Polynomial Rings/Fields/Division Rings

[nhartung]

Homework Statement

Let F be a field, F[x] the ring of polynomials in one variable over F. For a $$\in$$ F[x], let (a) be all the multiples of a in F[x] (note (a) is an ideal). If b $$\in$$ F[x], let c(b) be the coset of b mod (a) (that is, the set of all b + qa, where q $$\in$$ F[x]). F[x]/(a), then is the set of all such cosets of (a).
For F = Z/2 = {0,1} our field, which of the following are fields? (note: a field is just a commutative division ring, and Z/2 already has commutative multiplication. We just need to show the following are division rings.)

a) F[x]/(x² + 1)

b) F[x]/(x² + x + 1)

c) F[x]/(x³ + x² + 1)

Homework Equations

None

The Attempt at a Solution

I think I have a) done correctly and possibly b) but I want someone to take a look at them.

First let's find all of the elements in a) {0,1,x,x+1}
x² + 1 = 0 here so this also gives us x² = 1 since 1 + 1 also = 0.
I think the best way to determine that the following are division rings is to check that they all have inverses. So i'll make a multiplication table (excluding 0 and 1 as those are obvious).
(I'm having trouble formatting a table)
For x
* x = 1
* (x+1) = x^2 + x = x + 1

For x+1
* x = x+1
* (x+1) = x^2 + 2x + 1 = 0

Since x+1 does not have in inverse this is not a field.

Now on to b)
The elements in b are {1,x,x+1,x²,x²+1}
Note: x^2 + x + 1 = 0 so x^2 + x = 1.
These are some other things that we need to know for the multiplication table that I will prove here: x^3 = 1 and x^4 = x
(x^2 + x + 1)(x + 1) = 0 * 1 = x^3 + 2x^2 + 2x + 1 = x^3 + 1 = 0. Therefore x^3 = 1 since 1+1 = 0.
We know x^2 + x = 1 so (x^2 +x)^2 = 1 = x^4 + x^2 Now we can write 1 = 1 as x^2 + x = x^4 + x^2. Therefore x^4 = x.
Now for the multiplication table I find that not only does every element have an inverse but in fact has 2 elements that take it to the identity(except for 1 and 0). I'm wondering if this is a problem..
Again excluding 1 and 0 we have
For x:
* x = x^2
* (x+1) = 1
* (x^2) = x^3 = 1
* (x^2 + 1) = x^3+x = x+1

(Just going to post the simplified answer from here on)
For x+1:
* x = 1
* (x+1) = x^2 + 1
* (x^2) = x^2 + 1
* (x^2 + 1) = 1

For x^2:
* x = 1
* (x+1) = x^2 + 1
* (x^2) = x
* (x^2 + 1) = 1

For x^2 + 1:
* x = x+1
* (x+1) = 1
* (x^2) = 1
* (x^2 + 1) = x+1

I'm really just confused as to whether a field(or even just a group) can have 2 elements that take it to identity. If it can then b) is a field, if not then it's not.
For part c) I also find that the elements, excluding 1 and 0, also have 2 inverses each.

 

[micromass]

An easier way is to check if the polynomials, which you quotient out, are irreducible. If so, the ideals will be maximal and thus the corresponding ring will be a field.

For example a) is not a field, since $$(x^2+1)=(x+1)(x+1)$$ and is thus not irreducible...

 

[nhartung]

An easier way is to check if the polynomials, which you quotient out, are irreducible. If so, the ideals will be maximal and thus the corresponding ring will be a field.

This must be some theorem we haven't learned yet. In class I remember my professor specifically saying that we need to check these by creating a multiplication table and checking for inverses. However, if what you said is true, I've determined that b and c must be fields because the polynomials which we quotient out are irreducible. If I were to do this in the way my professor suggested would the fact that I am finding multiple inverses for elements mean anything?

 

[micromass]

Hmm, no, if you find two inverses then you did something wrong. An element should only have 1 inverse (in a field at least).

Let's see what you did wrong in b), you said that the elements in b were {0,1,x,x+1,x²,x²+1}. This is already incorrect. Since x²+x+1=0, we have that x²=x+1. Thus

$$x^2=x+1~\text{and}~x^2+1=x$$

so the elements in b only are {0,1,x,x+1}.

