Is $0\,\mathrm{m}$ dimensionless?
This is really not that big of a deal. In essence, though, the dimensions of zero are ill-defined, and $[0]$ (i.e. the dimensions of zero) is (a) not meaningfully defined, and (b) never used in practice.
Let me start by making one thing clear:
Would it be improper to drop the units of measurement of $0 \,\mathrm m$ like this in an academic paper?
Yes, this is perfectly OK, and it is standard practice.
The dimensionality map $[\,·\,]$ has multiple different conventions, but they all work in much the same way. The key fact is that physical quantities form a vector space under multiplication, with exponentiation (over the field $\mathbb Q$) taking the role of scalar multiplication. (This is the essential reason why dimensional analysis often boils down to systems of linear equations, by the way.) The different base dimensions - mass, length, time, electric charge, etc. - are assumed to be algebraically independent and to span the space, and the dimensionality map $[\,·\,]$ reads out the 'coordinates' of a given physical quantity in terms of some pre-chosen canonical basis.
This only works, however, if you exclude zero from the game. The quantity $1\,\mathrm m$ has a multiplicative inverse, but $0\,\mathrm m$ does not, so if you included it it would break the vector space axioms. This is in general OK - you're not forced to keep those properties - but it does preclude you from using the tools built on that vector space, most notably the dimensionality map. Thus $[0]$ doesn't map to anything.
Since you explicitly asked for it, here's one way you can formalize what I said above. (For another nice analysis, see this nice blog post by Terry Tao.)
- A positive physical quantity consists of a 8-tuple $(x,m,l,t,\theta,c,q,i) \in\mathbb R^\times\times \mathbb Q^7$, where $\mathbb R^\times=(0,\infty)$ is the real multiplicative group. This is usually displayed in the form $$ x\,\mathrm{kg}^m\mathrm{m}^l\mathrm s^t\mathrm K^\theta\mathrm A^c \mathrm{mol}^q \mathrm{cd}^i.$$
- The multiplication of two physical quantities $p=(x,m,l,t,\theta,c,q,i)$ and $p'=(x',m',l',t',\theta',c',q',i')$ is defined as $$pp'=(xx',m+m',l+l',t+t',\theta+\theta',c+c', q+q',i+i').$$ The multiplicative identity is $1=(1,0,0,0,0,0,0,0)$, and the multiplicative inverse of $p$ is $1/p=(1/x,-m,-l,-t,-\theta,-c,-q,-i)$.
- The exponentiation of a physical quantity $p$ to an exponent $r\in\mathbb Q$ is defined as $$p^r=(x^r,rm,rl,rt,r\theta,rc,rq,ri).$$
You can then easily check that these two operations satisfy the vector space axioms. The above construction is, in fact, a specific instantiation of the abstract vector space $\mathcal Q$ of physical quantities, but it suffices to take one specific example to show that this works.
As an aside, the choice for $\mathbb Q$ as the scalar field is because (a) it's essential for the vector space structure, and (b) it's still somewhat reasonable to have things like $\mathrm {m}^{-3/2}$ (e.g. the units of a wavefunction). On the other hand, things like $\mathrm {m}^{\pi}$ cannot be made to make sense.
The dimensionality map $[\,·\,]$ is, first and foremost, an equivalence relation, that of commesurability. That is, we say that for $p,p'\in\mathcal Q$, $$[p]=[p']\Leftrightarrow p/p'=(x,0,0,0,0,0,0,0)\text{ for some }x\in\mathbb R^\times.$$ This is in fact all you really need to do dimensional analysis, as I argued here, but it's still useful to go on for a bit.
The really useful vector space, if you want to do dimensional analysis, is the vector space of physical dimensions: the space of physical quantities once we forget about their numerical value. This is the quotient space of $\mathcal Q$ over the commesurability equivalence relation: $$\mathcal D=\mathcal Q/[\,·\,]=\{[p]\,:\,p\in\mathcal Q\}.$$ (Here I've abused notation slightly to make $[p]$ the equivalence class of $p$, i.e. the set of all physical quantities commensurable to $p$.) The vector space of physical dimensions, $\mathcal D$, has the same operations as in $\mathcal Q$:
- $[p][p']=[pp']$, and
- $[p]^r=[p^r]$.
It is easy to check that these definitions do not depend on the specific representatives $p$ and $p'$, so the operations are well-defined.
Dimensional analysis takes place in $\mathcal D$. From the definitions above, you can prove that the seven base units give rise to a basis $\{[1\,\mathrm {kg}], [1\,\mathrm m],\ldots,[1\,\mathrm{cd}]\}$ for $\mathcal D$. More physically, though,
- the seven base units are algebraically independent, which means that they cannot be expressed as multiples of each other, and
- they are enough to capture the dimensions of all physical quantities.
These are the key physical requirements on a set of base units for the abstract space $\mathcal Q$.
After this, you're all set, really. And it should be clear that there's no way to make zero fit into this scheme at all.
Is $0~\mathrm{m}=0~\mathrm{s}=0~\mathrm{kg}=0$?
No. Not according to the below. (Which has loads of citations.)
From http://www-ksl.stanford.edu/knowledge-sharing/papers/engmath.html:
For each physical dimension, the class of constant scalar quantities of a physical dimension forms an abelian group with the addition operator + and a zero identity element for that dimension (zeros of each dimension are different). The class of all scalars of any dimension, after removing the zero scalars, forms an abelian group with respect to multiplication. [My emphasis.]
From http://www-ksl.stanford.edu/knowledge-sharing/ontologies/html/physical-quantities/physical-quantities.lisp.html:
A zero quantity is one which, when multiplied times any quantity, results in another zero quantity (possibly the same zero). The class of zero quantities includes the number 0, and zero quantities for every physical dimension and order of tensor.
Q: Why not make one zero-thing which follows our [i.e. your] intuition?
A: We would have to make exceptions for all of our operators on quantities that depend on physical-dimension or tensor-order. [My emphasis.]
Additionally, from the BIPM ("Respect My Authoritah"):
The value of a quantity is generally expressed as the product of a number and a unit. The unit is simply a particular example of the quantity concerned which is used as a reference, and the number is the ratio of the value of the quantity to the unit. [My emphasis.]
I'm fairly sure that the only exception to 'generally' above is when the unit is the number one, $1$.
So, still no.