# Thread: algebra, a root of polynomial

1. ## algebra, a root of polynomial

Find $\lambda, \mu, v \in \{0,1\}
$
so that $(\alpha^2 + \alpha + 1)(\alpha^2 + 1) = \lambda\alpha^2 + \mu\alpha + v$

Solution from lecturer:

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

Therefore $\lambda = \mu = 1, v = 0$

I'm having real trouble understanding this solution...

I'm guessing this bit $\alpha^4 + \alpha^3 + \alpha+ 1$ comes from multiplying out and then removing any repeated elements? But why is this then equal to $\alpha^4 = \alpha^2 + \alpha$?

2. Hallo,

Use  tags

And be careful with { } :

{0,1}
Type \{0,1\}

3. Cheers for that Moo, got a bit mixed up there!

4. Originally Posted by hunkydory19
Find $\lambda, \mu, v \in \{0,1\}
$
so that $(\alpha^2 + \alpha + 1)(\alpha^2 + 1) = \lambda\alpha^2 + \mu\alpha + v$

Solution from lecturer:

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

Therefore $\lambda = \mu = 1, v = 0$

I'm having real trouble understanding this solution...

I'm guessing this bit $\alpha^4 + \alpha^3 + \alpha+ 1$ comes from multiplying out and then removing any repeated elements? But why is this then equal to $\alpha^4 = \alpha^2 + \alpha$?

You are talking of the field with generator polynomial $f(x) = x^3 + x+ 1$

If $\alpha$ is a root(the primitive element), then $\alpha^3 + \alpha+ 1 = 0$

Note that in $\mathbb{F}_2, -1 = 1$, since the characteristic is 2.

This means $\alpha^3 = - \alpha - 1 = \alpha + 1$

This also means $\alpha^4 = \alpha\cdot \alpha^3 = \alpha(\alpha + 1) = \alpha^2 + \alpha$

5. Arghhhh it all makes sense now! Thank you once again!

Just one more thing, as I am not familar with the signs of 1 and -1 being interchangable in F2, so was just wondering if this is a special case or does it happen in other fields as well e.g F3?

6. Originally Posted by hunkydory19
Arghhhh it all makes sense now! Thank you once again!

Just one more thing, as I am not familar with the signs of 1 and -1 being interchangable in F2, so was just wondering if this is a special case or does it happen in other fields as well e.g F3?
Actually, F2 means that you work with "equivalence classes" in Z/2Z. That is to say that any number will be written modulo 2.
-1 mod 2 = -1+2 mod 2 = 1 mod 2

In F3, it'll be mod 3.
-1 mod 3 = -1+3 mod 3 = 2 mod 3.

It's the same for the polynomial. You're working on a space called $\mathbb{R}[X]/(x^3+x+1)$ and you're working on polynomials "modulo f(x)", that is to say that you will always take the remainder in the division by f(x).

I hope it helps you understand...

7. Originally Posted by hunkydory19
Just one more thing, as I am not familar with the signs of 1 and -1 being interchangable in F2, so was just wondering if this is a special case or does it happen in other fields as well e.g F3?
The "Interchangability" comes from the fact that all our operations are modulo 2.

"-1" is the symbolic representation of additive inverse of 1. In modulo 2, the inverse of 1 is 1 itself. Thus -1 = 1.
In F3, -1 = 2, -2 = 1

Note: Dont worry about what I am about to say, but remember it.
Let $x \in \mathbb{F}$ of characteristic 2, then $-x = x$

8. Originally Posted by Isomorphism
Note: Dont worry about what I am about to say, but remember it.
Let $x \in \mathbb{F}$ of characteristic 2, then $-x = x$
This comes up while writing -x=(-1)x, and we're taken to the first case

9. If $\lambda, \mu, v, \lambda', \mu', v'$ are rational numbers find rationals numbers $\lambda'', \mu'', v''$such that:

$(\lambda\alpha^2 + \mu\alpha + v)(\lambda'\alpha^2 + \mu'\alpha + v') = \lambda''\alpha^2 + \mu''\alpha + v''$

$(\lambda\alpha^2 + \mu\alpha + v)(\lambda'\alpha^2 + \mu'\alpha + v') = \lambda \lambda'\alpha^4 + (\mu\lambda' + \lambda\mu')\alpha^3 + (v\lambda' + \mu\mu' + \lambda v')\alpha^2$ $+ (v \mu' + \mu v')\alpha + vv'$

Using the result in the previous posts $\alpha^4 = \alpha^2 + \alpha$
.
.
.

