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Closed surfaces and character varieties

The powerful character variety techniques of Culler and Shalen can be used to find essential surfaces in knot manifolds. We show that module structures on the coordinate ring of the character variety can be used to identify detected boundary slopes as well as when closed surfaces are detected. This approach also yields new number theoretic invariants for the character varieties of knot manifolds.


Introduction
Suppose that N is a compact, irreducible 3-manifold with torus boundary and that X is an irreducible algebraic component of the SL 2 C-character variety for N . It is known that the dimension of X is at least one [6]. It is also known that X contains a great deal of topological information about N .
Let X → X be a birational map from a smooth projective curve. This map is defined for all but finitely many points. These points are called ideal points. For γ ∈ π 1 N and χ ∈ X, define I γ (χ) = χ(γ). This determines a rational function I γ : X → C. This paper concerns the following landmark result of Culler and Shalen.
Theorem 1 (Culler-Shalen). Ifx is an ideal point, there is an associated non-empty essential surface Σ in N .
(1) Σ may be chosen to have empty boundary if and only if I α is regular atx for every peripheral element α ∈ π 1 N . (2) Otherwise, there exists a unique slope with the property that if α ∈ π 1 N represents this slope then I α is regular atx. In this case, every component of ∂Σ represents the slope corresponding to α.
When an essential surface arises from X and this theorem, we say that the surface is detected by X.
Since 1983, case (2) of Theorem 1 has been carefully studied. Many famous papers have provided both applications and insight for this case. Notably, the Culler-Shalen norm of [9] and the A-polynomial of [6] can tell us exactly which slopes are detected. These tools are remarkably effective and have been used in proofs of several famous theorems including the Smith Conjecture [22] and the Cyclic Surgery Theorem [9]. In contrast, there is very little in the literature concerning case (1) of Theorem 1.
It has also been difficult to interpret global properties of X in terms of the topology of N . This paper introduces new functions which relate detected boundary slopes and essential closed surfaces in the manifold to module structures of the coordinate ring C[X]. In the spirit of [7], these functions also contain number theoretic information about X.
The trace ring T (X) is the subring of the coordinate ring C[X] which is generated by {I γ | γ ∈ π 1 N }. Define T Q (X) to be the smallest Q-algebra in C[X] which contains T (X). Suppose that α ∈ π 1 N is primitive and peripheral. Then C[X], T Q (X), and T (X) have the structure of a C[I α ]-module, a Q[I α ]-module, and a Z[I α ]-module, respectively. Let Rk C X (α), Rk Q X (α), and Rk Z X (α) denote the ranks of these modules. Note that the functions I α are well-defined on slopes. Hence, these rank functions may be viewed as functions on S, the set of slopes for N . It is clear that (1) Rk C X ≤ Rk Q X ≤ Rk Z X . The following theorem is the main theorem of this paper.  (2) Otherwise, Rk X X (α) = ∞ if and only if X detects the slope α. We begin by proving the theorem for X = C. A straightforward application of this case yields a proposition of independent interest. Proposition 1.1. Let X • be the union of the irreducible components X of the SL 2 C character variety such that (1) X detects a closed essential surface if and only if X does, and (2) for each slope α, X detects a surface with boundary slope α if and only if X does.
Then X • is defined over Q.
This proposition, together with Theorem 1, also allows us to improve the result to include X = Q. As an application of this result, we prove another proposition. Proposition 1.2. If X contains the character of a non-integral representation which detects a closed essential surface, then X detects a closed essential surface.
Together with data compiled by Goodman, Heard, and Hodgson, this proposition gives many examples of hyperbolic knot manifolds for which the hyperbolic components of their character varieties detect closed essential surfaces. We also use this proposition in our proof that the main theorem holds in the case X = Z.
In addition to our main result, we begin an investigation of the basic properties of the functions Rk X X and include computations for several examples. It is easy to prove the following two propositions. Proposition 1.3. Suppose that N is a knot manifold and H 1 (N ; Z) ∼ = Z, generated by α. Let X A ⊆ X(N ) be the curve consisting of all abelian representations of π 1 N . Then Rk C X A (α) = Rk Z X A (α) = 1. Proposition 1.4. If N is a knot manifold and X is a norm curve component of X(N ) then Rk C X (α) ≥ 2 for every slope α. As a corollary to Theorem 2, we have the following. Corollary 1.5. Suppose N is the exterior of a two-bridge knot and µ, β | ωβ = µω is the standard presentation for π 1 N . If X ⊂ X(N ) is an irreducible algebraic component defined over Q then X is defined by an irreducible polynomial of the form The set {I j µβ } n−1 0 is a free basis for C[X] as a C[I µ ]-module, T Q (X) as a Q[I µ ]module, and T (X) as a Z[I µ ]-module.
