Simultaneously aligned bases

SIMULTANEOUSLY ALIGNED BASES

KEITH CONRAD

Let R be a PID, n be a positive integer, and M be a finite free R-module of rank n.

By the structure theorem for modules over a PID, for any submoduleM⁰ ofM also having rankn(to be called afull submoduleofM) we can find a basise₁, . . . , e_n ofM and nonzero a₁, . . . , a_ninR such thata₁e₁, . . . , a_ne_n is a basis ofM⁰. We call such a pair of bases ofM and M⁰ aligned.

Pick two full submodules of M, say M⁰ and M⁰⁰. If there is a basis e₁, . . . , e_n of M and two sets of nonzero a⁰₁, . . . , a⁰_n and a⁰⁰₁, . . . , a⁰⁰_n inR such that

M =

n

M

i=1

Rei, M⁰=

n

M

i=1

Ra⁰_iei, M⁰⁰=

n

M

i=1

Ra⁰⁰_iei

then we’ll sayM⁰ and M⁰⁰ admitsimultaneously aligned bases. Do such bases always exist?

Of course if R is a field then they do because the only full submodule of M is M, so the situation is trivial.

The following example shows simultaneously aligned bases need not exist in R² if R is not a field.

Example 1. Let R be a PID that is not a field, so R contains prime elements. Letπ be prime in R. Inside R² set

(1) M⁰ =R

1 0

+R

0 π²

= x

y

:y ≡0 modπ²

and

(2) M⁰⁰ =R

π 0

+R 1

π

= x

y

:y≡0 modπ, πx≡ymodπ²

.

First we determine an aligned basis for M⁰ and for M⁰⁰ as submodules of R². The first one is easy: M⁰=R ¹₀

+Rπ^{2 0}₁

, so we can use{ ¹₀ , ⁰₁

}as a basis ofR² and { ¹₀ , π^{2 0}₁

} as a basis of M⁰. For M⁰⁰, we rewrite it as

M⁰⁰=R 0

π²

+R 1

π

=Rπ² 0

1

+R 1

π

, so we can use { ⁰₁

, _π¹

} as a basis of R² and {π^{2 0}₁ , ¹_π

} as a basis of M⁰⁰. Using these aligned bases we see thatR²/M⁰ andR²/M⁰⁰ are both isomorphic to R/(π²).

Suppose there is some basis {e₁, e2} of R² and nonzero a1, a2, b1, b2 in R such that {a₁e₁, a₂e₂} is a basis of M⁰ and {b₁e₁, b₂e₂} is a basis of M⁰⁰. We are going to get a contradiction. Since R²/M⁰ ∼=R/(a₁)×R/(a₂) and R²/M⁰⁰ ∼= R/(b₁)×R/(b₂), from the known structure of R²/M⁰ and R²/M⁰⁰ we have

(3) (a₁a₂) = (π²), (b₁b₂) = (π²).

Writee1 = ^x_y¹

1

and e2 = ^x_y²

2

, so being a basis ofR² is equivalent to

(4) x₁y₂−x₂y₁∈R^×.

Granting (3), to have {a₁e1, a2e2} be a basis of M⁰ and {b₁e1, b2e2} be a basis of M⁰⁰ is equivalent to havinga₁e₁ and a₂e₂ lying in M⁰ and b₁e₁ and b₂e₂ lying in M⁰⁰.

1

2 KEITH CONRAD

Havinga₁e₁= ^a_a^1x1

1y1

anda₂e₂= ^a_a^2x2

2y2

inM⁰ is equivalent toa₁y₁, a₂y₂≡0 modπ². By (4),y₁ and y₂ can’t both be divisible byπ, so one ofa₁ ora₂ is divisible by π². Therefore by (3), {(a₁),(a2)} = {(1),(π²)}. So far the roles of e1 and e2 have been symmetric, so without loss of generality we can take

(a1) = (1), (a2) = (π²).

Thereforey₁ ≡0 modπ², soy₂ 6≡0 modπ (because y₁ and y₂ are relatively prime).

Having b₁e₁ = ^b_b^1x1

1y1

and b₂e₂ = ^b_b^2x2

2y2

in M⁰⁰ implies b₁y₁, b₂y₂ ≡ 0 modπ, so b₂ ≡ 0 modπ. It also implies, by (2), that πb1x1 ≡ b1y1 modπ² and πb2x2 ≡ b2y2 modπ². Since y1 is a multiple of π² and b2 is a multiple of π, these congruences mod π² become πb₁x₁ ≡0 modπ² and 0≡b₂y₂ modπ². Since y₂ is not a multiple ofπ,b₂ ≡0 modπ², so from (3) we have (b1) = (1) and (b2) = (π²). Thereforeπb1x1≡0 modπ² ⇒x1 ≡0 modπ.

