about affine variety

Question

let $ f(x,y,z) \in \mathbb{R}[x, y, z] $ such that :

$$ f(x,y,z) =x^{4}+x^{2} y^{2}+x^{2} z^{2}-x^{3} y-x^{3} z-x^{2} z y+\\ x^{2} y^{4}+y^{6}+y^{4} z^{2}-x y^{5}-x z y^{4}-z y^{5}+\\ x^{2} z^{6}+y^{2} z^{6}+z^{8}-x y z^{6}-x z^{7}-z^{7} y $$

i want to show that that $Z(f(x,y,z))$ is affine variety in $\mathbb{A^3}_{\mathbb{R}}$ .

we know $Z(f)=\left\{P \in \mathbb{A^3}_{\mathbb{R}} \mid f(P)=0\right. \}$

now $Z(f)$ is close and it's enough that we show $Z(f)$ is irreducible . on the other hand we know an algebraic set is irreducible if and only if its ideal is a prime ideal.so we have $I(Z(f))=\sqrt{f}$ so we must show that $\sqrt{f}$ is prime . also if we can show $f$ is irreducible then $Z(f)$ is irreducible.

Answer 1

Not a complete solution but I hope it helps.

Generally, when such a task is given to you (like homework), there is good hope there is some significant way how to make it much simpler.

1. In the first row you can factor out $x^2$ in second $y^4$ and in the third, you can factor out $z^6$. Observe that you have the same remainder in each row.

So we can write: $f(x,y,z)=(x^2+y^4+z^6)(x^2+y^2+z^2-yx-zx-zy).$

So $Z(f)=Z(x^2+y^4+z^6)\cup Z(x^2+y^2+z^2-yx-zx-zy)$.

But $Z(x^2+y^4+z^6)=(0,0,0)$ and that is also an element of $Z(x^2+y^2+z^2-yx-zx-zy)$ so you can simply it all to:

$Z(f)=Z(x^2+y^2+z^2-yx-zx-zy)$

2)Now one can guess that $x=y=z$ are zeroes. So the set contains affine line. If we can prove that no other points our in $Z(f)$ then points from this line, we have a solution, because line is irreducible over infinite field.