Polynomial Rings/Fields/Division Rings

  • Thread starter nhartung
  • Start date
  • #1
56
0

Homework Statement


Let F be a field, F[x] the ring of polynomials in one variable over F. For a [tex]\in[/tex] F[x], let (a) be all the multiples of a in F[x] (note (a) is an ideal). If b [tex]\in[/tex] F[x], let c(b) be the coset of b mod (a) (that is, the set of all b + qa, where q [tex]\in[/tex] 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.
 

Answers and Replies

  • #2
22,089
3,297
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 [tex](x^2+1)=(x+1)(x+1)[/tex] and is thus not irreducible...
 
  • #3
56
0
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?
 
  • #4
22,089
3,297
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

[tex]x^2=x+1~\text{and}~x^2+1=x[/tex]

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

I believe you made a similar mistake in c.
 
  • #5
56
0
Ah, I didn't see that. That should make life easier. Thanks
 
  • #6
56
0
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) = nxn-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) = nxn-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) = nxn-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 + ... + anxn, D(a) = a1 + 2a2x + 3a3x2 + ... + nanxn-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..
 
  • #7
Dick
Science Advisor
Homework Helper
26,260
619
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)).
 
Last edited:
  • #8
56
0
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.
 
  • #9
Dick
Science Advisor
Homework Helper
26,260
619
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.
 
  • #10
56
0
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.
 

Related Threads on Polynomial Rings/Fields/Division Rings

  • Last Post
Replies
1
Views
950
Replies
8
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
1K
Replies
2
Views
3K
  • Last Post
Replies
2
Views
923
  • Last Post
Replies
13
Views
584
Replies
1
Views
656
  • Last Post
Replies
1
Views
1K
Top