# The norm

To be precise, the norm structure we hope to have for Milnor K-groups
is a morphism \(N_{k'/k}:K_n^M(k')\rightarrow
K_n^M(k)\) for any finite extension \(k'/k\). When over \(\mathbb{P}_k^1\), this must behave like a
*reciprocity* law, that is, satisfies \[ \sum_\nu
N_{\kappa(\nu)/k}(\partial_\nu(\alpha))=0 \] for every \(\alpha\in K_{n+1}^M(k(t))\). Apart from
this, it has to satisfies some more properties, that is,

- functorial: \(N_{k'/k}N_{k''/k'}=N_{k''/k}\),
- projection formula: \(N_{k'/k}(x\cdot i_{k'/k}(y))=N_{k'/k}(x)\cdot y\) for every finite (simple) extension \(k'/k\). (Well we’d better stick to separable extensions at this time)

This norm construction appears firstly in [BT73], which is proved by the steps

- Construct the norm: For any finite extension \(k'/k\), embed \(k'\) in \(k(t)\) at valuation \(\mu\). Then we see the diagram

induces all the *norm*s, so take \(\nu\) to be \(\mu\). Clearly this satisfies the
reciprocity law, at least for \(k(t)\).

- This is the tricky part: prove those \(N_\nu\) do not actually depend on our
selection of the embedding, or to prove this is the
*only*correct norm.

To tell the truth, I don’t like this proof, which seems to be in much
chaos and is not so enlightening. It has long be known that we can
interpret Milnor K-groups (even algebraic K, but too far-reaching) in
the terms of intersection theory, and [MVW06] gives a construction
showing \[ H^{n,n}(k,\mathbb{Z})=K_n^M(k)
\] where \(H^{n,n}\) is the
motivic cohomolgy. However, his proof is not so satisfying, since his
construction of a pair of morphisms between motivic cohomology and
Milnor K is depending on the existence of the norm map and Weil
recprocity. So I decided to (try to) develop an alternative apporach
that does not really requiring *any* preliminary knowledge about
the Milnor K. Wish me good luck and forward!

# Singular homology

Following [SV96], fix a base field \(k\) and all the schemes are considered
being locally of finite type over \(k\), which means that dimensions behave
well. Define the \(n\)-simplex \(\Delta_k^n\) as the hypersurface \(\mathbb{A}_k^{n+1}/(T_0+\cdots+T_n-1)\).
With the obvious face \(\partial_i\)
and degeneracy \(s_i\) maps this turns
out to be a cosimplicial object. For any \(X\rightarrow Y\) over \(k\), define \(C_0(X/Y)\) to be the abelian group
generated by cycles on \(X\) which are
finite and surjective over \(Y\).
Furthermore we let assign a simplicial object \(C_\bullet(X/k)\) to \(X/k\) namely \(C_n(X/k)=C_0(X\times\Delta^n/\Delta^n)\).
Finally we assign obvious complex \(C_*(X/k)\) (with differentials as
alternating face sums) to \(X/k\) and
our **singular homology** \(H_i^{sing}(X/k)\), or \(H_i(X/k)\) if there’s no danger of
confusion, is defined to be the homology of such complex. The base
scheme \(S\) in \(C_\bullet(X/S)\) controls the dimension
really well and for a cover \(U\coprod
V\rightarrow X\) we have Mayer-Vetoris

and

We now define \(\mathbb{G}_m\) to be \(\mathbb{A}_k^1\backslash\{0\}\), and \((\mathbb{G},*)\) be the virtual scheme which maps to the cokernel of

We may say that \(\mathbb{G}_m\) is
*pointed* (in this case, by \(1:*\rightarrow\mathbb{G}_m\)) and \((\mathbb{G},*)\) is *punctured*.

### Calculation of \(H^{sing}_i(\mathbb{G}_m,*/k)\)

From [SV96] we know that \(H_i(\mathbb{G}_m/k)\) is \(\mathbb{Z}\oplus k^*\) when \(i=0\) and \(0\) otherwise. We may just use this conclusion since the higher homology group is not so straightforward to calculate, but rather do some small observations on \(H_0\). By definition \(H_0(\mathbb{G}_m/k)\) is the cokernel of