I believe you made a similar mistake in c.

 

[nhartung]

Ah, I didn't see that. That should make life easier. Thanks

 

[nhartung]

Alright a new question here:

For F a field, define the derivative map D: F[x] --> F[x] as the unique function with the following properties:

D(a + b) = D(a) + D(b)
D(a * b) = [D(a) * b] + [a * D(b)] and
D(x) = 1.

My professor also added F must be Z/p (p a prime) and F = Q (the rational numbers).

a) Show that D(xⁿ) = nx^n-1 (n an integer) (Note: a full induction proof is not necessary. Just show the equation holds true for a general n.)

I think I have this solved pretty easily, here's my proof.

xⁿ is really just, x * x * x * ... * x, n times.
So x² can be written x * x. So D(x²) is just D(x * x). This gives us:
(1 * x) + (x * 1) = 2x which holds true to D(xⁿ) = nx^n-1. Now we have a defined D(x²).
Now try x³, which is really just x² * x. This gives us:
(2x * x) + (x² * 1) = 2x² + x² = 3x², which holds true to D(xⁿ) = nx^n-1. We could keep doing this to prove this holds for any integer n.

Now b is where I'm getting stuck.

b) Show that if a = a₀ + a₁x + a₂x² + ... + a_nxⁿ, D(a) = a₁ + 2a₂x + 3a₃x² + ... + na_nx^n-1.

Basically we need to define a coefficient derivative.
First we need a constant rule. Defining one for the integers is easy as 1 = x⁰ so D(1) = D(x⁰) = 0 * x^-1 = 0. Then we can extend it to any integer since 2 = 1 + 1 and D(2) = D(1) + D(1) = 0, 3 = 2 + 1 etc...
However, my professor said that F is the rational numbers and I'm having a tough time proving that constant rational numbers go to zero.

If I had the constant rule defined for the rational numbers then I could easily finish my coefficient derivative proof:

D(a₀) goes to 0 as it is just a constant then for all other elements like D(a₁x) = D(a₁ * x) = [D(a₁) * x] + [a₀ * 1] ...
... D(a₂x²) = D(a₂ * x²) = [D(a₂) * x²] + [a₂ * 2x]
These would be true if I could prove the constant rule..

 

[Dick]

Apply your second rule to D(1*1). What do you conclude about D(1)? Can you figure out how to extend that to show D(c)=0 for all c in F, if F=Z/p or Q? It should be easy to show D(n)=0 where n is 1+1+...+1 n times. For the rationals, now think about D(n*(1/n)).

 

[nhartung]

Aha! I think I have it now. We know D(1) = 0. We also know 1 can be written as the product of any n and it's inverse 1/n. (Every n but 0 has an inverse here since this is a field). With this we can prove that the derivative of any rational number is 0 as follows:
1 = (4 * 1/4) = D(1) = D(4 * 1/4) = (0 * 1/4) + [4 * D(1/4)] = (4 * D(1/4) = 0 which means D(1/4) must equal zero. We can show this for 2/4 and 3/4 as they are just 1/4 + 1/4 and 1/4 + 1/4 + 1/4 respectively. We can show this for any n | n ≠ 0. which is all we need.

 

[Dick]

Aha! I think I have it now. We know D(1) = 0. We also know 1 can be written as the product of any n and it's inverse 1/n. (Every n but 0 has an inverse here since this is a field). With this we can prove that the derivative of any rational number is 0 as follows:
1 = (4 * 1/4) = D(1) = D(4 * 1/4) = (0 * 1/4) + [4 * D(1/4)] = (4 * D(1/4) = 0 which means D(1/4) must equal zero. We can show this for 2/4 and 3/4 as they are just 1/4 + 1/4 and 1/4 + 1/4 + 1/4 respectively. We can show this for any n | n ≠ 0. which is all we need.

Yes, I think you have it now. You do need F=Z/p or Q to make this work, I think. In other fields you might need a separate assumption that F(c)=0 for c in F.

 

[nhartung]

Thanks a lot, you and micromass have both helped me out immensely this semester. This is my last problem set this semester so hopefully I won't be back.