Hey so I am having trouble with this problem:



I'm not sure if there's a better way to show it's a field other than checking all the field axioms

Thanks

I think this is what you're looking for part (a)?

Proof:

(=>) Suppose p = a + b with ab ≡ 1 (mod p) and consider the polynomial

([1]px + [a]p) ([1]px + p) .

Expanding this polynomial we get
(x + [a]p) (x + p) = [1]px2 + ([a]p + p) x + [a]pp
= [1]px2 + [a + b]px + [ab]p
= [1]px2 + [0]px + [1]p
= [1]px2 + [1]p

And so we have factorized [1]px2 + [1]p in Zp[x].

(<=) Now suppose f = [1]px2 + [1]p is reducible by contradiction. Then there must be a nontrivial factorization of f. Since f
has degree 2, the most general form of this factorization is
(8) [1]px2 + [1]p = (cx + d) (ex + f)
with c, d, e, f ∈ Zp. Expanding the right hand side of (??) and identifying the coefficients of like powers of
x, we find
(9) ec = [1]p
(10) cf + de = [0]p
(11) df = [1]p
Let a, b ∈ Z be any integers such that [a]p = cf, and p = de. Then (??) implies
[a]p + p = cf + de = [0]p ⇒ a + b ≡ 0 (mod p)
[a]pp = (cf)(de)=(cf)(de) = [1]p[1]p = [1]p ⇒ ab ≡ 1 (mod p)

QED

This post was edited by Arkil on Mar 5 2016 01:32pm

Why do you suppose that p = a+b with ab = 1 mod p?

a)
Let [f(x)] denote the congruence class of f(x) modulo p(x)
The addition and multiplication in E are
[f(x)+g(x)] = [f(x)] + [g(x)]
[f(x)g(x)] = [f(x)] [g(x)]
Then we see that the rings axioms that hold F[x] must also hold in E. If you know F[x] is a commutative ring with 1(if not this is really straight forward to prove), then all you need to show is that every element in E has a multiplicative inverse.
First lets show that for f(x) in F[x] non-zero polynomial with degree < n, that the only common factor between f(x) and p(x) is a polynomial with degree 0, ie a non-zero constant.
So suppose d(x)|f(x) and d(x)|p(x) => d(x)a(x)=f(x) and d(x)b(x)=p(x) for some a(x),b(x) in F[x].
Since p(x) is irreducible, we see that either d(x) or b(x) has degree 0. If deg(b(x)) = 0 then we have that deg(d(x)) = deg(p(x)) = n. This is because then you multiply 2 polynomials, you add their degree.
But this is a contradiction that deg(f(x)) < n. Thus we get that deg(d(x)) = 0, ie f(x) and p(x) are coprime.
Then because F[x] is a euclidean domain, there is the notion of a division algorithm => we have Euclidean algorithm => we have extended Euclidean algorihm
ie there exists a(x) and b(x) in F[x] such that a(x)f(x) + b(x)p(x) = 1
=> a(x)f(x) = 1 - b(x)p(x)
=> a(x)f(x) = 1 mod p(x)
ie the inverse of f(x) in E is a(x)
Thus E is a field.

b ) Define vector addition and multiplication as in E, with scalars t in F. Since E is a field, it satisfies all vector space axioms, so all we need to check is that the dimension of this vector space is n.
Consider B = {1,x,x²,...,x^(n-1)}. If we can show that this is a basis, we are done (need to show they span E and they are linearly independent)
Suppose f(x) ∈ F[x]. By the division algorithm, we know that f(x) is congruent to some polynomial with degree < n modulo p(x).
Thus we have f(x) = a_(n-1)x^(n-1) + ... + a_1x + a_0 mod p(x).
So B spans E. To show B is linearly independent, we show a_(n-1)x^(n-1) + ... + a_1x + a_0 = 0 mod p(x) => a_(n-1) = .. = a_1 = a_0
Suppose a_(n-1)x^(n-1) + ... + a_1x + a_0 = 0 mod p(x) , then we have
p(x)d(x) = a_(n-1)x^(n-1) + ... + a_1x + a_0 for some d(x) ∈ F[x]
Again using the property that when multiplying polynomials, the degrees are added; if the RHS is not equal to zero then, we get a contradiction since deg(p(x)) = 0
Thus we conclude that the RHS = 0 => a_(n-1) = ... = a_1 = a_0 = 0. ie B is linearly independent and is a basis for E. Thus E is a vector space over F with dimension n.

c) Since each vector is dependent on the choices of a_(n-1), .... , a_1, a_0 and each a_i has p choices, we have p^n elements.