# Abstract algebra Polynomials and Prime

• RJLiberator
In summary, the conversation discusses a problem involving a polynomial g(x) belonging to the set of polynomials with integer coefficients. It is given that g(x) has a degree of at least 2 and a prime number p is also given. The conditions (i), (ii), and (iii) for g(x) are specified, and three questions (a, b, and c) are asked.The summary then goes on to explain the meaning of Z[x], the conditions for g(x), and the questions in detail. It also suggests a different way of writing g(x) to make the problem easier to understand. Finally, it points out a major issue with the question and suggests resolving it.
RJLiberator
Gold Member

## Homework Statement

Let g(x) ∈ ℤ[x] have degree at least 2, and let p be a prime number such that:
(i) the leading coefficient of g(x) is not divisible by p.
(ii) every other coefficient of g(x) is divisible by p.
(iii) the constant term of g(x) is not divisible by p^2.

a) Show that if a ∈ ℤ such that [a]_p ≠ [0]_p, then [g(a)]_p ≠ [0]_p.

b) Show that if b ∉ ℤ such that _p = [0]_p, then [g(b)]_p^2 ≠ [0]_p^2.

c) Conclude that g(x) has no roots in ℤ

## The Attempt at a Solution

Well, the problem I am having here is understanding the question at all.

So g(x) exists in Z[x].

What does Z[x] exactly mean?

It has a degree of at least 2, and we have p as a prime number.

We know the leading coefficient of g(x) is NOT divisible by p.

So this means we have C*x^n+...
Where c is not divisible by p.

Every other coefficient of g(x) is divisible by p.
So we have C*x^n + g*x^(n-1)+...
where g IS divisible by p.

The constant term fo g(x) is not divisible by p^2. So we now have
C*x^n + g*x^(n-1)+...+ z
where z is not divisible by p*p.

So we look at question a. If a exists in the integers such that [a]_p cannot equal [0]_p then [g(a)]p cannot equal [0]p.
So we have
C*a^2+g*a+z as an example polynomial.

and here is where I start to break down.
Can't decide what to do from here.

Any idea on this? :)

Perhaps we can take a step back and instead just look at part a

a) Show that if a exists in the integers such that [a]_p cannot equal [0]_p then [g(a)]_p cannot equal [0]_p.

Logically, this seems consistent and almost obvious.

Let's try this, using the division algorithm:

We take the first term of some polynomial
Q*a^n = dp+r

Q is NOT divisible by p by (i).
So if we assume, for a contradiction, that r = 0.
Then Q*a^n/p cannot equal the integer d because
a) Q is not divisible by p
b) [a]_p =/= [0]_p

Therefore, [g(a)]_p =/= [0]_p.

RJLiberator said:

## Homework Statement

Let g(x) ∈ ℤ[x] have degree at least 2, and let p be a prime number such that:
(i) the leading coefficient of g(x) is not divisible by p.
(ii) every other coefficient of g(x) is divisible by p.
(iii) the constant term of g(x) is not divisible by p^2.

a) Show that if a ∈ ℤ such that [a]_p ≠ [0]_p, then [g(a)]_p ≠ [0]_p.

b) Show that if b ∉ ℤ such that _p = [0]_p, then [g(b)]_p^2 ≠ [0]_p^2.

c) Conclude that g(x) has no roots in ℤ

## The Attempt at a Solution

Well, the problem I am having here is understanding the question at all.

So g(x) exists in Z[x].

What does Z[x] exactly mean?

It has a degree of at least 2, and we have p as a prime number.

We know the leading coefficient of g(x) is NOT divisible by p.

So this means we have C*x^n+...
Where c is not divisible by p.

Every other coefficient of g(x) is divisible by p.
So we have C*x^n + g*x^(n-1)+...
where g IS divisible by p.

The constant term fo g(x) is not divisible by p^2. So we now have
C*x^n + g*x^(n-1)+...+ z
where z is not divisible by p*p.

So we look at question a. If a exists in the integers such that [a]_p cannot equal [0]_p then [g(a)]p cannot equal [0]p.
So we have
C*a^2+g*a+z as an example polynomial.

and here is where I start to break down.
Can't decide what to do from here.

a) Show that if a ∈ ℤ such that [a]_p ≠ [0]_p, then [g(a)]_p ≠ [0]_p.
should be
a) Show that if ##a∈ℤ: a ≠ 0 \mod{p} ⇒ g(a) ≠ 0 \mod{p}##

and
b) Show that if b ∉ ℤ such that _p = [0]_p, then [g(b)]_p^2 ≠ [0]_p^2.
should be
b) Show that if ##b ∉ ℤ: b = 0\mod{p} ⇒ g(b) ≠ 0 \mod{p^2}##

That is my decoding of your input. Unfortunately ##b ∉ ℤ## generates two major problems:
(i) Where is ##b## supposed to be in? ##ℚ##?, ##ℝ##? or ##ℂ##?
(ii) If ##b ∉ ℤ## then ##b \mod{p}## doesn't make any sense.

For the rest: I would write ##g(x) = a_n x^n + a_{n-1} p^{r_{n-1}} x^{n-1} +... + a_{1} p^{r_{1}} x + a_0 p ##
where none of the ##a_i ∈ ℤ ## is divisible by ##p##, ## n ∈ℕ, n>1## and ##r_j ∈ ℕ, r_j>0##.

That should make things easier to talk about. However, please resolve the ##b## mystery!

## 1. What is abstract algebra?

Abstract algebra is a branch of mathematics that studies algebraic structures such as groups, rings, and fields. These structures are defined in an abstract way rather than using specific numbers or symbols.

## 2. What are polynomials?

Polynomials are expressions that consist of variables and coefficients, combined using the operations of addition, subtraction, and multiplication. They can have one or more terms, and the variables can have different powers.

## 3. What is a prime number?

A prime number is a positive integer that is only divisible by 1 and itself. For example, 2, 3, 5, and 7 are all prime numbers.

## 4. How are abstract algebra, polynomials, and prime numbers related?

In abstract algebra, polynomials are often studied in the context of polynomial rings, which are algebraic structures that follow certain rules and properties. Prime numbers also play a role in these structures, as they can be used to create fields and other important concepts in abstract algebra.

## 5. What are some applications of abstract algebra in real life?

Abstract algebra has many real-life applications, including cryptography, coding theory, and computer graphics. It is also used in physics, chemistry, and engineering to solve complex problems and model systems. Additionally, abstract algebra is essential in understanding and developing advanced mathematical theories and concepts.

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