Decomposition of the algebraic variety into irreducible components I want to decompose the algebraic variety $v(\mathfrak{a}) \subset \mathbb{A}^3(\mathbb{C}),$ where $\mathfrak{a}$ is an ideal, $\mathfrak{a} = (xy+yz, x^3y^3 + x^2y^2),$ into irreducible components, and find $\mathbf{rad}(\mathfrak{a}).$
I understand in common how to manage with this, and after some effort I have got
$v(\mathfrak{a}) = v(y) \cup v(x,z) \cup v(x+z, xy +1).$
But I can not prove that the ideal $(x+z, xy+1)$ is prime (unlike easier cases with other two ideals).
I tried to proof that $(x+z, xy+1)$ is a kernel of a map $f: \mathbb{C}[x,y,z] \to \mathbb{C}(t),$ where
$x \to t, y \to \frac{-1}{t}, z \to -t.$ But I had no success. May be I'm wrong and this ideal is not prime. In this case there must be another decomposition.
My approach for finding $\mathbf{rad}(\mathfrak{a})$ is also using that $(x+z, xy +1)$ is prime ideal.
 A: First, please note that $(x + z, xy + 1)$ is not the kernel of a map $f : \mathbb C[x,y,z] \to \mathbb C(t)$. It is the kernel of a map $f: \mathbb C[x, y, z] \to \mathbb C[t]_t$ (where $\mathbb C[t]_t = \mathbb C[t, t^{-1}]$).
So I suggest you break down the proof into two steps.
Step 1: You can show that $\mathbb C[x, y, z]/(x + z) \cong \mathbb C[x, y]$. You can then show that, under this isomorphism, the ideal $(x + z, xy + 1)/(x + z) \subset \mathbb C[x, y, z]/(x + z)$ maps to the ideal $(xy + 1) \subset \mathbb C[x, y]$. Hence, $\mathbb C[x, y, z]/(x + z, xy + 1) \cong \mathbb C[x, y]/(xy + 1)$ by the Third Isomorphism Theorem.
[Edit: It may help to use the fact that $\mathbb C[x, y, z] \cong \mathbb C[x, y, u]$ via $u := z + x$. It is then easy to see that $\mathbb C[x, y, u]/(u) \cong \mathbb C[x, y]$.]
So to prove that $(x + z, xy + 1)$ is a prime ideal in $\mathbb C[x, y, z]$, you merely need to prove that $(xy + 1)$ is prime in $\mathbb C[x, y]$. (This is because $\mathbb C[x, y, z]/(x + z, xy + 1)$ is an integral domain iff $\mathbb C[x, y]/(xy + 1)$ is an integral domain.)
Step 2: You can show that $\mathbb C[x, y]/(xy + 1) \cong \mathbb C[t]_t$ using the universal property of localisations (see here). And clearly $\mathbb C[t]_t$ is an integral domain.
Alternatively, you can show that $(xy + 1)$ is prime in $\mathbb C[x, y]$ directly. Since $\mathbb C[x, y]$ is a unique factorisation domain, it suffices to show that $xy + 1$ is irreducible, and you can do that by elementary methods.

Edit: As for finding ${\rm rad}(\mathfrak a)$...
By the Nullstellensatz, ${\rm rad}(\mathfrak a)$ consists of all polynomials that vanish on $V(\mathfrak a)$.

*

*The polynomials that vanish on $V(y)$ are the elements of ${\rm rad}(y) = (y)$.

*The polynomials that vanish on $V(x, z)$ are the elements of ${\rm rad}(x,z) = (x,z)$.

*The polynomials that vanish on $V(x + z, xy + 1)$ are the elements of ${\rm rad}(x + z, xy + 1) = (x + z, xy + 1)$.

So the polynomials that vanish on $V(\mathfrak a) = V(y) \cup V(x, z) \cup V(x + z, xy + 1)$ are the elements of $(y) \cap (x, z) \cap (x + z, xy + 1)$.