We point out that, with the appropriate definitions, these results also hold for the PSL 2 C-character variety. In what follows, we use X to indicate that we are working in the SL 2 C setting and Y in the PSL 2 C setting. When N ⊂ S 3 is a knot exterior, we reserve λ, µ ∈ π 1 N to be a longitude, meridian pair. We write X A and Y A for the algebraic sets of abelian characters. When N is hyperbolic, we use the notation X 0 and Y 0 to indicate algebraic components of the character varieties which contain a discrete faithful character.
Below, we list the results of our calculations in five different examples. The last example is of particular interest. It shows that the inequalities (1) need not be equalities and that, although T (X) must always be torsion free as a Z[I σ ]-module, it need not be a free module.
(1) Suppose N is the exterior of a trefoil knot. Then Rk Z X0 (µ) = 1. This shows that the converse to Proposition 1.3 does not hold. (2) Suppose N is the exterior of the figure-eight knot. We have and λ = (γηγ) −1 ηγη. Then µ and λ are primitive, peripheral, and generate a peripheral subgroup of N . We have Note that, since Y 0 is a norm curve, these ranks achieve their minimum possible value. In contrast to examples (1) and (2), the PSL 2 C-rank is strictly smaller than the SL 2 C-rank (at µ). As with examples (1) and (2), all Z[I µ ]-modules are free. (4) Suppose N is the exterior of the knot 8 20 . N is hyperbolic, so we consider X 0 . The diagram in Figure 1 gives a Wirtinger presentation for π 1 N which reduces to µ, γ | µγ(µγµ) −1 γ 3 (µγµ) −1 γµγ −2 where γ = ησ.
The functions I µ , I γ , and I γµ give an embedding of X 0 into C 3 . I γ and  (5) Suppose N is the once punctured hyperbolic torus bundle from Section 5 of [11]. Then The elements τ and λ = [α, β] form a basis for the peripheral subgroup of N and the functions give an embedding of X(N ) into C 7 . For ∈ {0, 1}, we have an irreducible algebraic component X ⊂ X(N ) which contains a discrete faithful character.

The equality
Rk Q X (λ) = 2 · Rk C X (λ) reflects the fact that X is not defined over Q.
1.1. Acknowledgements. We thank Steve Boyer, Brendan Hasset, and Kelly McKinnie for helpful conversations and suggestions.

Algebraic Geometry
Throughout this section we take k to be an algebraically closed field. All varieties are defined over k. When X is an irreducible variety we write k[X] and k(X) to denote the ring of regular functions on X and the function field for X, respectively. If p ∈ X, define O X,p to be the ring of germs of functions which are regular on neighborhoods of p. We have a surjective homomorphism k[X] → k given by f → f (p). Let m p be the maximal ideal which is the kernel of this map. By Theorem 3.2 part (c) of [13], O X,p is isomorphic to the localization k[X] mp . We use this isomorphism to identify these two rings and think of O X,p as a subring of k(X).
This equation is called an integral dependence relation of b over A.
By Corollary 5.3 of [1], the integral closure of A in B is a ring. The following characterization of integral closures is useful.
Theorem 3 (Corollary 5.22 of [1]). Suppose A is a subring of a field K. The integral closure of A in K is the intersection of all valuation rings of K that contain A.
Definitions 2.2. Suppose that X is an affine variety.
(1) X is a normal variety if k[X] is integrally closed in k(X).
(2) A normalization of X is an irreducible normal variety X ξ and a regular birational map ξ : If X is an affine variety and p ∈ X then the quotient m p /m 2 p has the structure of a k-vector space.
Theorem 1 in Chapter II, Section 5.1 of [21] states that non-singular affine varieties are normal.
It is clear that if m p is principal then dim m p /m 2 p = 1. Theorem 2 in Chapter II, Section 5.1 of [21] implies that if X is a normal affine algebraic curve and p ∈ X then m p is principal. Thus, we have the following well-known theorem.
Theorem 4. Suppose X is an irreducible affine algebraic curve. X is non-singular if and only if X is normal.
The following theorem is a direct consequence of Theorems 4 and 5 in Chapter II, Section 5.2 in [21] (and the first part of the proof of Theorem 4).
Theorem 5. If X is an irreducible affine algebraic curve then X has a normalization ξ : X ξ → X. Moreover, the normalization is unique, affine, and its coordinate ring is the integral closure of ξ * (k[X]).