But x₁ and y₁ can’t both be multiples of π since they are relatively prime, so we have a contradiction.

We now seek a criterion on pairs of full submodules that determines when they have simultaneously aligned bases. WhenM is a finite freeR-module andM⁰ is a full submodule with aligned bases {e₁, . . . , en} for M and {a₁e1, . . . , anen} for M⁰, the linear operator A:M → M where A(e_i) = a_ie_i has image M⁰ and detA = a₁· · ·a_n 6= 0. Conversely, if A:M →M is a linear operator with nonzero determinant, then A(M) is a full submodule of M with (detA) = (c₁· · ·c_k) as ideals, where M/A(M) has the cyclic decomposition R/(c1)× · · · ×R/(ck). Therefore the full submodules ofM are the same thing as images of linear operatorsA:M →M with nonzero determinant, and detAis determined up to unit multiple by the structure ofM/A(M) as anR-module. Writing a full submoduleM⁰ ofM asA(M) for some linear operator Aon M, how much does M⁰ determineA?

Lemma 2. If A₁ and A₂ are two linear operators on M with nonzero determinant, then A1(M) =A2(M) if and only if A1=A2U for some U ∈GL(M).

Proof. Let e1, . . . , en be a basis of M. IfA1(M) =A2(M) then A1(ei) = A2(fi) for some f_i ∈M. Let U: M → M be the linear map satisfying U(e_i) =f_i for all i. Then A₁(e_i) = A2(U(ei)) =A2U(ei), so by linearityA1(m) =A2U(m) for allm∈M, and thusA1 =A2U. From A₁(M) = A₂(M) we get M/A₁(M) = M/A₂(M), so detA₁ and detA₂ are equal up to unit multiple. Then the condition detA1 = (detA2)(detU) implies detU ∈R^×, so U ∈GL(M).

Conversely, ifA1 =A2U withU ∈GL(M) thenA1(M) =A2(U(M)) =A2(M).

By this lemma, if we write a full submodule ofM asA(M) for some A∈End(M), then A is determined byA(M) up to right multiplication by an element of GL(M).

Pick two full submodules ofM, sayA(M) and B(M), with simultaneously aligned bases:

there is a basise1, . . . , enofM and two sets ofnnonzeroa1, . . . , anandb1, . . . , bninR such that

M =

n

M

i=1

Re_i, A(M) =

n

M

i=1

Ra_ie_i, B(M) =

n

M

i=1

Rb_ie_i.

LetD:M →M andD⁰:M →M be the linear maps defined byD(e_i) =a_ie_i andD⁰(e_i) = biei. Written as matrices with respect to the basis e1, . . . , en, both D and D⁰ become diagonal matrices, so Dand D⁰ are diagonalizable operators onM. EasilyA(M) =D(M) and B(M) =D⁰(M), so D=AU and D⁰ =BV for some U and V in GL(M). Obviously Dand D⁰ commute, soAU and BV commute. We now show the converse is true too.

SIMULTANEOUSLY ALIGNED BASES 3

Theorem 3. Choose A and B in End(M) with detA 6= 0 and detB 6= 0. Suppose there are U and V in GL(M) such that AU and BV commute and are diagonalizable. Then the submodules A(M) and B(M) of M have simultaneously aligned bases.

Proof. Set A⁰ = AU and B⁰ = BV, so A⁰(M) = A(M) and B⁰(M) = B(M). Since A⁰ is diagonalizable, there is a basis e1, . . . , en of M and nonzero a1, . . . , an in R such that A⁰(ei) =aiei for alli. Then

M =

n

M

i=1

Re_i, A⁰(M) =

n

M

i=1

RA⁰(e_i) =

n

M

i=1

Ra_ie_i.

Letλ1, . . . , λ_k be the distinct values amonga1, . . . , anand setMj ={v∈M :A⁰(v) =λjv}

(this is theλ_j-eigenspace ofA⁰). Eache_iis in someM_j, soM =M₁+M₂+· · ·+M_k. Elements from differentMj’s are linearly independent (same as proof in vector spaces that eigenvectors for different eigenvalues of a linear operator are linearly independent). Therefore

M =M₁⊕ · · · ⊕M_k.

