Algebraic curve

Remarks

Conventions differ on whether a curve must be irreducible, reduced, projective or smooth; each text states its choice. For an integral variety $X$ over $k$, the dimension equals the transcendence degree of its function field over $k$. So $X$ is a curve exactly when $k(X)$ has transcendence degree $1$. Example: the plane curve $X \colon y^2 = x^3 - x$ in $\mathbb{A}^2_k$ is a curve. Its function field $k(x)[y]/(y^2 - x^3 + x)$ is a finite extension of $k(x)$, so it has transcendence degree $1$. Non-example: the affine plane $\mathbb{A}^2$ has dimension $2$, and a finite set of points has dimension $0$, so neither is a curve. Let $k$ be algebraically closed. Then each finitely generated field extension of $k$ of transcendence degree $1$ is the function field of exactly one smooth projective curve, up to isomorphism.