# Giving a hypersurface by higher degree polynomials

Suppose a hypersurface in $\mathbb{P}^n$ is given by an equation $F(Z) = 0$. It is easy to show that polynomials $Z_i F$ ($i = 0,\ldots,n$) give the same hypersurface, but I have trouble demonstrating that $K[Z_0,\ldots,Z_n]/I \cong K[Z_0, \ldots, Z_n]/I'$ where $I = (F)$, $I' = (Z_i F \mid i = 0,\ldots,n)$.

I tried to construct an isomorphism explicitly, but I got stuck: intuitively, $\varphi: G \mapsto \sum_{i=0}^n Z_i G$ seems like it could be right, it is defined correctly because if $G = FH$, then $\varphi(G) = \sum_i Z_i F H \in I'$.

But then I have to show that $\varphi$ is surjective, and I'm stuck: I can't find a way to show that for any polynomial $G$ we have $G \equiv \sum_i Z_i F \: (\mathrm{mod}\ I')$.

Am I on the right track? Is there an easier way to show that $K[Z]/I \cong K[Z]/I'$?

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The reason you have trouble showing your algebras are isomorphic is that they are not isomorphic!

For example take $n=1$ and $F=z_0$.
The algebras $K[z_0,z_1]/(z_0)$ and $K[z_0,z_1]/(z_0^2,z_1z_0)$ are not isomorphic because the first one is reduced (meaning its only nilpotent is zero) whereas in the second one the class of $z_0$ is nilpotent.

In scheme theory, a more advanced version of algebraic geometry, one would even say that the locus of zeros of the ideals $(z_0)$ and$(z_0^2,z_1 z_0)$ are different subschemes of $\mathbb P^1_K$, even though their underlying sets are the same, namely the singleton set $\lbrace [0:1]\rbrace$.

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But in his "AG: The First Course" Harris argues that a projective variety is isomorphic to its image under Veronese embedding, and in constructing the equations for this image he is using this trick to raise degrees of the original equations. How is this supposed to work then? – Alexei Averchenko Jun 9 '12 at 8:49
Dear Alexei, there is no contradiction: $I$ and $J$ have the exact same zero sets. Your question was about algebras and Harris makes no mention of these. The point to keep in mind is that there is no bijective correspondence between subvarieties and ideals (or algebras). The importance of the Nullstellensatz is that it addresses this discrepancy. – Georges Elencwajg Jun 9 '12 at 8:57
One last thing: Harris writes that two projective varieties are called projectively equivalent if there is a projective transformation that takes one to the other, or equivalently if their homogeneous coordinate rings are isomorphic as graded algebras. But in my example surely there is a projective transform ($\mathrm{id}$) that takes one to another, how come the rings are not isomorphic then? – Alexei Averchenko Jun 9 '12 at 15:14
Dear Alexei, this again is a subtle point. Namely: two projectively equivalent varieties in $X,X'\subset \mathbb P^n$ indeed have isomorphic algebras.But what is the algebra of $X$ ? It is the quotient $S(X)=K[z_0,...,z_n]/I(X)$ where $I(X)$ is the set of all polynomials vanishing on $X$. And that is the crucial point: it may very well be the case (as in my example) that a smaller ideal $J \subsetneq I(X)$ also has $X$ as its zero-set. Harris makes no claim about the quotient $K[z_0,...,z_n]/J$, even if $X$ is projectively normal. – Georges Elencwajg Jun 9 '12 at 16:41
Thank you very much for your time 8) I really mean it. – Alexei Averchenko Jun 10 '12 at 4:56

The problem is that rings correspond to working with affine schemes. Inside $\mathbf{A}^{n+1}$, the ideals $I = (F)$ and $J = (x_0,x_1,\ldots,x_n)I$ cut out different schemes, because one has an embedded point at the origin $(0,\ldots,0)$. Asking that the projective hypersurfaces are the same is equivalent to asking whether $$\mathrm{Proj}(K[Z]/I) = \mathrm{Proj}(K[Z]/J),$$ which is weaker than asking that $K[Z]/I = K[Z]/J$. For the former, one can check equality on the affines $x_i \ne 0$ which cover $\mathbf{P}^n$, but the latter is false in this case, because $F \notin J$ (just by considering degrees, for example).

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