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How can I find the maximal ideals in F_2[x]/(x^8-x), that is, the finite field with 2 elements modulo the ideal generated by x^8-x?

My attempt at a solution:
The irreducible factorization of x^8-x in F_2[x] would be x^8-x = x^8+x = (x)(x+1)(x^3+x+1)(x^3+x^2+1), so we can use the remainder theorem to say that
F_2[x]/(x^8-x) is isomorphic to the direct product of F_2[x]/(x), F_2[x]/(x+1), F_2[x]/(x^3+x+1), and F_2[x]/(x^3+x^2+1).
I feel pretty certain that this should be used in figuring out what the maximal ideals are, but I'm not sure exactly how to employ this.

Other comments:
Since F_2 is a field, F_2[x] is a Euclidean Domain, Principal Ideal Domain, and a Unique Factorization Domain. If (x^8-x) is a prime ideal, then the quotient F_2[x]/(x^8-x)should be a Principal Ideal Domain as well, which would mean the nonzero prime ideals would be all of the maximal ones. On that note, is (x^8-x) a prime ideal?

Okay, let R = F_2[x] and I = (x^8-x).

First of all, I is NOT a prime ideal! If it were, it would be maximal as you said, and if it were maximal, then R/I would be a field, which it very clearly is not!

Likewise, how can it be prime (same as irreducible in a PID) if you just gave me its factorization?? You have:

I = I1 ∩ I2 ∩ I3 ∩ I4

where I1 = (x), I2 = (x+1), I3 = (x^3+x+1), and I4 = (x^3+x^2+1), just like you said.

But you're on the right track, really. All you really need to do to solve this problem.

First of all, what is R/I? We can figure it out. Each component can be figured out.

Well, what is R/I1 ? It corresponds to the map from R to F where f(x) becomes f(0).

So R/I1 is F_2.

Likewise, R/I2 is F_2, just where the map is f(x) becomes f(1).

What about the other two factors? Well, all of the factors (including the two I've mentioned, is R / (some irreducible polynomial). That is always a field.

You should know, then, that R/I3 and R/I4 are also fields. They are both fields of size 8, of which there is only one (F_8), up to isomorphism.

So, R is isomorphic to R/I1 x R/I2 x R/I3 x R/I4, which is isomorphic to F2 x F2 x F8 x F8.

Hopefully this gives you some idea of what R really is.

Now, back to the question, what are the maximal ideals?

Well, the maximal ideals in a polynomial ring can be characterized in many ways:

1. All ideals such that R/I is a field.
2. All prime ideals in R.
3. All ideals with no larger ideal between it and R (the definition of maximal).

Item 1 we've already mentioned -- so those irreducible components will give us the right ideals. And number 3 is just the definition of maximal. But let's try it thinking about item 2, which you mention:

These ideals need to be prime, so you really want to look to the prime factors of the original polynomial x^8-x, which you've already done.

So, the maximal ideals are (x), (x+1), (x^3+x+1), and (x^3+x^2+1).

All other ideals are intersections of those, like (x^2+x) or (x^4+x^2+x).

So those four prime ideals are maximal, and that's it. You already had the factorization of x^8-x, you just needed to hash it out a little.

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Clyde Oliver


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