The projective coordinates on projective space P n give distinguished open affine subsets {U i } n 0 which cover P n . So if X ⊂ P n is an irreducible projective variety then we have the distinguished affine open subsets U i ∩ X which cover X. We say that X is non-singular if U i ∩ X is non-singular for every i = 0, . . . , n.
Definition 2.4. Suppose that X is an irreducible affine curve and let ξ : X ξ → X the normalization of X. As in [13,Chapter 6], there is a smooth projective curve X (unique up to isomorphism) so that X ξ is isomorphic to an open set in X. The projective variety X is called the smooth projective model for X.
Henceforth, we identify X ξ with its image in X and we let ι : X X ξ be the rational map which is the identity on X ξ . Definition 2.5. The ideal points of X are the points in the set I(X) = X − X ξ . It follows from the uniqueness of normalizations and smooth projective models that I(X) is well-defined.
We have dominant birational maps . Now suppose that Y is an affine variey and ϕ : X → Y is a regular map with ϕ(X) = Y . Then ϕ * : k(Y ) → k(X) is a field monomorphism and ϕ * (k[Y ]) ⊆ k[X].
∈ O X,p . Also, if ϕ is a birational map, then ϕ is surjective if and only if ϕ has no hole.
Proof. Take a distinguished open affine set U i ⊂ P n withp ∈ U i . Set Xp = U i ∩ X. By Theorem 4, Xp is a non-singular affine curve so k[Xp] is integrally closed. Recall that O Xp,p is isomorphic to the localization k[Xp] mp . Proposition 5.13 of [1] gives that this localization is integrally closed. The inclusion induced isomorphism k( X) → k(Xp) restricts to an isomorphism O X,p → O Xp,p . Hence O X,p is integrally closed.
The pointp is a hole in ϕ :  Proof. By Lemma 2.7, we need only show that if ϕ has no hole then k[X] is integral over ϕ * (k[Y ]).
Assume, to the contrary, that k[X] is not integral over ϕ * (k[Y ]). Since ξ * is injective, this implies that k[X ξ ] is not integral over (ϕξ) * (k[Y ]). Taking integral closures in k(X ξ ) we have ).
Using the claim, The claim follows from Corollary 6.6 of [13] since it gives an open set X R in X and a pointp ∈ X R such that ι * is a finitely generated f * (k[Y ])-module (see for example Proposition 5.1 and Corollary 5.2 of [1]). Hence, we have the following immediate corollary.   3. Character varieties, boundary slopes, and closed essential surfaces Definition 3.1. A knot manifold is a connected, compact, irreducible, orientable 3-manifold whose boundary is an incompressible torus.
Let N be a knot manifold and Γ = π 1 (N ). Denote the set of SL 2 C-representations of Γ as R(N ) and the set of characters of representations in R(N ) as X(N ). Let t : R(N ) → X(N ) be the map which takes representations to their characters. In [10], it is shown that R(N ) and X(N ) are affine algebraic sets defined over C and the map t is regular. We will refer to R(N ) and X(N ) as the representation variety and character variety for N .
Culler and Shalen have revealed deep connections between essential surfaces in N and the character variety X(N ). We outline some of their results in what follows. For more background, see [10], the survey article [23], or Chapter 1 of [9].
Suppose that X is a non-empty algebraic subset of X(N ).
(1) Given γ ∈ Γ, the trace function for γ on X is the regular function called the trace ring for X.
The following proposition gives that, as a ring, T (X) is finitely generated.
The smallest C-algebra containing T (X) is the coordinate ring C[X]. Therefore, a generating set {γ i } n 1 for Γ gives an embedding of X into C 2 n −1 by taking the functions I V to be coordinate functions. It is straightforward to see that under this embedding, X(N ) is cut out by polynomials with coefficients in Z.
We have a regular map ∂ : X(N ) → X(∂N ) from X(N ) to the character variety for the peripheral subgroup (well-defined up to conjugation) of Γ given by restricting characters. For a non-empty algebraic subset X of X(N ), let ∂X denote the Zariski closure of ∂(X).

Definitions 3.4.
(1) The unoriented isotopy class of an essential simple closed curve in ∂N is called a slope. We denote the set of slopes on N as S. (2) A boundary slope is a slope which is realized as a boundary component of an essential surface in N .
Each slope corresponds to a pair {±a} ⊂ H 1 (∂N ; Z). The inverse of the Hurewicz isomorphism is an isomorphism H 1 (∂N ; Z) → π 1 ∂N . This gives a monomorphism e : H 1 (∂N ; Z) → Γ which is well-defined up to conjugation. Since traces are invariant under inverses and conjugation, the function {±a} → I e(a) is well-defined. When α = {±a} is a slope we write I α = I e(a) .
