Is $V=\mathbb{Z}^{4}_{7}$ a inner product space?

Question

Is $V=\mathbb{Z}^{4}_{7}$ a inner product space?

$<v,w>=v_1w_1+v_2w_2+v_3w_3+v_4w_4, \mod{7}$

We know that $\mathbb{Z}_7=\{0,1,2,3,4,5,6\}$

Let's consider $<v,v>$. We want to show that $<v,v>=0$ iff $v=0$ is false.

Let $v=[3,1,2,0]$

Then $<v,v>=3(3)+1(1)+2(2)+0(0), \mod{7}$

$=14\mod{7}=0$

But $v\neq 0$, therefore this is not an inner product space.

Is this correct? I'm not sure if the mod applies while adding each piece (like each $v_iw_i$, or in the end? Because in $\mathbb{Z}_7$, $4+4=1$, and so did I do this correctly?

Answer 1

You are totally correct.

Recall that $\mathbb{Z}_7 \cong \mathbb{Z}/7\mathbb{Z}$ with a natural projection morphism $f: \mathbb{Z} \rightarrow \mathbb{Z}_7$ acting via $f(n) = n \pmod 7$. The fact that $f$ is a ring homomorphism means that it doesn't matter whether you apply $\operatorname{mod}$ before or after adding and multiplying: you get the same answer.

To be specific with your example:

$$0 \pmod 7 = f(14) = f(3\cdot3+1\cdot1+2\cdot2+0\cdot0) = \\ f(3)\cdot f(3)+f(1)\cdot f(1)+f(2)\cdot f(2)+f(0)\cdot f(0) = 0$$