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?
 A: 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
$$F_a=e_1^4-5e_1^2e_2+4e_1e_3+2e_2^2-(4+a)e_4,$$
where $e_i$ denotes the $i$-th elementary symmetric polynomial in $k[x_0,x_1,x_2,x_3]$. These polynomials are easily seen to be irreducible in $k[e_1,e_2,e_3,e_4]$ as they are linear in $e_3$. They are in fact irreducible in $k[x_0,x_1,x_2,x_3]$ as well; for any factor of $F_a$ its symmetric conjugates are also factors of $F_a$. So if $F_a$ has a quadratic factor then this factor has precisely two conjugates, hence it is stabilized by $A_4\subset S_4$. Then it is of the form
$$c_1(x_0^2+x_1^2+x_2^2+x_3^2)+c_2e_2=c_1(e_1^2-2e_2)+c_2e_2,$$
and so it is in fact symmetric, a contradiction. Similarly, if $F_a$ has a linear factor then this factor has precisely four conjugates, hence it is stabilized by some $S_3\subset S_4$. Therefore it is conjugate to
$$c_1x_0+c_2(x_1+x_2+x_3),$$
for some constants $c_1$ and $c_2$. Comparing coefficients quickly yields a contradiction.
