Abstract algebra Polynomials and Prime

[RJLiberator]

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

Homework Equations

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.

 

[RJLiberator]

Any idea on this? :)

 

[RJLiberator]

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.

 

[fresh_42]

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

Homework Equations

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.

I read it this way:

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!