- The other irreducible polynomial in
**Z**_{3}[*x*] is*x*^{2}+*x*+ 2.

The map*x*↦*y*+ 2 maps**Z**_{3}[*x*]/ <*x*^{2}+ 1 > isomorpically to**Z**_{3}[*y*]/ <*y*^{2}+*y*+ 2 > .The map

*x*↦*y*+ 1 maps**Z**_{2}[*x*]/ <*x*^{3}+*x*^{2}+ 1 > isomorpically to**Z**_{2}[*y*]/ <*y*^{3}+*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*p*^{2}quadratic polynomials altogether and any reducible quadratic polynomial is a product of linear factors, there are*p*^{2}-*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*p*^{3}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)(*p*^{2}+ 1)

^{1}/_{6}*p*(*p*- 1)(*p*+ 1)(*p*^{3}+*p*- 1)

^{1}/_{7}*p*(*p*- 1)(*p*+ 1)(*p*^{4}+*p*^{2}+ 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}=*a*^{p}+_{p}*C*_{1}*a*^{p-1}*b*+ ... +_{p}*C*_{p-1}*ab*^{p-1}+*b*^{p}=*a*^{p}+*b*^{p}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_{p}*C*_{r}=*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 2*x*have order 4; the rest have order 8.Since

*x*^{3}+ 2*x*= -1 we have 2*x*(*x*^{2}+ 2) = 1 and so*x*^{-1}= 2*x*^{2}+ 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}=*x*^{12}. Since*x*^{3}=*x*+2, we have*x*^{6}=*x*^{2}+*x*+ 1 and*x*^{12}=*x*^{2}+ 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
*x*^{2}+*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

**Z**_{p}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*p*^{k}. (You don't even need the fact that the group is cyclic!)