(adsbygoogle = window.adsbygoogle || []).push({}); 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 [itex]\neq[/itex] {0}

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

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

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)[itex]\geq[/itex]1

Now we let f(x)[itex]\in[/itex]I and apply the division algorithm.

f = qg + r, where q,r[itex]\in[/itex]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[itex]\in[/itex]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...

Please help!

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# Principal Ideal, Polynomial generators

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