Hey so I'm working on an assignment right now and am having trouble with a question.



So what I'm having trouble is showing that the generator of the ideal, call it I, is the minimal polynomial of A.

What I have so far is that if a polynomial ideal is generated by some g in I, we know that for f in I we have f = gq for some q in K[x].
This also tells us that g divides f.
From what I understand though (probably wrong..), doesn't it being the minimal polynomial of A come from the definition of a generator?
g is in I so g(A) = 0 and we got g by assuming it was of least degree, as shown by the following proof from my textbook:

Let g be a polynomial in I that is nonzero and of least degree. By the division algorithm, for any f in I we have:
f = gq + r for some q,r in K[x] with deg r < deg g. Then we have r = f - gq. We can see r is also in I.
Since deg r < deg g (I also don't really understand this step), we must have r = 0. Hence f = gq and g divides f.

SO since we assume g is of least degree isn't it automatically the minimal polynomial of A? Would I just say that it follows from the definition of g that it is the minimal polynomial of A?

Any help is appreciated, thanks

This post was edited by Bloo_Guardian on Feb 20 2015 10:39pm

Yes, the minimal polynomial of A is obviously the generator of your ideal I.
Actually you could use both definitions of the minimal polynomial :
- the polynomial of least degree that cancel A,
- the generator of the aforementioned ideal.

Actually, the division algorithm is used one step before, to prove that the ideal in K[X] is generated by a specific polynomial (ie : to prove that K[X] is a principal ideal domain).

the minimal polynomial is the monic(leading coefficient equal to 1) polynomial of least degree such that P(A) = 0
so the minimal polynomial would actually be g/a where a is the leading coefficient.



deg r < deg g because if deg r > deg g then you could divide g into r at least one more time.

can show using
r = f - gq
that r = 0

we know f,g ∈ I
using properties of polynomials we can show
f - gq ∈ I
=>r ∈ I
from above we said deg r < deg g; deg r is smaller than the least degree of a nonzero element of I. Therefore r must be 0