Let $K$ be a field. While doing an exercise I am trying to find the ideal $I:=\langle x^2+y^2-1,y-x^2+1\rangle$ in $K[x,y]$. I am guessing that the ideal is principal since otherwise the exercise would not make much sense. As a first step one can consider the relation from the second generator $y=x^2-1$ and we get $$I = \langle x^2+(x^2-1)^2-1,y-x^2+1 \rangle = \langle x^4-x^2, y-x^2+1 \rangle $$ But now I do not see a way to simplify that any further.
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$\begingroup$ The ideal is not principal. $\endgroup$– Sassatelli GiulioCommented Jul 7 at 7:51
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$\begingroup$ why do you say the exercise wouldn't make much sense if the ideal wasn't principal? $\endgroup$– Jackozee HakkiuzCommented Jul 7 at 7:53
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$\begingroup$ @SassatelliGiulio And why is that? $\endgroup$– Flynn FehreCommented Jul 7 at 7:56
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$\begingroup$ @JackozeeHakkiuz I am asked to determine the scheme theoretic intersection of the two affine schemes $\mathrm{Spec}K[x,y]/\langle x^2+y^2-1 \rangle$ and $\mathrm{Spec}K[x,y]/\langle y-x^2+1 \rangle$ which is the quotient by the sum of the two given ideals. If it were not principal I thought there would be no point in calculating it. $\endgroup$– Flynn FehreCommented Jul 7 at 8:05
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1$\begingroup$ @Flynn: why not? The quotient is much easier to work with than the ideal itself, since the second polynomial lets you get rid of $y$. $\endgroup$– Qiaochu YuanCommented Jul 7 at 8:45
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1 Answer
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This ideal is not principal as the intersection $\text{Spec}(K[x,y]/I)\cong\text{Spec}(K[x]/(x^4-x^2))$ is zero-dimensional, while a principal ideal in $\mathbb{A}^2$ should be of purely codimension 1.
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$\begingroup$ Why is the intersection already $\mathrm{Spec}K[x]/(x^4-x^2)$? I mean, I am trying to calculate that intersection. $\endgroup$ Commented Jul 7 at 8:45
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2$\begingroup$ @FlynnFehre Consider the ring homomorphism $K[x]\hookrightarrow K[x,y]\rightarrow K[x,y]/I$. You can find that this is a surjective homomorphism with kernel $(x^4-x^2)$. $\endgroup$ Commented Jul 7 at 8:48