Precise definition of affine, smooth, and irreducible A book which I'm reading now says that "the Drinfeld curve 
$$ \mathbf{Y} = \{\, (x, y) \in \mathbf{A}^2(\mathbb{F}) \mid xy^q - yx^q = 1 \,\}$$
is affine, smooth, and irreducible." Here $p$ is an odd prime, $\mathbb{F}$ is the algebraic closure of the field of $p$ elements $\mathbb{F}_p$, and $q$ is a power of the prime $p$. Since I have no solid foundation of algebraic geometry, I would like to make sure that what the precise definitions of those words in this context are.
(Some definitions I see on the internet are overwhelming.)
From the proof I guess the following definitions. Are these correct?
A subset $\mathbf{V}$ of $\mathbf{A}^n(\mathbb{F})$ is 


*

*affine: a closed subspace of $\mathbf{A}^n(\mathbb{F})$ with respect to the Zariski topology 

*smooth: (if $\mathbf{V}$ is affine) for defining polynomials $f_i \in \mathbb{F}[X_1, \dotsc, X_n]$ Jacobi matrix $[\partial f_i/\partial X_j]$ with respect to the formal partial derivative is not zero matrix at all points in $\mathbf{V}$

*irreducible: $\mathbf{V}$ is equal to $V(I) = \{\, x \in \mathbf{A}^n(\mathbb{F}) \mid \forall f \in I,\ f(x) = 0 \,\}$ for some prime ideal $I$ of $\mathbb{F}[X_1,\dotsc,X_n]$

 A: *

*Your definition of affine is correct. 

*This is not correct. You need a stronger condition than saying that the Jacobian matrix is nonzero at each point: it should in fact have maximal rank (i.e. rank equal to the codimension of $V$) at each point. Of course, if the codimension is 1, like in your curve example, those two things are the same. 

*This is correct (in the world of varieties, at least). Maybe a more concrete way to say it, though, is that $I(V)$ is prime, where $I(V)$ means all polynomials vanishing at all points of $V$.

A: Yes it is affine and irreducible since it is cut out from the affine $2$-space by an irreducible polynomial. So you have to show that the polynomial $x y^q - y x^q - 1$ is irreducible.
Notice that $\frac{\partial}{\partial x} (x y^q - y x^q - 1) = y^q$ and $\frac{\partial}{\partial y}(x y^q - y x^q - 1) = -x^q$. Thus, if $(\alpha,\beta)$ lies on the curve and is in the kernel of Jacobi matrix, we get $\beta^q=0$, $\alpha^q=0$ and then $1=\alpha \beta^q - \beta \alpha^q=0$, contradiction. This shows that the $1 \times 2$-Jacobi matrix has full rank and therefore the curve is smooth. (In general, it is not enough to test that the Jacobi matrix does not vanish, but one needs that it has maximal rank.)