For v∈M_j,A⁰(B⁰v) =B⁰(A⁰v) = B⁰(λ_jv) = λ_j(B⁰v), so B⁰(M_j) ⊂M_j for all j. Let d_j be the rank ofM_j. Since M_j is a finite free R-module, the structure theorem for modules over a PID says there is a basise1j, . . . , e_djj of Mj and nonzero c1j, . . . , c_djj inR such that

M_j =Re_1j⊕ · · · ⊕Re_djj, B⁰(M_j) =Rc_1je_1j ⊕ · · · ⊕Rc_djje_djj. Then

M =

k

M

j=1

Mj =

k

M

j=1 dj

M

`=1

Re_`j,

B(M) =B⁰(M) =

k

M

j=1

B⁰(M_j) =

k

M

j=1 dj

M

`=1

Rc_{`j</sub>e<sub>`j},

and

A(M) =A⁰(M) =

k

M

j=1

A⁰(M_j) =

k

M

j=1

λ_jM_j =

k

M

j=1 dj

M

`=1

Rλ_je_`j.

We have found simultaneously aligned bases forA(M) andB(M) in M. Let’s consider now any finite number of full submodules, not just two. The definition of simultaneously aligned bases for more than two full submodules of a finite freeR-module is clear: a basis for the whole module that can be scaled to a basis of each of the submodules.

Example 4. If we view the ring of integers of a number field as a Z-module, any finite set of nonzero ideals in it has simultaneously aligned Z-bases. This is proved in [1], where Example1 also appears for the caseR=Zand π= 3.

Corollary 5. For r ≥ 2 and A₁, . . . , A_r in End(M) with nonzero determinants, the sub- modules A1(M), . . . , Ar(M) of M have simultaneously aligned bases if and only if there are U1, . . . , Ur inGL(M)such thatA1U1, . . . , ArUr are diagonalizable and pairwise commuting.

In particular, if A₁, . . . , A_r are diagonalizable and pairwise commuting in End(M) with nonzero determinants then the submodules A1(M), . . . , Ar(M) of M have simultaneously aligned bases.

Proof. If there are simultaneously aligned bases forA₁(M), . . . , A_r(M), then the same ar- gument as before leads toU1, . . . , Urin GL(M) such thatA1U1, . . . , ArUrare diagonalizable and pairwise commuting.

4 KEITH CONRAD

Conversely, suppose there are U₁, . . . , U_r in GL(M) such that A₁U₁, . . . , A_rU_r are diag- onalizable and pairwise commuting operators on M. Set A⁰₁ =A1U1, . . . , A⁰_r =ArUr. We want to show the submodules A₁(M), . . . A_r(M) have simultaneously aligned bases in M.

Since A⁰₁(M) =A₁(M), . . . , A⁰_r(M) =A_r(M), we can replace A₁, . . . , A_r with A⁰₁, . . . , A⁰_r: to showA⁰₁(M), . . . , A⁰_r(M) have simultaneously aligned bases when A⁰₁, . . . , A⁰_r are diago- nalizable and pairwise commuting, we will proceed by the same inductive argument that is used to show a set of commuting diagonalizable operators on a finite-dimensional vector space are simultaneously diagonalizable.

Since A⁰₁ is diagonalizable, there is a basis e1, . . . , en of M and nonzero a1, . . . , an inR such thatA⁰₁(e_i) =a_ie_i for all i, so

M =

n

M

i=1

Rei, A⁰₁(M) =

n

M

i=1

RA⁰₁(ei) =

n

M

i=1

Raiei.

Letλ1, . . . , λ_k be the distinct values amonga1, . . . , an. Then as before, M =M₁⊕ · · · ⊕M_k,

where M_j = {v ∈ M : A⁰₁(v) = λ_jv} (and M_j 6= {0}). As before, each M_j is preserved by A⁰₂, . . . , A⁰_r and the restrictions of these operators¹ to Mj are pairwise commuting with nonzero determinant. Once we show the restrictions of A⁰₂, . . . , A⁰_r toM_j are each diago- nalizable, then by induction on the number of operators there are simultaneously aligned bases forA⁰₂(Mj), . . . , A⁰_r(Mj) as submodules ofMj (that is, eachMj has a basis that can be scaled termwise to provide a basis of those submodules). All elements ofM_j are eigen- vectors forA⁰₁, so by stringing together bases of M1, . . . , Mk to give a basis of M we have a simultaneously aligned basis for A⁰₁(M), . . . , A⁰_r(M) in M, and then we’d be done (since A⁰₁(M) =A1(M), . . . , A⁰_r(M) =Ar(M)).

References

[1] H. B. Mann and K. Yamamoto, “On canonical bases of ideals,” J. Combinatorial Theory2(1967), 71–76.

1We have no reason to expectA2, . . . , Arpreserve theMj’s.