\(C_0(\mathbb{G}_m/k)\) is very easy
since every integral subscheme of \(\mathbb{G}_m\) of codimension \(1\) is indeed represented by \(f(t)=t^n+\cdots a_0\in k[t]\) where \(a_0\not=0\). What about \(C_1(\mathbb{G}_m/k)\)? Thanks to the
smoothness, a codim \(1\) cycle \(Z\) of \(\mathbb{G}_m\times\mathbb{A}^1\) can be
represented by some \(g(t,x)=b_n(x)t^n+\cdots
b_0(x)\in k[t,x]\) where \(b_0(x)b_n(x)\) is not zero. The requirement
that \(Z\) being surjective over \(\mathbb{A}^1\) is equivalent to say that
those \(b_i(x)\) cannot be trivial at
the same time, and being *finite* requires V.Voedosky’s small
trick: we may view \(\mathbb{G}_m\times\mathbb{A}^1\) as an open
subscheme of \(\mathbb{G}_m\times\mathbb{P}^1\), and \(Z\) being finite is equivalent to being
*closed* in those two schemes both, or to say \(Z\) coincides with it’s own projective
closure. The only two suspecious points to check is \(t=0\), which means \(b_0(x)\in k^*\), and \(t=\infty\), which means that \(b_n(x)\in k^*\). We may further assume that
\(b_n(x)=1\), then \(\partial_0-\partial_1\) gives \[
D(t^n+\cdots+b_0)\sim D(t^n+b_0)
\] This means that the *only* remaning information in
\(H_0(\mathbb{G}_m/k)\) coming from
\(C_0(\mathbb{G}_m/k)\) is just degree
and the coefficient of constant term, that is, \(\mathbb{Z}\oplus k^*\). We’d rather take
the second term to be \((-1)^na_0\) to
match the intuition of the product of all positions of the roots.

What about the punctured one \(H_0(G_m,*/k)\)? We know that the point object \(C_\bullet(k/k)\) is constant as \(\mathbb{Z}\), and the point takes \(n\) to \((x-1)^n\). By modding out this degree \(1\) term in the exact sequence \[ \cdots\rightarrow H_0(k/k)\rightarrow H_0(\mathbb{G}_m/k)\rightarrow H_0(\mathbb{G}_m,*/k)\rightarrow 0 \] we see immediately that \(H_0(\mathbb{G}_m,*/k)=k^*\). The same long exact sequence also suggests that higher homology groups of \((\mathbb{G}_m,*)\) vanishes, just as \(\mathbb{G}_m\)’s do.

# The ring construction

It’s already clear that \(H_0(\mathbb{G}_m,k)\) is our basic piece
\(k^*\), and now we’re gonna construct
the higher part of Milnor K. For a pair of schemes \(X\) and \(Y\) over \(S\) and assuming all the smoothness we
required, there’s a homomorphism between abelian groups \[
prod:C_0(X/S)\otimes_\mathbb{Z}C_0(Y/S)\rightarrow C_0(X\times Y/S)
\] namely sending a pair \((Z\hookrightarrow X,W\hookrightarrow Y)\)
to \((X\times W)\cdot(Z\times Y)\)
where \(\cdot\) is the intersection
product over \(X\times Y\), which turns
out to be supported on the zeroth \(Tor\) part \((X\times W)\times(Z\times Y)\) which is
finite and surjective over \(S\),
therefore well defined. *(Surjectiveness needs some further
treamtment here, but mostly from cycle-theoretic intersection theory.
I’ll refine the related note later.)* As a result, a further
morphism is induced, namely

Eilenberg-Zilber theorem suggests that \(\phi\) is a quasi-isomorphism (and even a chain homotopical equivalance) so we get a derived morphism, the dotted arrow in \(D^-(Ab)\). Taking homology group we get a “cup product” \[ \cup:H_i(X/S)\otimes H_j(Y/S)\rightarrow H_{i+j}(X\times Y/S) \] with cycles representing homology classes goes in the inverse direction against \(\phi\)

It’ll be quite complicated dealing with the higher homologies.
However, a good news is we only need to deal with the zeroth part of the
theory. Namely, the \(n\)-fold cup
product at zero degree \[
H_0(X_1/S)\otimes\cdots\otimes H_0(X_n/S)\rightarrow
H_0(X_1\times\cdots\times X_n/S) \] is determined by the related
production of zero cycles. For two punctured (virtual) schemes \((S,*)/k\) and \((T,*)/k\), we see from our construction
that there’s a well defined **smash product** \(C_*((S\wedge T,*)/k)\) quasi-isomorphic to
\[ 0\rightarrow C_*(*)\rightarrow
C_*(S)\oplus C_*(T)\rightarrow C_*(S\times T)\rightarrow 0. \]
The interesting part of this is the punctured point representing \(0\) in the zeroth relative homology groups
kills the cup product, namely \[ *\cup
c'=c\cup *=0. \] This suggests that we should consider the
\(n\)-fold smash product of \((\mathbb{G}_m,*)/k\), and then there’s the
ring \(M(k)\) defined to be \(\oplus_{n\geq
0}H_0(\wedge^n(\mathbb{G}_m,*)/k)\). The production on \(M(k)\) is straightforward as the cup
product, and this is our candidate for Milnor K-theory of the field
\(k\). To prove this *is* Milnor
K indeed, we need to see the following exact sequence \[ 0\rightarrow I\rightarrow\otimes
H_0((\mathbb{G}_m,*)/k)\rightarrow M(k)\rightarrow 0 \] where
\(I\) is the two-sided ideal generated
by degree \(2\) elements like \(c\otimes c'\) where \(c+c'=1\).

