linear algebra - 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