On \(\PP^1\), any two points satisfy \([a] \sim [b]\) by \(f(z) = (z-a)/(z-b)\). Degree-zero divisors are sums of such differences, so \(\Pic^0(\PP^1) = 0\) and \(\Pic(\PP^1) \cong \ZZ\) by degree.
On an elliptic curve \(E\), distinct points \(p \neq q\) are never linearly equivalent: \(p - q = \div(f)\) would make \(f : E \to \PP^1\) a degree-one map, hence an isomorphism, contradicting \(g(E) = 1 \neq 0 = g(\PP^1)\). Instead \(\Pic^0(E) \cong E\) by \(p \mapsto [p - O]\), so \(\Pic(E) \cong E \oplus \ZZ\).
The contrast is the whole content: \(\Pic^0\) measures how far the curve is from being rational, and it vanishes exactly for \(\PP^1\). A curve is rational precisely when it admits two distinct linearly equivalent points, which is the same as admitting a degree-one map to \(\PP^1\).
The degree argument is the one to reproduce: a principal divisor \(p - q\) gives a rational function with one simple zero and one simple pole, so the induced map to \(\PP^1\) has degree one. For a non-rational example, \(y^2 = x^3 - x\) homogenised to \(y^2 z = x^3 - xz^2\) is the standard witness, and the group law on it is exactly the isomorphism \(E \cong \Pic^0(E)\).
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PR-Y5S7VThe degree of a divisor on a projective curveuses