Homework Help: Multiplicative Inverse of Polynomials with Integer Coeff.

1. Feb 7, 2016

RJLiberator

1. The problem statement, all variables and given/known data
If ℤ[x] denotes the commutative ring consisting of all polynomials with integer coefficients, list all the elements in ℤ[x] that have a multiplicative inverse in ℤ[x].

2. Relevant equations
Multiplicative inverse if rs = 1 where rs ∈ R (rs are elements of the ring).

Polynomials are of the form a_n*x^n+a_(n-1)*x^(n-1)+...+a_1*x+a_0
in this case, the a coefficients are all integers.

3. The attempt at a solution

At first glance, this is an overwhelming question. They want me to list ALL the elements that have a multiplicative inverse?
This fact leads me to believe that there is, in fact, a very special rule for this to occur which limits the amount that can be inverses OR they just want a general rule.

Yeah, I'm really lost on this one.
I realize I don't have much of an answer here, so I am not looking for a big hint yet. I know PF policy is to show your work prior to help.

Just give me something to get me going...

2. Feb 7, 2016

HallsofIvy

This is much easier than you think it is! First do you see that if a polynomial has $a_0= a\ne 0$ while $a_i= 0$ for i> 0, then the 'inverse polynomial" is the polynomial with $a_0= \frac{1}{a}$ and $a_i= 0$ for i> 0?

And use the fact that if one polynomial has a term $x^n$ with non-zero coefficient and the other has $x^m$ with non-zero coefficient, then their product has $x^{n+m}$ with non-zero coefficient.

3. Feb 7, 2016

Staff: Mentor

I think this part is misleading.
But this part should help.

4. Feb 7, 2016

RJLiberator

As for the first line, regarding 1/a, this number is no longer in the integers. I would think this would not work, correct? (perhaps this is what mfb is alluding to)

As for the second line, I'm not sure how this helps. It kinds puts me in the same position that I have been in.

I can have all these polynomials in my head, but now I observe multiplication over them.
Is it that we need m+n = 0 so that x^0 = 1 ?

In which case x^3*x^(-3) would be an inverse.
So that would be all the polynomials with opposite degrees to themselves.

5. Feb 7, 2016

geoffrey159

if $RS = 1$, then both are non-zero and $0 = \text{deg}(1) = \text{deg}(RS) = \text{deg}(R) + \text{deg}(S)$.
So what is the degree of $R$ and $S$ ?
What is the only possibility (EDIT: 2 possibilities) left so that $R,S\in \mathbb{Z}[X]$?

6. Feb 7, 2016

Staff: Mentor

Is x^(-3) a polynomial?

Note: this is a matter of definition.

7. Feb 7, 2016

RJLiberator

Damn, good call. x^(-3) is NOT a polynomial.

So the sum of degree must be = 1.
deg(r)+deg(s) = 1.
2 possibilities: Either r is 0 and s = 1 or s is 0 and r is 1?

You are saying that the degree must be = 1? How is this so? If we have a degree of 1, I can't imagine a situation where x^1*s = 1. (where s is an element of polynomials, in this case a constant integer)

8. Feb 7, 2016

Why 1?

9. Feb 7, 2016

RJLiberator

That's what I initially was thinking.
That the sum of the degrees must actually be 0.

This kinda threw me off that direction:
But I understand now that if the sum of the degrees must be = 0.
Then degree of r and s must both be 0.

Giving that he said that there were 2 possibilities, hmmm...

1. either r and s must both be 0.
2. Or either they do not have degrees?

1 and 2 are the same tho, hm.

Is this the right train of thought?

10. Feb 7, 2016

geoffrey159

OK so both are constant polynomials of $\mathbb{Z}[X]$.
So $R = a$ and $S = b$, $(a,b) \in \mathbb{Z}$ such that $ab = 1$
What are the divisors of 1 in $\mathbb{Z}$ ?

11. Feb 7, 2016

RJLiberator

So we have 1*1=1
and -1*-1=1

as the two possibilities?

That's really all there is over the polynomials?

I understand your argument here. Where degree has to be 0 because anything more will not result in 1 being a possible answer. This makes sense to me.
And as we stated, there can't be negative degrees, so that means they have to be constant in the integers.
And since we have a*b = 1 where a and b are integers, then the only possibilities are 1 and -1.

12. Feb 7, 2016

Staff: Mentor

Correct.
As you now see, the list is short enough to make that reasonable.

13. Feb 7, 2016

RJLiberator

Indeed, I cannot make a counter argument to it.