Solution f="https://forum.tuteehub.com/tag/1">1. The second part is also very simple. Indeed, recall that if k=k¯¯¯ is an algebraically closed field every prime ideal of k[X,Y] is of the form ⟨X−x,Y−y⟩ or of the form (f(X,Y)) with f(X,Y)∈k[X,Y] irreducible. If you take k[X,Y]/I for some ideal I, we have that the set of the prime ideals of k[X,Y]/I is simply the set of the primes in k[X,Y] of the primes which contain I. Now, since the polynomial X2−Y2−f="https://forum.tuteehub.com/tag/1">1
manpreet
Best Answer
2 years ago
My idea:
(1) The first part is easy. Since x2−y2−com/tag/1">1∈C[y][x]x2−y2−com/tag/1">1∈C[y][x] is Eisenstein with respect to the prime y+iy+i, it is irreducible. Hence the ideal generated by it is a prime ideal.
(2) I am struggling with this part. I have one approach (not sure if correct).
Approach : Let I=⟨x2−y2−com/tag/1">1⟩I=⟨x2−y2−com/tag/1">1⟩. Then R=C[x,y]/IR=C[x,y]/I. Any ideal of RR is of the form J/I
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