I was trying to understand affine varieties, and I considered something that I quickly deduced was false:
- False: If the vanishing ideal has even one irreducible generator, then the zero set is irreducible.
So I considered $I=(x,yz)\in \Bbb C[x,y,z]$.
I saw that $$V(x,yz)=V(x)\cap V(yz)=V(x)\cap(V(y)\cup V(z))= (V(x)\cap V(y))\cup (V(x)\cap V(z))$$ $$=V(x,y)\cup V(x,z)$$ Now at this step I was happy. I can see that our variety is just the $y$ and $z$ axis.
- Question 1: Can I break up the variety by intersection whenever I see a comma?
- Question 2: Can I break up the variety by union, whenever I see a product?
With these two, which I have used above, I can then see that $V(x,yz)=V(x^2,xz,xy,yz)$, but the ideal inside the bracket on the right, cannot generate $x$? My ideal isn't equal to its radical anymore, which it was originally?
- Question 3: I suppose we can't distinguish between $V(x,yz)$ and $V(x^2,yz)$ can we?
- Question 4: $V(I)$ is only irreducible, when every generator is?
For $Q4$, my thought is that if there is a single reducible generator, say $f_1$ in $I=(f_1,\cdots,f_n)$ and all $f_i$ for $i>1$ are irreducible, then by $Q1$ I have: $$V(I) = V(f_1)\cap \cdots \cap V(f_n),$$ and say $f_1=g_1g_2$, then I have: $$V(I) = (V(g_1)\cup V(g_2))\cap V(f_2)\cap \cdots\cap V(f_n)$$ $$= \mathbf{(}V(g_1)\cap V(f_2)\cap \cdots \cap V(f_n)\mathbf{)}\cup \mathbf{(}V(g_2)\cap V(f_2)\cap \cdots\cap V(f_n)\mathbf{)}$$ and thus we have shown $V(I)$ is reducible?