Irreducible components of $x_1x_4 -x_2x_3 = x_1x_3 - x_2^2=0$ I am going through introductory course on commutative algebra.
Let us have an algebraic set $X$ in $\mathbb{C}^4$ that is described as 
$x_1x_4 -x_2x_3 (f) = x_1x_3 - x_2^2 (g) =0$. 
I am asked to find irreducible subsets of X, and to compare $I = (f,g)$, the ideal $I(X)$ which is all the polys that $h|_X$=0 and the intersection of ideals of the irreducible components.
So far
$f$ is a cone with the origin at $(0,0,0,0)$. 
One of the components seem to be $x_1=x_2=0$

Here is a similar task that gives a way to make a problem easier in case there are elements that are reducible.
The task is taken from here and I have solved the same question for 
 $(x_1x_2-x_3^2=x_3-\lambda(x_1+x_2)=0)$ for $\lambda \in \mathbb{C}$ which is an intersection of a plane and a cone. But I can't imagine 4-dimensional space. I used the fact that homogeneous polynomial of second power is always reducible. May be that could help here.
Update. I have found three related questions:
Checking one of prime ideals of X,
Krull dimension of I(X),
Groebner base for the system.
thanks in advance
 A: Assume $P=(a_1,a_2,a_3,a_4)\in X$ satisfies $a_1=0$. Then $a_2=0$ and $a_3,a_4\in\mathbb{C}$ can be arbitrary. This shows that $X_1=V(x_1,x_2)\subseteq X$, where $X_1$ is irreducible: $(x_1,x_2)$ is a prime ideal of $\mathbb{C}[x_1,x_2,x_3,x_4]$, because the factor ring $\mathbb{C}[x_1,x_2,x_3,x_4]/(x_1,x_2)\cong\mathbb{C}[x_3,x_4]$ is a domain. Note that a priori this doesn't imply that $X_1$ is an irreducible component of $X$, because $X_1$ could be properly contained in an irreducible component.
There are points $P\in X\setminus X_1$, for example points $P$ with $a_2\neq 0$. Take such a point $P$ and let $X_2$ be the irreducible component of $X$ containing $P$. Let $I_2:=I(X_2)$, a prime ideal minimal among the prime ideals containing $(f,g)$. 
Multipliying the first polynomial by $x_3$ and substituting one gets 
$x_1x_3x_4-x_2x_3^2=x_2^2x_4-x_2x_3^2=x_2(x_2x_4-x_3^2)$.
One has $x_2(x_2x_4-x_3^2)\in (f)\subseteq I_2$. Since $I_2$ is prime and $x_2\not\in I_2$ one gets $x_2x_4-x_3^2\in I_2$, that is $(x_2x_4-x_3^2,f,g)\subseteq I_2$. According to the second of the posts you are citing the ideal $(x_2x_4-x_3^2,f,g)$ is prime, hence $I_2=(x_2x_4-x_3^2,f,g)$.
Note that $X_1\not\subseteq X_2$, since $I_2\not\subseteq (x_1,x_2)$. Hence one can hope that $X=X_1\cup X_2$ is the decomposition of $X$ into irreducible components: let $P\in X\setminus X_1$, then $a_1\neq 0$. One thus gets
$a_2a_4=\frac{a_2^2a_3}{a_1}=\frac{a_1a_3^2}{a_1}=a_3^2$,
so that $P\in X_2$.
