Effective divisor on a curve

Question

I was reading the online courseware by MIT on Arithmetic Geometry and I came across this paragraph:

Let $C/k$ be a (smooth, projective, geometrically irreducible) curve of genus $1$ over a perfect field $k$. Let $n$ be the least positive integer for which Div$_k C$ contains an effective divisor $D$ of degree $n$ (such divisors exist; take the pole divisor of any non-constant function in $k(C)$, for example). If $C$ has a $k$-rational point, then $n = 1$ and $C$ is an elliptic curve.

I do not understand why does such an $n$ exist when Div$_kC$ is the free abelian group generated by the points of $C$, and thus we can obtain any effective divisor of any degree. Furthermore, what does the parenthesized part mean? And the final sentence seems to suggest that if $C$ does not have a $k$-rational point, then $n$ must be greater than $1$.

Answer 1

Q1: $Div(C)$ is generated by the closed points of $C$.

For example if $p$ is a closed point of $C$ with residue field $L(\supset k) $, then $\deg[p]=[L:k].$ So $\deg[p]>1$ if $[L:k]>1.$

Q2: Let $f\in k(C)^*$, a non-constant rational function on $C$, then the $$div(f)=(f)_0-(f)_{\infty}.$$ In particular, $(f)_{\infty}$ is the divisor (hence pole divisor) associated to the poles of the rational function $f$, which is effective. Now $(f)_{\infty}\neq 0$, precisely because $f$ is non-constant. So you can take its degree that will be a positive integer.

Q3: If $C$ does not have a $k$-rational point, then (as seen in answer to Q1 above) for a closed point $p$ with residue field $L\supsetneq k$, $\deg[p]=[L:k]>1$.

Answer 2

No, $\operatorname{Div}_{\overline{k}}(C)$ is the free abelian group generated by the $\overline{k}$-points of $C$; $\operatorname{Div}_{{k}}(C)$ consists of those divisors in $\operatorname{Div}_{\overline{k}}(C)$ that are invariant under the action of $\operatorname{Gal}(\overline{k}/k)$. For instance, consider the affine plane curve $C: x^2 + y^2 = -1$ over $\mathbb{Q}$. Since $C$ has no $\mathbb{Q}$-points, then there is no divisor of degree $1$ in $\operatorname{Div}_{\mathbb{Q}}(C)$. However, the degree $2$ divisor $(0, i) + (0, -i)$ is in $\operatorname{Div}_{\mathbb{Q}}(C)$.