Question

can anyone help me construct some polynomials that satisfy f(1,1)=0=f(1,2).

Answer 1

i already have x-1, but need at least one with both variables

Answer 2

TedG: Would you please make certainly you've copied this problem down correctly. Here's why I ask: your f(1,1)=0=f(1,2) implies that at x=1, y can be either 1 or 2. If that's the case, your relationship is definitely not a function.

Answer 3

Being a function requires that for each x, there is only one corresponding y.

Answer 4

@mathmale The f should have a line above it and is given as \[\tilde{f}: \mathbb{A}^{m} \rightarrow K\] where the \[\mathbb{A}^{m}\] denotes the m dimensional affine space. if that changes anything.

Answer 5

Basically the question i am trying to complete is about finding a finite basis to an ideal. I know how to complete it, but i first need to find some polynomials where if i were to plug in (1,1) and (1,2) they would equal 0.

Answer 6

I see. While still standing behind what I first wrote, I must admit that you're ahead of me in this particular course work and cannot be of much further help to you. Sorry.

Answer 7

an example: for f(1,1)=0=f(-1,2), possible polynomials could be, x^2-1 and 2y+x-3

Answer 8

\[
f(x,y)=(x-1) (y-2) (y-1)=x y^2-3 x y+2 x-y^2+3
y-2
\]

Answer 9

Thanks, Ted. I'm with you there, but please note that you initially typed
f(1,1)=0=f(1,2), which is unacceptable if f is a function.

Answer 10

Why f(1,1)=0=f(1,2 is not acceptable? See my function above

Answer 11

@mathmale that is the question that i have been given

Answer 12

@TedG your question is fine

Answer 13

@eliassaab does it help if i mention that this is over Q[X,Y], is it possible for a simple polynomial with rational coefficients?

Answer 14

more simple*

Answer 15

This is the answer to your question
\[f(x,y)=(x-1) (y-2) (y-1)=x y^2-3 x y+2 x-y^2+3
y-2
\]

Answer 16

Apologies. Finally, I see you're dealing with a function of TWO variables. In this case, I'm all wet. Proceed, gentlemen.

Answer 17

NP