Vector spaces - Multiplying by zero scalar yields zero vector

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The following proof is solely based on vector space related axioms. Axiom names are italicised. They are defined in Wikipedia (see vector space article).

Vector spaces - Multiplying by zero scalar yields zero vector

\begin{array}{lrll} \text{Let} & \dots & \text{be} & \dots \\ \hline & F && \text{a field.} \\ & V && \text{a vector space over $F$.} \\ & 0 && \text{an identity element of addition of $F$.} \\ & \mathbf{0} && \text{an identity element of addition of $V$.} \\ & \mathbf{v} && \text{an arbitrary vector in $V$.} \\ \end{array}

$$\text{Then, }0\mathbf{v} = \mathbf{0}.$$

Proof. We will denote by $1$ an identity element of scalar multiplication; we will denote by $(-\mathbf{v})$ an additive inverse of $\mathbf{v}$. \begin{align*} 0\mathbf{v} &= 0\mathbf{v} + \mathbf{0} && \text{by }\textit{Identity element of vector addition} \\ &= 0\mathbf{v} + (\mathbf{v} + (-\mathbf{v})) && \text{by }\textit{Inverse elements of vector addition} \\ &= (0\mathbf{v} + \mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Associativity of vector addition} \\ &= (0\mathbf{v} + 1\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Identity element of scalar multiplication} \\ &= ((0 + 1)\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Distributivity of scalar multiplication (field addition)} \\ &= ((1 + 0)\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Commutativity of field addition} \\ &= (1\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Identity element of field addition} \\ &= \mathbf{v} + (-\mathbf{v}) && \text{by }\textit{Identity element of scalar multiplication} \\ &= \mathbf{0} && \text{by }\textit{Inverse elements of vector addition} \\ \end{align*} QED

To shorten the proof, we may write as suggested by André Nicolas,

Proof. Let $(V,+,\cdot)_F$ be a vector space over the field $F$. We wish to show that $\forall v\in V$ one has $0\cdot v=\mathbf{0}$, where $0$ is the zero scalar and $\mathbf{0}$ is the zero vector. Let $v$ be an element of the vector space $V$;

By one of the axioms of field addition, $$0\cdot v=(0+0)\cdot v.$$ Since scalar multiplication is distributive over addition, $$(0+0)\cdot v=0\cdot v+0\cdot v.$$ From the previous two equalities we conclude that $$0\cdot v=0\cdot v+0\cdot v.$$ Adding to both sides the inverse element for addition of $0\cdot v$, which we'll denote by $-0\cdot v$: $$0\cdot v+(-0\cdot v)=0\cdot v+0\cdot v+(-0\cdot v).$$ By the inverse axiom, $$\mathbf{0}=0\cdot v+\mathbf{0},$$ hence by the identity axiom, $$0\cdot v=\mathbf{0}.\tag*{$\square$}$$