So we put:

$\lambda'' = \lambda\lambda' + \mu\mu' + \lambda v$

$\mu'' = \lambda\lambda' + \mu\lambda' + \lambda\mu' +v\mu' + \mu v'$

$v'' = \mu\lambda' + \lambda\mu' + vv'$

Is this the correct method for this question...I'm not sure since the result I used was in {0,1} whereas the question did not specify this...?

10. Originally Posted by hunkydory19
If $\lambda, \mu, v, \lambda', \mu', v'$ are rational numbers find rationals numbers $\lambda'', \mu'', v''$such that:

$(\lambda\alpha^2 + \mu\alpha + v)(\lambda'\alpha^2 + \mu'\alpha + v') = \lambda''\alpha^2 + \mu''\alpha + v''$

$(\lambda\alpha^2 + \mu\alpha + v)(\lambda'\alpha^2 + \mu'\alpha + v') = \lambda \lambda'\alpha^4 + (\mu\lambda' + \lambda\mu')\alpha^3 + (v\lambda' + \mu\mu' + \lambda v')\alpha^2$ $+ (v \mu' + \mu v')\alpha + vv'$

Using the result in the previous posts $\alpha^4 = \alpha^2 + \alpha$
.
.
.

So we put:

$\lambda'' = \lambda\lambda' + \mu\mu' + \lambda v$

$\mu'' = \lambda\lambda' + \mu\lambda' + \lambda\mu' +v\mu' + \mu v'$

$v'' = \mu\lambda' + \lambda\mu' + vv'$

Is this the correct method for this question...I'm not sure since the result I used was in {0,1} whereas the question did not specify this...?
Yes...but you should understand why it happens:

If you have $\lambda\alpha^2 + \mu\alpha + v = 0$ and $\{1,\alpha,\alpha^2\}$ is a basis. Then any linear combination that is a 0, implies that all the coefficients are zero.

11. Thanks Iso...2 more bits I'm unsure about and then I'll be feeling a lot more confident over this section of the exam!

Let $\alpha \in R$ be a root of g and $E = Q(\alpha)$. Find a basis of E considered as a vector space over Q.

This is just ${1, \alpha, \alpha^2}$ right?

Then lastly whats the process for finding out whether g has more than one root in E?

Cheers again, really really appreciate all this help.

12. Originally Posted by hunkydory19
Let $\alpha \in R$ be a root of g and $E = Q(\alpha)$. Find a basis of E considered as a vector space over Q.

This is just ${1, \alpha, \alpha^2}$ right?
I dont know which g you are talking about...

If you are talking about the original $x^3 + x + 1 = 0$, then you are right.

I will just tell you the general theory.There are a lot of constraints here.

* What I am saying holds only if g is irreducible in Q.

*It depends on the degree of the polynomial.If the degree of the polynomial is 'm' and the polynomial is $g(x) = g_0 + g_1x + g_2 x^2 + g_3 x^3 +.... +g_m x^m$, then $\{1,\alpha,\alpha^2,.....\alpha^{m-1}\}$ is the basis.

Note: Naturally $g(\alpha) = 0 \Rightarrow g_0 + g_1\alpha + g_2 \alpha^2 + g_3 \alpha^3 +.... +g_m\alpha^m = 0$. This shows a non-trivial linear combination is 0. However if $\alpha$ is a root of any irreducible polynomial with degree less that m, then it contradicts the minimality of the degree of g.

Originally Posted by hunkydory19
Then lastly whats the process for finding out whether g has more than one root in E?
This could be tricky. I know what to do in finite fields.
However since you are talking about Q, I dont think that this is an easy question to answer. If there was really any method to generally get other roots given one root, we wouldnt have had Abels' impossibility theorem.

In finite fields, if the characteristic is q, and $\alpha$ is a root then $\alpha^q$ is also a root