Is the cylinder an algebraic variety?

Question

$$ \cases { x^2 + y^2 = 1 \\ z = z } $$ where the 2nd equation can be regarded as $\varnothing$.

Is this an algebraic variety?

If yes, is the surface defined by the intersection of 2 (perpendicular) cylinders also an algebraic variety?

Thanks.

Answer 1

$V(x^2+y^2-1, x^2 + z^2 - 1) = \{(x,y,z) \in k \mid x^2 + y^2 = 1 \wedge x^2 + z^2 = 1\}$ is the set of points defined by the intersection of two perpendicular cylinders. So is $W(x^2+y^2-1, y^2 + z^2 - 1)$.