Question

Please do both problems. They are short.

I will rate for sure. Thanks!

Question# 1

Let p(x) = x^3 + 3x + 1 = (x + 3)^2 (x + 4) in Z5[x].

(a) Perform the following computations in Z5[x]/(p(x)). Give your answers in the form [r(x)] where r(x) has degree as small as possible.

i. [4x] + [3x^2 + x + 2]

ii. [x^2 ][2x^2 + 1]

(b) Show that Z5[x]/(p(x)) has zero divisors

Question #2

Let p(x) = x^3 + x + 1 in Z2[x], and let R = Z2[x]/(p(x)).

(a) Explain briefly how you know that R has 8 elements.

(b) Is R is a field?

(c) Write out the multiplication table for R. (Don’t write out the addition table.) I suggest that you omit the brackets.

Answer 1

Solution: Question #2 (a)

Definition: A polynomial of degree with coefficients in a field F is
said to be irreducible over F if it cannot be written as a product
of two non constant polynomials over F of degree less than n.

Let in and let .

. Since , polynomial of degree 3

has no roots in , so it cannot be written as a product
of two non constant polynomials over of degree less than 3.

Hence is irreducible over .

Since p(x) is irreducible, (p(x)) is maximal and is a field.

Since degree of p(x) is 3, every element of can be expressed

as .

Since each is in , there are 2 options for each coefficient in the coset
representative. Thus there are possible standard representatives.

Also it can be seen that each is unique.

Therefore R has 8 elements.

(b) Hence is irreducible over .

Since p(x) is irreducible, (p(x)) is maximal and is a field.

Therefore R is a field.

(c) Here

Table

*	p(x)	1+p(x)	x+p(x)	x^2+p(x)	1+x+p(x)	1+x^2+p(x)	x+x^2+p(x)	1+x+x^2+p(x)
p(x)	p(x)	1+p(x)	x+p(x)	x^2+p(x)	1+x+p(x)	1+x^2+p(x)	x+x^2+p(x)	1+x+x^2+p(x)
1+p(x)	1+p(x)
x+p(x)	x+p(x)
x^2+p(x)	x^2+p(x)
1+x+p(x)	1+x+p(x)
1+x^2+p(x)	1+x^2+p(x)
x+x^2+p(x)	x+x^2+p(x)
1+x+x^2+p(x)	1+x+x^2+p(x)

Note: Result: Suppose that and f(x) is irreducible over where p is a prime. If deg(f(x)) = n, Then that is a field with elements.