Multiplicative Inverse of Polynomials with Integer Coeff.

  • #1
RJLiberator
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Homework Statement


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].

Homework 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.



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...
 

Answers and Replies

  • #2
HallsofIvy
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This is much easier than you think it is! First do you see that if a polynomial has [itex]a_0= a\ne 0[/itex] while [itex]a_i= 0[/itex] for i> 0, then the 'inverse polynomial" is the polynomial with [itex]a_0= \frac{1}{a}[/itex] and [itex]a_i= 0[/itex] for i> 0?

And use the fact that if one polynomial has a term [itex]x^n[/itex] with non-zero coefficient and the other has [itex]x^m[/itex] with non-zero coefficient, then their product has [itex]x^{n+m}[/itex] with non-zero coefficient.
 
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  • #3
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This is much easier than you think it is! First do you see that if a polynomial has [itex]a_0= a\ne 0[/itex] while [itex]a_i= 0[/itex] for i> 0, then the 'inverse polynomial" is the polynomial with [itex]a_0= \frac{1}{a}[/itex] and [itex]a_i= 0[/itex] for i> 0?
I think this part is misleading.
And use the fact that if one polynomial has a term [itex]x^n[/itex] with non-zero coefficient and the other has [itex]x^m[/itex] with non-zero coefficient, then their product has [itex]x^{n+m}[/itex] with non-zero coefficient.
But this part should help.
 
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  • #4
RJLiberator
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This is much easier than you think it is! First do you see that if a polynomial has [itex]a_0= a\ne 0[/itex] while [itex]a_i= 0[/itex] for i> 0, then the 'inverse polynomial" is the polynomial with [itex]a_0= \frac{1}{a}[/itex] and [itex]a_i= 0[/itex] for i> 0?

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

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
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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]##?
 
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  • #6
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In which case x^3*x^(-3) would be an inverse.
So that would be all the polynomials with opposite degrees to themselves.
Is x^(-3) a polynomial?

Note: this is a matter of definition.
 
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  • #7
RJLiberator
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Is x^(-3) a polynomial?

Note: this is a matter of definition.

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

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]##?

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)
 
  • #9
RJLiberator
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That's what I initially was thinking.
That the sum of the degrees must actually be 0.

This kinda threw me off that direction:
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]##?

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
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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}## ?
 
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  • #11
RJLiberator
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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}## ?

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
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Correct.
They want me to list ALL the elements that have a multiplicative inverse?
As you now see, the list is short enough to make that reasonable.
 
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  • #13
RJLiberator
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Indeed, I cannot make a counter argument to it.

This thread is solved.

Thank you guys for your help.
 

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