Is this definition of linear code redundant?

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A subset $C ⊆ F^n_q$ is a linear code if

1.

$u + v ∈ C$ for all $u, v ∈ C$

(if $u = (u_1, u_2, . . . , u_n)$, $v = (v_1, v_2. . . . , v_n)$ then $u + v = (u_1 +_q v_1, u_2 +_q v_2 . . . , u_n +_q v_n))$

2.

$au ∈ C$ for all $u ∈ C$, and all $a ∈ GF(q)$ if $u = (u_1, u_2, . . . , u_n)$, then $au = (au_1, au_2, . . . , au_n))$

Isn't this definition redundant? $au = u + u +u ... + u$ (added up $a$ times). I think the second part of the definition is basically the same as the first one.

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This definition is in general not redundant.

While it is true that if we are working over a field $GF(p)$ where $p$ is prime, any element of the field can be written as $1+1+\cdots+1$, in which case the closure under addition is all that is needed.

However, we can have fields $GF(q)$ where $q$ is a prime power, and not prime; here we need the second condition, since if $a$ is not in the base field, then it is not possible to write $a=1+1+\cdots+1$.

For example, the field $GF(4) = \{0,1,\alpha,\alpha+1$ where $\alpha$ satisfies $\alpha^{2}+\alpha+1=0$. Then the code $C=\{(0,0),(1,1)\}$ is closed under addition (since we still have $1+1=0$ in $GF(4)$), but is not closed under multiplication by $\alpha$ or $\alpha+1$.