mist on Nostr: Cawfee is down so I made an alt. I think you asked a good question. As you said, ...

Cawfee is down so I made an alt. I think you asked a good question.

As you said, N_\alpha is generated by one element, which we can write as (x - \alpha) g(x) for some g(x) \in K[x].

Now suppose that (x - \alpha) h(x) \in F[x], for some other h(x) \in K[x].

Since (x - \alpha) h(x) is a polynomial in F[x] which has \alpha as a root, it lies in the ideal N_\alpha by definition. Therefore:

(x - \alpha) h(x) = p(x) (x - \alpha) g(x)

for some p(x) \in F[x]. (This is what it means for (x - \alpha) g(x) to be a generator.)

Rewrite this as

(x - \alpha) (h(x) - p(x) g(x)) = 0.

Since K is an integral domain, so is K[x]. Now since (x - \alpha) is clearly nonzero, we conclude that h(x) - p(x) g(x) = 0.

This is what you wanted: h(x) is a multiple of g(x).

Let me know if this looks right - I'm happy to elaborate