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Consider the algebraic set $V(X^2-YZ,X-XZ)$. Find the irreducible components of this set and show that $I(V)=(X^2-YZ,X-XZ)$.

I reasoned that $X-XZ=0$ iff $X=0$ or $Z=1$. If $X=0$, we get $Y=0$ or $Z=0$ from the other equation. If $Z=1$ we get $X^2-Y=0$. Hence the irreducible components are $V(X,Y), V(X,Z), V(Z-1,X^2-Y)$. Is this correct?

I'm not sure how to do the next question. Of course automatically we get one inclusion, but how about the other?

Thanks

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Yes, your argument regarding the irreducible components is correct. Of course, you need to be able to see that these components are indeed irreducible, i.e. the ideals $(X,Y),(X,Z)$ and $(Z-1,X^2-Y)$ are prime.

Since we have $V = V(X,Y) \cup V(X,Z) \cup V(Z-1,X^2-Y)$ it follows that $I(V) = (X,Y) \cap (X,Z) \cap (Z-1,X^2-Y)$. Lets compute this intersection of prime ideals on the right hand side.

First observe that since $(X,Z) + (Z-1,X^2-Y)$ contains $1$, we must have that $J:=(X,Z) \cap (Z-1,X^2-Y) = (X,Z)(Z-1,X^2-Y)= (X(Z-1),X(X^2-Y),Z(Z-1),Z(X^2-Y))$

Now, $X^2 - YZ = -X X(Z-1) + Z(X^2-Y)$ and so $X^2 - YZ \in J$. Moreover, $Z(X^2-Y) = X X(Z-1) + (X^2-YZ)$ and similarly $X(X^2-Y) = Y X(Z-1) + X (X^2-YZ)$ and so $J = (X(Z-1),X^2-YZ,Z(Z-1))$.

Next take $f \in (X,Y) \cap J$. Then $f = f_1 X(Z-1) + f_2 (X^2-YZ) + f_3 Z(Z-1)$. Since $f, f_1 X(Z-1) + f_2 (X^2-YZ) \in (X,Y)$ we must have that $f_3 Z(Z-1) \in (X,Y)$. But $(X,Y)$ is prime and so $f_3 \in (X,Y)$, which gives that $(X,Y) \cap J = (X(Z-1),X^2-YZ,YZ(Z-1))$.

Finally, note that $YZ(Z-1) = -X X(Z-1) + (Z-1) (X^2-YZ)$ and so $I(V) = (X,Y) \cap J = (X(Z-1),X^2-YZ))$.

Notice that the above argument does not make use of Hilbert's Nullstellensatz (which in turn would require the underlying field to be algebraically closed), as would the argument of trying to show that the ideal $(X(Z-1),X^2-YZ)$ is radical.

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    $\begingroup$ Thanks. How would you show that $(Z-1,X^2-Y)$ is prime? $\endgroup$ – user137090 May 12 '14 at 17:59
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    $\begingroup$ @user137090 You can see a proof in my answer to another question posted by you. $\endgroup$ – user26857 May 12 '14 at 19:50
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    $\begingroup$ @user137090: the proof in user26857's answer quoted in the comment above is exactly the one i had in mind. So please read that. $\endgroup$ – Manos May 12 '14 at 20:40
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Your proof is correct, the variety that you obtain is the union of the varieties given by $$X=Y=0, X=Z=0, Z=1,Y=X^2.$$ It is then easy to see that each polynomial that vanishes on this set belongs to $$I=(X^2-YZ,X-XZ).$$ Indeed, modulo $I$ you can replace $YZ$ with $X^2$ and $XZ$ with $X$ so every polynomial is equivalent to $$P(Z)+Q(X,Y)$$ for some polynomials $P,Q$. Because the polynomial vanishes on $X=Y=0$, we get that $P$ is a constant and can assume $P=0$. Because the polynomial vanishes on $Z=1,Y=X^2$ and on $X=Z=0$, $Q(X,Y)$ is a multiple of $X(Y-X^2)$. It remains to see that $X(Y-X^2)\in I$, which is because $$X(Y-X^2)=X(Y-YZ)=Y(X-XZ)=0\pmod I.$$

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