# Real norms on vector spaces over finite fields

I am interested in functions of the form $\psi: F^n \to \mathbb{R}^+$, where $F$ is a finite field, that have norm-like properties, e.g., $\psi(x+y) \le \psi(x) + \psi(y)$. Does anybody know if there is any literature on this area?

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Exactly which properties do you want? If you want something that behaves like the usual $\lVert \alpha\mathbf{v}\rVert =|\alpha|\,\lVert\mathbf{v}\rVert$ for scalars $\alpha$, then the "absolute value" function is going to have to map everything to $1$. Then it's just a matter of mapping the basis vectors and making sure any vector that has nonzero $i_1,\ldots,i_k$ coordinates has image less than $\psi(\mathbf{e}_{i_1})+\cdots+\psi(\mathbf{e}_{i_k})$, and that would satisfy the conditions. Seems like there is just too much freedom, so there would be very little to be said. – Arturo Magidin Aug 27 '11 at 0:28
@Arturo: I seem to have gotten that positive homogeneity yields $\psi(x)=0$ for all $x\in F$. Are you mixing up identities, or did I mess up? – robjohn Aug 27 '11 at 0:39
@robjohn: I may not be considering all of them. But positive homogeneity is derived from several of the properties of the norm so I don't know which ones we would have "on hand". – Arturo Magidin Aug 27 '11 at 3:10
@Arturo: you said that $\lVert \alpha\mathbf{v}\rVert =|\alpha|\,\lVert\mathbf{v}\rVert$ implies that "absolute value" maps everything to $1$. I got that $\psi(rx)=|r|\psi(x)$ implies that $\psi(x)=0$. The difference between $0$ and $1$ was my concern. – robjohn Aug 27 '11 at 3:45
@robjohn: Dratted comments. Originally, I had that since $\alpha^{p^n}=\alpha$ for some $n$, then it follows that you will have $|\alpha|^{p^n}=|\alpha|$ for whatever function $|\cdot|$ will represent. This forces $|\alpha|=0$ or $|\alpha|=1$, and if you want non-triviality, one needs the latter (but it got lost in the shuffle). Note that you cannot really talk about arbitrary $r\in\mathbb{Q}$ in what you do, because $rx$ only makes sense for "rationals" with denominator prime to $p$... – Arturo Magidin Aug 27 '11 at 3:47

If you are truly only interested in the triangle inequality, then there is the Hamming weight $w(x_1,x_2,\ldots,x_n)=m$ where $m$ is simply the number of non-zero components. This gives you a metric. Mind you, the space $F^n$ is finite, so any metric on it is going to give you the discrete topology. Adding any kind of norm-like requirements (on top of the triangle inequality) is problematic for several reason, as others have pointed out.
The triangle inequality might allow some interesting functions, but positive homogeneity doesn't. A finite field has a prime characteristic, $p$. Positive homogeneity says $\psi(rx)=|r|\psi(x)$ for $r\in\mathbb{Q}$. To satisfy this property, $\psi(0)=\psi(px)=p\psi(x)$, which first says $\psi(0)=0$ and then $\psi(x)=0$ for all $x\in F$.
As Arturo Magidin notes, we can't use all $r\in\mathbb{Q}$, just the ones whose denominator is not divisible by $p$. – robjohn Aug 27 '11 at 4:12