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## 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]/(x

^{2}+ 1)

b) F[x]/(x

^{2}+ x + 1)

c) F[x]/(x

^{3}+ x

^{2}+ 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}

x

^{2}+ 1 = 0 here so this also gives us x

^{2}= 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,x

^{2},x

^{2}+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.