- #1
nugget
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Homework Statement
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)> for some g(x)[itex]\in[/itex]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)[itex]\in[/itex]R[x] that has minimal degree among all non-zero polynomials in I?
Homework Equations
Theorems, the division algorithm.
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!