Irreducibility of an explicit complex projective variety

Question

Let $Y\subset \mathbb P^n_\mathbb C$ be a subvariety defined by a series of homogeneous polynomials $f_1, \ldots, f_t$. Is there an effective way to determine the irreducibility of $Y$ as an algebraic variety?

It seems to me that computers are not able to do such calculations since we are in $\mathbb C$ (compare to the possibility to determine the irreducibility if the base field was $\mathbb Q$).

In particular, the variety that interests me is the surface $\Sigma$ in $\mathbb P^4$ (using $x_1, \ldots, x_5$ as coordinates) defined by $$ x_3^2 + 2x_1x_4 - 4x_2x_4 +2x_2x_5 - 4x_1x_5 = 0$$ and $$ x_4(x_1^2 - 2x_2^2) + x_5(x_2^2 - 2x_1^2) = 0. $$

Using Macaulay2, we managed to show that the variety is irreducible over $\mathbb Q$:

Macaulay2, version 1.22
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems,
               Isomorphism, LLLBases, MinimalPrimes, OnlineLookup,
               PrimaryDecomposition, ReesAlgebra, Saturation, TangentCone

i1 : R = QQ[x1, x2, x3, x4, x5]

o1 = R

o1 : PolynomialRing

i2 : q = x3^2 + 2*x1*x4 - 4*x2*x4 +2*x2*x5 - 4*x1*x5

       2
o2 = x3  + 2x1*x4 - 4x2*x4 - 4x1*x5 + 2x2*x5

o2 : R

i3 : g = x4*(x1^2 - 2*x2^2) + x5*(x2^2 - 2*x1^2)

       2        2        2       2
o3 = x1 x4 - 2x2 x4 - 2x1 x5 + x2 x5

o3 : R

i4 : I = ideal(q, g)

              2                                        2        2        2    
o4 = ideal (x3  + 2x1*x4 - 4x2*x4 - 4x1*x5 + 2x2*x5, x1 x4 - 2x2 x4 - 2x1 x5 +
     ---------------------------------------------------------------------------
       2
     x2 x5)

o4 : Ideal of R

i5 : primaryDecomposition I

               2                                        2        2        2    
o5 = {ideal (x3  + 2x1*x4 - 4x2*x4 - 4x1*x5 + 2x2*x5, x1 x4 - 2x2 x4 - 2x1 x5 +
     ---------------------------------------------------------------------------
       2
     x2 x5)}

o5 : List

i6 : isPrimary I

o6 = true

i7 : isPrime I

o7 = true

Is there any possibility to see if the surface $\Sigma$ is irreducible over $\mathbb C$?

Answer 1

Let me explain how to show that the projective surface $\Sigma$ is geometrically irreducible (see also the comments above).

First, we know that $\Sigma$ is irreducible (this was checked by the OP).

Also, a routine calculation should reveal that the point $(1:2:0:0:0)$ is nonsingular (the OP also checked this).

As the point (1:2:0:0:0) is obviously $\mathbb{Q}$-rational, it follows from the following lemma that $\Sigma$ is geometrically irreducible (i.e., $\Sigma_{\mathbb{C}}$ is irreducible).

Lemma. Let $X$ be an irreducible finite type scheme over a field $k$. Suppose that there is a point $P\in X(k)$ such that $X$ is smooth at $P$. Then $X$ is geometrically irreducible.

Proof. Let $S\subset X$ be the smooth locus of $X$. Note that $S$ is a dense open subscheme. (It is always open. Density follows from the fact that it is non-empty and that $X$ is irreducible: any non-empty open of an irreducible scheme is dense.) Since $S$ is a dense open of an irreducible scheme, it is itself irreducible. Now, by assumption $S(k)\neq \emptyset$. Therefore, $S$ is geometrically connected. (To see this, use that Galois action of the absolute Galois group permutes the connected components of $S_{\overline{k}}$, but fixes $P$.) Since $S$ is smooth and geometrically connected, it is geometrically irreducible. It follows that $S_{\overline{k}}$ is an integral dense open subscheme of $X_{\overline{k}}$. Thus $X$ is geometrically irreducible. QED