affine variety from ideal I have a problem with this excercise:
*Take a field $F$, algebraic closed, $charF\neq 2$.
1)Consider the following ideal $I=(xy+yz+zx-\frac{1}{2}u^2 , x+y+z+u)$ in $F[x,y,z,u]$.
Is $Z(I)$ affine variety in $A^4$?
2)Consider the following ideal $I=(x^2+y^2+z^2,x+y+z)$ in $F[x,y,z]$. Is $Z(I)$ affine variety in $A^3$*\
How can I check, that it is variety?
I think, that my solutions are incorect, or noncomplete. For now I have:\
1)$u=-x-y-z$, then $F[x,y,z,u]/I=F[x,y,z]/(x^2+y^2+z^2)$, polynomial $x^2+y^2+z^2$ is irreducible, so $Z(I)$ must be variety.\
2)$x=-y-z$, then $F[x,y,z]/I=F[y,z]/(2(y^2+yz+z^2))$
 A: (2)The answer is no in characteristics other than 3,for which we have to show I is radical,but not prime.For this it's enough to show K[x,y,z]/I has no nilpotent ,but is not an Integral Domain.
   We observe the ring homomorphism K[x,y,z] →K[x,y] induced by x → x,y → y,z → -x-y has kernel (x,y,z) and thus it induces an isomorphism.
  K[x,y,z]/I=K[x,y]/(x^2+y^2+(-x-y)^2) ,Since, z=-x-y
            =K[x,y]/(x^2+xy+y^2)
If Characteristic is not 3,the x^2+xy+y^2 is a product of two distinct linear factors,which shows the quotient is reduced but not Integral,thus Z(I) is not variety.
If characteristic is 3,then (x^2+xy+y^2) is equal to (x-y)^2,so the radical of Ideal it generates is generated by x-y .The quotient by this radical is isomorphic to K[x] ,so is an Integral Domain,hence I(Z(I)) is prime and Z(I) is a variety.
A: (1) We have,
$$K[x,y,z,u]/I=K[x,y,z]/(xy+yz+zx-(1/2)*(x+y+z)^2)$$ 
Since $$u=-x-y-z
            =K[x,y,z]/(x^2+y^2+z^2)$$
and as we have established $x^2+y^2+z^2$ is obviously irreducible in Field theory,so the ideal it generates is prime,showing the quotient is Integral Domain and so $Z(I)$ is a affine variety.
