- The other irreducible polynomial in Z3[x] is x2 + x + 2.
The map x ↦ y + 2 maps Z3[x]/ < x2 + 1 > isomorpically to Z3[y]/ < y2 + y + 2 > .
The map x ↦ y + 1 maps Z2[x]/ < x3 + x2 + 1 > isomorpically to Z2[y]/ < y3 + y + 1 > .
- There are p quadratic polynomials of the form (x - α)2 and p(p - 1)/2 of the form (x - α)(x - β). Since there are p2 quadratic polynomials altogether and any reducible quadratic polynomial is a product of linear factors, there are p2 - p - p(p - 1)/2 = 1/2 p(p - 1) irreducible ones.
Play the same game with the cubic polynomials:
There are p of the form (x - α)3, p(p - 1) of the form (x - α)(x - β)2 and p(p - 1)(p - 2)/6 of the form (x - α)(x - β)(x - γ).
The other reducible ones are of the form (x - α)q(x) with q(x) an irreducible quadratic.
We've just counted those and so there are p × p(p - 1)/2 like this.
Take these away from the total number p3 of all cubic monic polynomials to get 1/3 p(p-1)(p+1) irreducible ones.
The corresponding formulae for irreducible polynomials of degrees 4, 5, 6 and 7 are:
1/4 p(p - 1)(p + 1)p
1/5 p(p - 1)(p + 1)(p2 + 1)
1/6 p(p - 1)(p + 1)(p3 + p - 1)
1/7 p(p - 1)(p + 1)(p4 + p2 + 1)
- Let f (for Frobenius!) be the map. Then it is easy to see that f(ab) = f(a)f(b).
For the addition: f(a + b) = (a + b)p = ap+ pC1ap-1b + ... + pCp-1abp-1 + bp = ap + bp as required since all the binomial coefficients vanish mod p.
To see that the binomial coefficient is divisible by p if p is prime, observe that pCr = p! (p - r)! /r! and there is nothing on the bottom to cancel out the p in the p! on top
- If J is an ideal of S then it is easy to check that f-1(J) is an ideal of R which certainly contains f-1(0) = I.
If K is an ideal of R containing I then f(K) is an ideal of S. (You need the fact that f is onto to verify that the image is closed under multiplication by elements of S.)
These two correspondences are inverse to each other:
It is easy to see that starting with J you get back to J. Starting with K an element r ∈ K maps to an element s ∈ S and then under f-1 back to a collection of elements of R which all differ from r by elements of the kernel I. Hence they are all in the ideal K and we have our original set.
- Starting with any non-zero matrix and multiplying on the left and right by other "elementary matrices" (ones with one 1 and 3 zeros) will let you get a non-zero entry in any of the four positions. Then multiplying by scalar matrices and adding will let you get any matrix. Hence an ideal which contains any non-zero matrix must contain them all. So the ring has only two ideals {0} and R.
A similar proof shows that any matrix ring with field entries is simple. (Where did we use the fact that the entries were from a field?)
- Try a few. The element 1 has order 1; 2 has order 2; x and 2x have order 4; the rest have order 8.
Since x3 + 2x = -1 we have 2x(x2 + 2) = 1 and so x-1 = 2x2 + 1.
Since the group has order 26 the order of an element is either 1, 2, 13 or 26 (by Lagrange's theorem). So we need only show that x does not have order 13. If x has order 13 then x-1 = x12. Since x3 = x +2, we have x6 = x2 + x + 1 and x12 = x2 + 1 ≠ x-1.
So x has order 26.
In a cyclic group of order 26, the number of generators is the number of elements 1, 2, ..., 25 which are coprime to 26. There are 12 of these.
- Substitute to verify that x2 + x has roots 0, 2, 3 and 5. This does not contradict the result that a polynomial of degree k has at most k roots because that was for polynomials over a field.
- By FLT the polynomial gives zero on substituting x = 0, 1, 2, ... , p - 1. Hence it factorises into p linear factors.
Proving FLT uses the fact that the multiplicative group of the field Zp has p - 1 elements and that the order of any element divides the order of the group. The same proof works for the multiplicative group of the field of order pk. (You don't even need the fact that the group is cyclic!)