Hallo,
Use [tex] [/tex] tags
And be careful with { } :
Type \{0,1\}{0,1}
Find so that
Solution from lecturer:
Therefore
I'm having real trouble understanding this solution...
I'm guessing this bit comes from multiplying out and then removing any repeated elements? But why is this then equal to ?
Please could someone please explain this simply?
Thanks in advance!
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 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...
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 of characteristic 2, then
If are rational numbers find rationals numbers such that:
Using the result in the previous posts
.
.
.
So we put:
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...?
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 be a root of g and . Find a basis of E considered as a vector space over Q.
This is just 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.
I dont know which g you are talking about...
If you are talking about the original , 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 , then is the basis.
Note: Naturally . This shows a non-trivial linear combination is 0. However if is a root of any irreducible polynomial with degree less that m, then it contradicts the minimality of the degree of g.
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 is a root then is also a root