A routine yet very important tool islocalization. The origin of all difficulties in dealing with general rings is that nonzero elements do not always have multiplicative inverse; one cannot easily solve linear equations. The localization is a powerful technique to get around this problem. As we build rational numbers from integers bydeclaring that nonzero numbers have multiplicative inverse, the localization enlarges a given ring and formally allows certain elements to be invertible. It is necessary and sometimes desirable not to invert all nonzero elements, in order for the localization to be useful. For a consistent definition, we need a multiplicatively closed subset S containing 1, but not containing 0, of a ringR and declare that the elements ofS is invertible. The new ring is written as S−1R, in which a usual formula r1
s1 +
r2
s2 =
r1s2+r2s1
s1s2 holds. The original ring naturally
maps intoS−1Rasφ:r7→ r
1. The localization means that one views all data as defined overS
−1R
via the natural mapφ.6
A localized ring, by definition, has more invertible elements, and hence has less nontrivial ideals. In fact, our Laurent polynomial ring is a localized ring of the polynomial ring by inverting monomials, e.g.,{xiyj|i, j ≥0}. Nontrivial ideals such as (x) or (x, y) in the polynomial ring become the unit
ideal (1) in the Laurent polynomial ring. Further localizations in this thesis are with respect to prime ideals. In this case, we say the ring islocalized at a prime ideal p. A prime idealp has a defining property that ab /∈p whenevera /∈p and b /∈p. Thus, the set-theoretic complement of
p is a multiplicatively closed set containing 1. In (R\p)−1R, denoted by R
p, any element outside
p is invertible, and therefore p becomes a unique maximal ideal ofRp. Moreover, the localization
sometimes simplifies the generators of an ideal. For instance, ifR=F[x, x−1] and p= (x−1), the
ideal ((x−1)(x5−x+ 1))⊆R localizes to (x−1)
p ⊆Rp sincex5−x+ 1 is an invertible element
ofRp.
An important fact about the localization is that a module is zero if and only if its localization at every prime ideal is zero. Further, the localization preserves exact sequences. So we can analyze a complex by localizing at various prime ideals. For a thorough treatment about localizations, see Chapter 3 of [80]. The term ‘localization’ is from geometric considerations where a ring is viewed as a function space on a geometric space.
Lemma 3.4.1. LetI be the associated ideal of an exact code Hamiltonian, andm be a prime ideal of R. Then,I6⊆mimplies that the localized homology
K(L)m= ker(L)m / im(σL)m
is zero for all L≥1.
It is a simple variant of a well-known fact that a module over a local ring is free if its first non-vanishing Fitting ideal is the unit ideal [79, Chapter 1 Theorem 12].
Proof. Recall that the localization and the factoring commute. By assumption,
(Iq())m= (Iq(σ))m= (1) =Rm =:S.
6 Recall that the local ringS has the unique maximal idealm, and any element outside the maximal ideal is a unit. If every entry ofis inm, thenIq()⊆m6=S. Therefore, there is a unit entry, and
by column and row operations,is brought to
∼= 1 0 0 0
where0 is a submatrix. It is clear thatIq−1(0)⊆Iq() since anyq−1×q−1 submatrix of0 can
be thought of as aq×qsubmatrix ofwhere the first column and first row have the unique nonzero entry 1 at (1,1). It is also clear thatIq−1(0)⊇Iq() since anyq×q submatrix ofcontains either
zero row or column, or the (1,1) entry 1 of . Hence, Iq−1(0) = (1), and we can keep extracting
unit elements into the diagonal by row and column operations [79, Chapter 1 Theorem 12]. Afterq steps,t×2qmatrixbecomes precisely
∼= idq 0 0 0
where idq is theq×qidentity matrix. Since localization preserves the exact sequenceG→P →E,
σmaps to the lowerqcomponents ofP with respect to the basis whereis in the above form. Since Iq(σ) = (1), we must have (after basis change)
σ∼= 0 0 idq 0 .
Therefore, even after factoring by the proper ideal bL, the homologyK(L) = kerL / imσL is still
zero.
Corollary 3.4.2. The associated ideal of an exact code Hamiltonian is the unit ideal, i.e.,Iq(σ) =R,
if and only if
K(L) = kerL / imσL = 0
Proof. IfI(σ) =R, I(σ) is not contained in any prime idealm. The above lemma saysK(L)m = 0.
Since a module is zero if and only if its localization at every prime ideal is zero, K(L) = 0 for all L≥1.
For the converse, observe that if F is any extension field ofF2, for any F2-vector spaceW, we
have dimFF⊗F2W = dimF2W. We replace the ground field F2with its algebraic closureF
a to test
whetherK(L)6= 0. IfIq(σ) is not the unit ideal, then it is contained in a maximal idealm(R. By Nullstellensatz, m= (x1−a1, . . . , xD−aD) for someai ∈Fa. Since in R any monomial is a unit, we haveai6= 0. Therefore, there exists L≥1 such thataLi = 1 and 2-L. The equation x
L−1 = 0
has no multiple root.
We claim that K(L) 6= 0. It is enough to verify this for the localization atm. Since anything outsidemis a unit inRm and eachxLi −1 contains exactly one xi−ai factor, we see (bL)m=mm.
Therefore, (L)m =m/(bL)m and (σL)m =σm/(bL)m is a matrix over the field R/m =Fa. Since Iq(σ) ⊆ m, we have Iq(σL)m = 0. That is, rankFa(σL)m < q. It is clear that dimFaK(L)m =
dimFaker(L)m/im(σL)m ≥1.
This corollary says that in order to have adegenerateHamiltonianH(L), one must have a proper associated ideal. We shall simply speak of adegeneratecode Hamiltonian if its associated ideal is proper.