### The first step: surjectiveness

We now prove that \(\otimes^nH_0((\mathbb{G}_m,*)/k)\rightarrow H_0(\wedge^n(\mathbb{G}_m,*)/k)\) is a surjection. Fix a smooth scheme \(X/k\), and denote the presheaf sending any smooth \(Y/k\) to \(C_0(X\times_k Y/Y)\) by \(\mathbb{Z}_{tr}(X)\) (In fact, this is the same signature used in [MVW06]), then we know that \(\mathbb{Z}_{tr}(X)\) is an abelian sheaf on \(Sm/k\) with etale topology. This is proved in [MVW06] as a fundemental theorem for etale motives. Consider the sheaf \(\mathbb{Z}_{tr}(\bullet)\) restricted to \(Et/k\), we have the product discussed above \[ prod:C_0(X/k)\otimes C_0(Y/k)\rightarrow C_0(X\times Y/k). \] Passing to affine case, it’s easily seen from the following diagram

that intersecting with a closed point is exact, hence the production of closed points with residue field \(l\) and \(l'\) respectively is a formal sum of closed points on \(X\times Y\) with terms coincide with the closed points of \(l\otimes_k l'\). This suggests that if our base scheme \(k\) is algebraically closed, then \(C_0(X/k)\otimes C_0(Y/k)\) should always equal to \(C_0(X\times Y/k)\). Since \(\mathbb{Z}_{tr}(X)\) and \(\mathbb{Z}_{tr}(Y)\) are sheaves on \(Et/k\), we see that \[ C_0(X/k)\otimes_{\mathbb{Z}}^{et} C_0(Y/k)\rightarrow C_0(X\times Y/k) \] is an isomorphism, where \(\otimes_{\mathbb{Z}}^{et}\) is the tensor product in the category of \(Ab(Et/k)\). Now use this on the punctured \(\mathbb{G}_m\), we have

where \(\wedge^nC_0(\mathbb{G}_m,*)\) is the \(n\)-fold tensor product of complex of presheaves representing \(C_0(\mathbb{G}_m,*)\), and the second term is the sheafication(already being a seperated presheaf so just take the zeroth Cech cohomology), which is quasi-isomorphic to the third term by our comments above. Remember what we’re gonna proof is that the morphism \[ \otimes^nk^*=\otimes^nH_0(\mathbb{G}_m,*)\rightarrow H_0(\wedge^n(\mathbb{G}_m,*)) \] is surjective, which is given by the sheafication

being surjective on global section. Namely, a global section of \((\wedge^nC_0(\mathbb{G}_m,*))^+(k)\) is determined uniquely by a section \(\wedge^nC_0(\mathbb{G}_m,*)(k')\) where \(k'/k\) is galois. Now the remaning problem is to check that the \(Gal(k'/k)\)-module \(\otimes^n(k')^*\) has a fixed subgroup \(\otimes^nk^*\). (Something is going wrong about this part. I’ll alter my approach.)

# References

[SV96] Suslin, Andrei, and Vladimir Voevodsky. “Singular Homology of Abstract Algebraic Varieties.” Inventiones Mathematicae 123, no. 1 (December 1996): 61–94. https://doi.org/10.1007/BF01232367.

[MVW06] Mazza, Carlo, Vladimir Alexandrovich Voevodsky, and Charles A. Weibel. Lecture Notes on Motivic Cohomology. Clay Mathematics Monographs, volume 2. Providence (R.I.): American mathematical society, 2006.

[BT73] H. Bass and J. Tate, The Milnor ring of a global field, Algebraic K-theory II (Battelle Memorial Inst., Seattle, Washington, 1972), Lecture Notes in Math., vol. 342, Springer-Verlag, New York, 1973