# Does $x_0^4+x_1^4+x_2^4+x_3^4-ax_1x_2x_3x_4$ really define a surface?

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?

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 $$a\in k$$.
• Why does being homogeneous of degree 4 mean it defines a subvariety in $\mathbb P^3$? The polynomial is in 5 variables... – Arbutus Apr 3 at 1:43
• @Arbutus It concerns a family of polynomials, each homogeneous of degree $4$, parametrized by $a\in k$. I have edited to clarify. – Inactive - avoiding CoC Apr 3 at 1:46
• 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 $V(F)\subseteq \mathbb P^4$? – Arbutus Apr 3 at 1:51
• 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 $x_1,\ldots,x_{n+1}$. – Asal Beag Dubh Apr 3 at 10:18