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I'm working through Shafarevich's Basic Algebraic Geometry book and in one of the exercises (number II.1.15 in my edition), he asks "for what values of $a$ does the surface $x_0^4+x_1^4+x_2^4+x_3^4-ax_1x_2x_3x_4=0$ have singular points?". From the definitions I'm working with, a surface is variety of dimension 2. (Note that Shafarevich's definition of a variety does not assume irreducibility.) If $X=V(F)\subseteq \mathbb P^4$, where $F$ is the polynomial above, why does $X$ have dimension 2? I feel like this is just not true: for example, if $a=0$ then $F$ is irreducible, so $X$ is an irreducible hypersurface, and therefore $X$ should have codimension 1 in $\mathbb P^4$ by Theorem 6.1.2 in Shafarevich ("every irreducible component of a hypersurface in $\mathbb P^n$ has codimension 1). Is Shafarevich just abusing the term "surface" here or am I making some mistake?

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EDIT: I overlooked the fact that the last monomial is $x_1x_2x_3x_4$, and read it as $x_0x_1x_2x_3$. With this in mind, my best guess is that this is a typo; either it should say hypersurface or it should say $x_0x_1x_2x_3$.


Original answer: The question concerns a family of polynomials $\{F_a\}_{a\in k}$ where $$F_a:=x_0^4+x_1^4+x_2^4+x_3^4-ax_0x_1x_2x_3.$$ Each polynomial $F_a\in k[x_0,x_1,x_2,x_3]$ defines a subvariety of codimension $1$ in $\Bbb{P}^3$, not in $\Bbb{P}^4$. This means $F_a=0$ is a surface for each value of $a\in k$.

As an aside, note that these polynomials are symmetric and satisfy $$F_a=e_1^3-5e_1^2e_2+4e_1e_3+2e_2^2-(4+a)e_4,$$ where $e_i$ denotes the $i$-th elementary symmetric polynomial in $k[x_0,x_1,x_2,x_3]$. This clearly shows that $F_a$ is irreducible for all $a\in k$.

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  • $\begingroup$ Why does being homogeneous of degree 4 mean it defines a subvariety in $\mathbb P^3$? The polynomial is in 5 variables... $\endgroup$ – Arbutus Apr 3 at 1:43
  • $\begingroup$ @Arbutus It concerns a family of polynomials, each homogeneous of degree $4$, parametrized by $a\in k$. I have edited to clarify. $\endgroup$ – Inactive - avoiding CoC Apr 3 at 1:46
  • $\begingroup$ I get that, I'm just confused by how you're considering its locus as lying in $\mathbb P^3$. In your answer, it looks like you've just ignored the $x_4$ term by considering $F_a\in k[x_0,x_1,x_2,x_3]$, in which case I get that it's a surface, but are we not supposed to think of $F_a$ as lying in $k[x_0,x_1,x_2,x_3,x_4]$, in which case $V(F)\subseteq \mathbb P^4$? $\endgroup$ – Arbutus Apr 3 at 1:51
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    $\begingroup$ No worries. The polynomial is correct from the book, but curiously the exercise is absent from a later edition (I just checked this...). Hm.. Maybe he just meant to write hypersurface and I should stop obsessing over it.. $\endgroup$ – Arbutus Apr 3 at 1:57
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    $\begingroup$ Just to chip in, I agree that this is a typo. My feeling is that the last monomial should be $x_0x_1x_2x_3$; from experience I know it is easy to forget whether you are calling your homogeneous cooordinates $x_0,\ldots,x_n$ or $x_1,\ldots,x_{n+1}$. $\endgroup$ – Asal Beag Dubh Apr 3 at 10:18

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