In my lecture notes we have the following:
Mordell proved the following:
Let $C$ be a nonsingular cubic curve with rational coefficients. Then the abelian group of rational points on $C$ is finitely generated.
$$E|_{\mathbb{Q}} \ \ \ \ y^2=x^3+ax+b$$ $ \left( a, b \in \mathbb{Z} \ \ , \ \ \ D(f) \neq 0 \right )$
$$E(\mathbb{Q})=\{(\alpha , \beta ) \in \mathbb{Q} \times \mathbb{Q} | \beta^2=\alpha^3+a \alpha +b \} \cup \{ \left [0, 1, 0\right ]\}$$
That means that $$E(\mathbb{Q})=E(\mathbb{Q})_t \times \mathbb{Z}^r, r \geq 0$$
$$E(\mathbb{Q})_{\text{torsion}}=\{P \in E(\mathbb{Q}) | P \text{ of finite order }\}$$
$\mathbb{Z}^r$ is the direct product of cyclic groups of finite order.
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Can you explain why $E(\mathbb{Q})$ is the union of $\{(\alpha , \beta ) \in \mathbb{Q} \times \mathbb{Q} | \beta^2=\alpha^3+a \alpha +b \} $ and $\{ \left [0, 1, 0\right ]\}$ ???
The group contains the rational points of a non-Singular cubic curve, so it contains the set $\{(\alpha , \beta ) \in \mathbb{Q} \times \mathbb{Q} | \beta^2=\alpha^3+a \alpha +b \} $ . But why does it also contain the set $\{ \left [0, 1, 0\right ]\}$ ???
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How do we conclude to $E(\mathbb{Q})=E(\mathbb{Q})_t \times \mathbb{Z}^r, r \geq 0$ ??? What does this mean ???
There are two somewhat unrelated questions:
The "exceptional" point ought to be the "point at infinity" (though you do not give enough details to be certain); to have a nice group-law on the curve you need to consider it in projective space not just the affine part.
$E(\mathbb{Q})$ is a commutative group. For any commutative group $G$ one can consider $G_t$ the subset of elements of finite order. This forms a subgroup called the torsion subgroup of $G$. Every finitely generated commutative group is of the form $G_t \oplus \mathbb{Z}^r$ where $G_t$ is the torsion subgroup, an in this case finite commutative group. (This is a result from group theory and does not really depend on the specific of this situation.)