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# Polynomial Question watch

1. I was wondering about this: are there always finitely many polynomials (up to scalar multiples) of a fixed degree (say n) such that a complex number z is a root of these polynomials?

Maybe this is obvious?
2. Could you clarify - from what I can make out, it seems obvious, but perhaps I'm misunderstanding you - clearly the number of polynomials of degree (n-1) isn't finite. Just multiply any of these by (w-z) to get a poly of degree n which has z as a root.
3. Suppose there are finite polys P(n) which has z is one of their roots.
Then
Let P(x) = {P1, P2, ....} be set of all poly P(n) which has root z.
Let X = {z, z1, z2, ....,} be the set of all the roots of all the polys P(n)
As number of P(n) is finite, and each P(n) has at most n roots, then X is finite set.
-> Suppose z* is not in Z.
Take P(w) = (w - z)n-1(w - z*).
So P(w) has degree of n, has 1 root is z, but P(w) is not in P(x)

well, I don't see it right
4. (Original post by Wrangler)
Actually, re-reading your post, I don't think that's what your asking! Could you clarify - from what I can make out, it seems obvious, but perhaps I'm misunderstanding you - clearly the number of polynomials of degree (n-1) isn't finite. Just multiply any of these by (w-z) to get a poly of degree n which has z as a root.
Yes that's all I could make of the question, in which case it's obviously not so.
5. Well, I was showing that rt(3)+rt(-5) is algebraic using a deg 4 polynomial. That is easy. But I was wondering how many polynomials other than the one I found would work to show that its algebraic--i.e. for how many polynomials of deg 4 (say) is rt(3)+rt(-5) a root?
6. Am I just horribly over simplfying this but couldn't you just think about it like :

P(x) = (x-a)(x-b)(x-c)(x-d)

Let , then since you are free to pick c and d to be any real numbers or pair of complex conjugate roots there are uncountably infinite different 4th degree polynomials whose coefficents are real but who have as a root.
7. (Original post by J.F.N)
Well, I was showing that rt(3)+rt(-5) is algebraic using a deg 4 polynomial. That is easy. But I was wondering how many polynomials other than the one I found would work to show that its algebraic--i.e. for how many polynomials of deg 4 (say) is rt(3)+rt(-5) a root?
You are talking about minimal polynomials I think. Given an algebraic number z there is a unique monic polynomial of least degree which has z as a root. This poly is called its minimal polynomial. Of necessity it is irreducible.
8. (Original post by Neapolitan)
You are talking about minimal polynomials I think. Given an algebraic number z there is a unique monic polynomial of least degree which has z as a root. This poly is called its minimal polynomial. Of necessity it is irreducible.
Irreducible over Q? Why is that obvious?
9. If it weren't then we can factor it and one of its factors would have z as a root yet have degree less than the minimal polynomial.
10. (Original post by dvs)
If it weren't then we can factor it and one of its factors would have z as a root yet have degree less than the minimal polynomial.
I see.

Also, another question: if F is a field, then the units in the polynomial ring F[x] are just all non-zero constant elements of the field, right?
11. Yeah

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