# Principal Ideal, Polynomial generators

1. Sep 9, 2011

### nugget

1. The problem statement, all variables and given/known data

Suppose R is an integral domain and I is a principal ideal in R[x], and I $\neq$ {0}

a) Show I = <g(x)> for some g(x)$\in$R[x] that has minimal degree among all non-zero polynomials in I.

b) Is it necessarily true that I = <g(x)> for every g(x)$\in$R[x] that has minimal degree among all non-zero polynomials in I?

2. Relevant equations

Theorems, the division algorithm.

3. The attempt at a solution

For (a) we can take a g(x) of minimal degree as our generator; if deg(g) = 0, then g(x) is a unit in I and I = R[x], so we can assume deg(g)$\geq$1

Now we let f(x)$\in$I and apply the division algorithm.

f = qg + r, where q,r$\in$R[x] and deg(r)<deg(g)

This means that r = f - qg which is an element of I. By our choice of g having minimal degree, this means that r = 0. Otherwise r would be an element of I, and could be written as r = pg, p$\in$R[x]. (I'm not quite sure how best to explain the reasoning behind the fact that r must equal 0)

finishing the proof after this is fine.

b) I'm confused for this question, but I assume its a contradiction type answer, although I can't imagine a g(x) that wouldn't generate a principal ideal...

2. Sep 10, 2011

### micromass

Staff Emeritus
Why??

Did you prove that the division algorithm holds when R is simply an integral domain?

Consider $\mathbb{Z}$ and take the ideal generated by X+2. Then 2X+4 is in the ideal and is of minimal degree.

3. Sep 11, 2011

### nugget

Hey,

regarding that deg(g)=0 bit, I guess I just assumed that a constant would be a unit in I. This would only be the case if R was a field, right?

Now I'm really confused for this question too; don't know how to prove the division algorithm, let alone for an integral domain.

I think I understand what you're saying for b)

Z is an integral domain so we can use that for R. b) is not true because <x+2> generates I and is of minimal degree, but <2x+4> can't generate I (it can't generate x+2, as Z doesn't contain fractions) but is also of minimal degree. Hence not every g(x) in R[x] of minimal degree generates I.

4. Sep 11, 2011

### micromass

Staff Emeritus
The trick for (a) is that you already know that I is principal. That is, you already know that I=<g(x)> for some g. The only thing you need to prove is that g is of minimal degree.