An normed space is defined as a vector space V plus a norm operation over $V$.

Is it meaningful to generalize this notion to modules, where one is dealing with rings instead of fields?

What I'm actually interested is to find the shortest non-zero "vector" over a subspace of, say $\mathbb{Z}_6^3$. As the addition and scalar multiplication are performed over a ring rather than a field, is defining a Euclidean norm $\|v\|_2=\sqrt{v\cdot v}$ meaningful?

PS: I'm actually asking this to better understand a comment raised by Daniel Lichtblau on another question of mine on Mathematica.SE.


To find something which is "shortest" you need at the very least some means of comparison, so your "norm" would need to have values in some (hopefully linearly) ordered set.

The expression $\sqrt{v\cdot v}$ doesn't make much sense, either, until you give some meaning to $\sqrt\cdot$, which sounds troublesome, especially in cases like ${\bf Z}_6$ when you have zero divisors.

In any case, trying to assign something like norm to modules over rings which aren't even domain seems rather hopeless, as you will have vectors which multiplied by a zero divisors yield zero, which is bad. Similar problem occurs in case of any rings of positive characteristic, including fields, where you can add a vector to itself a few times to get zero.

You could, perhaps, thing something up in case of domains of characteristic zero, but that's not that far from a subfield of complex numbers case discussed in Wikipedia (up to adding quotients; a problem could arise in case and overly large cardinality (larger than $\mathfrak c$), which is probably not what you have in mind).

That said, you certainly can assign some value within some total order to an element of a module over arbitrary ring, but I don't think you can find something that would deserve being called a norm in general.

  • $\begingroup$ Reading your answer, I first tried to define the norm as a function over $\mathbb{R}$, rather than the underlying ring. For instance, for $(1,4,5) \in \mathbb{Z}_6^3$, the norm will be $\sqrt{1^2+4^2+5^2}=\sqrt{42} \approx 6.5$. However, such definition will not satisfy a property of norms: $\|cv\|=|c|\cdot\|v\|$ for any $c$ in the underlying ring and any $v$ in the module. This property does not seem to be satisfied even when the underlying ring is actually a field like $\mathbb{Z}_7$. $\endgroup$ – M.S. Dousti Jul 30 '13 at 18:18

If you have a normed ring, then it makes good sense to study normed modules over the normed ring (with some compatibility between the norms on the ring and on the module). This is a standard consideration in those parts of functional analysis that are closer to algebra (and in particular, in non-archimedean functional analysis).

On a finite field, the only sensible norm is the discrete norm, defined by $|a| = 0$ if $a = 0$ and $|a| = 1$ otherwise. But this becomes slightly more interesting in your case, because $\mathbb Z_6$ is not a field. Rather, by the Chinese remainder theorem, we have $\mathbb Z_6 = \mathbb Z_2 \times \mathbb Z_3$, the product of two finite fields, and so we can give it a kind of product norm. Actually, we have two choices, and $l_1$ version and an $l_{\infty}$ version.

In the former, we add up the norms on the components; in the latter, we take the max. of the norms of the two components.

So concretely, for the $l_1$ version, if $x = (a,b) \in \mathbb Z_2 \times \mathbb Z_3$, we define $|x| = 0$ if $a = b = 0$, $|x| = 1$ if $a$ or $b = 0$, but not both, and $|x| = 2$ if neither $a$ nor $b = 0$.

In the $l_{\infty}$ version, $|x| = 0$ if $x = 0$ and $|x | = 1$ otherwise.

The $l_{1}$ version looks more interesting to me, since it is more sensitive to the product structure.

You could now extend this to $\mathbb Z_6^3$, again in an $l_1$-way, i.e. $|(x_1,x_2,x_3)| = |x_1| + |x_2| + |x_3|$.

Basically, this measures how different $(x_1,x_2,x_3)$ is from zero, by measuring how far each component $x_i$ is from zero, both mod $2$ and mod $3$.

I didn't look at the mathematica post you link in your question, so I don't know if this will help you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.