11b + 12b\alpha^2 - 28b\alpha$

$= (-12b \alpha^2 + 12b\alpha^2)+(6b\alpha + 22b\alpha - 28b\alpha) - 11b$

$= 0+0 - 11b = -11b = 1$

Hence proved

14. Thank you so much Iso!! Absolute legend.