Problem:
Solution:
For part (a), we show the set is closed from addition and multiplication.
Suppose f,g is in I(S), h∈k[x1,⋯,xn] then for all x∈S, f(x)=g(x)=0.
Now (f+g)(x)=0 and also (hg)(x)=0, so f+g and hg are both in I(S).
Therefore I(S) is an ideal.
For part (b), we do exactly the same as in Exercise 10. Now we have f(x,y)=h(x,y)(x−y)+r(x), f(a,a)=0⟹r(a)=0, ∀a≠1. so now we still have r(x)=0 and therefore the ideal is still ⟨x−y⟩.
For part (c), we already know polynomials that vanish on the whole integer grid is necessarily the zero polynomial, so I(Z) is simply {0}.
Solution:
For part (a), we show the set is closed from addition and multiplication.
Suppose f,g is in I(S), h∈k[x1,⋯,xn] then for all x∈S, f(x)=g(x)=0.
Now (f+g)(x)=0 and also (hg)(x)=0, so f+g and hg are both in I(S).
Therefore I(S) is an ideal.
For part (b), we do exactly the same as in Exercise 10. Now we have f(x,y)=h(x,y)(x−y)+r(x), f(a,a)=0⟹r(a)=0, ∀a≠1. so now we still have r(x)=0 and therefore the ideal is still ⟨x−y⟩.
For part (c), we already know polynomials that vanish on the whole integer grid is necessarily the zero polynomial, so I(Z) is simply {0}.
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