Assume now that X ⊆ X(N ) is an irreducible affine curve and let be the corresponding maps and varieties as defined in Section 2. The following theorem is well known and fundamental, see Theorem 2.2.1 and Proposition 2.3.1 of [10]. It is a translation of Theorem 1 into the language of Sections 2 and 3 of this paper.
Theorem 7 (Culler-Shalen). For every ideal pointx of X, there is an associated non-empty essential surface Σ in N .
(1) Σ may be chosen to have empty boundary if and only if (2) Otherwise, there exists a unique slope α such that I α ∈ O X,x . In this case, every component of ∂Σ represents the slope α.
Theorem 7 is broadly applicable. For instance, if N is the exterior of a nontrivial knot in S 3 then Kronheimer and Mrowka show that X(N ) contains a curve with infinitely many irreducible characters [16]. Also, if the interior of N admits a complete hyperbolic structure with finite volume and χ ρ is the character of a discrete faithful representation then there is a unique algebraic component X 0 ⊆ X(N ) which contains χ ρ . Furthermore, X 0 is a curve (see [24] and [23]). We refer to such a curve as a hyperbolic curve. (1) Suppose that X ⊆ X(N ) is an irreducible affine curve andx is an ideal point of X. A non-empty essential surface Σ ⊂ N is associated tox if it is contained in a surface given byx and Theorem 7.
(2) Suppose that X is an algebraic subset of X(N ). A surface Σ is detected by X if there is an ideal pointx of an irreducible affine curve in X so that Σ is associated tox. (3) If an essential surface Σ ⊂ N is detected by X (associated tox) and ∂Σ = ∅ the boundary slope represented by a component of ∂Σ is detected by X (associated tox). The slope is strongly or weakly detected (associated) depending on whetherx satisfies case (1) or case (2) of Theorem 7, respectively.

Remark 2.
It is natural to ask if, whenever N contains a closed essential surface, there is a closed essential surface in N detected by X(N ). The author is not certain if the answer is known to be no, however there are compelling reasons to believe that the answer is no, see for example [5] or Remark 5.1 in [11].
We conclude this section with some applications of the work in Section 2.
Theorem 8. Suppose N is a knot manifold and X is an irreducible algebraic subset of X(N ). The following are equivalent.
(1) X does not detect a closed essential surface.
) and so condition (3) cannot hold. The inequality dim(X) > dim(∂X) also implies that X contains a curve C to which Theorem 7 applies. This yields a closed essential surface detected by X. Hence, conditions (2) and (3) both imply that dim(X) = dim(∂X).
We have established that each of the four conditions imply that dim(X) = dim(∂X). By Proposition 2.4 of [6], dim(X) ≥ 1. Also dim(∂X) ≤ 1, otherwise for any slope α on ∂N , there is a curve in X to which Theorem 7 can be applied to give an essential surface in N with boundary slope α. This contradicts Hatcher's theorem [15] that N has only finitely many boundary slopes. So, if any condition (1) through (4) holds, then dim(X) = dim(∂X) = 1.
The theorem now follows from Theorems 6, 7, and Corollary 2.8.
Given a trace function I γ ∈ C[X], let C[I γ ] denote the C-subalgebra of C[X] generated by I γ . Consider the regular map I γ : X → C and the induced map Theorem 9. Suppose N is a knot manifold, X is an irreducible algebraic subset of X(N ), and α is a slope. The following are equivalent.
(1) X does not detect a closed essential surface and α is not detected by X.
is a finitely generated and free.
Proof. As with conditions (3) and (4) of the previous theorem, we know that (3) implies (2) and that (2) It remains to show that (1) and (2) are equivalent. Assume first that (2) holds. Theorem 8 shows that X does not detect a closed essential surface and dim(X) = 1. Hence I α is not constant. We have the regular map I α : X → C and I α (X) = C. Since C[X] is integral over C[I α ], Theorem 6 shows that I α : X → C does not have a hole. By Theorem 7, α is not strongly detected by X. Since X does not detect a closed essential surface, α cannot be weakly detected by X.
Now suppose that (1) holds. As in the proof of Theorem 8, we have that dim(X) = 1. Consider the regular map I α : X → C. Either I α (X) = C or I α is constant on X. But since dim(X) = 1, Theorem 7 implies that if I α is constant on X then α is strongly detected by X or X detects a closed essential surface. So we must have I α (X) = C.
By Theorem 6, it suffices to show that the regular map I α does not have a hole. Again, we appeal to Theorem 7 and notice that if I α has a hole then either α is strongly detected or X detects a closed essential surface. Definition 3.6. Let X ⊆ X(N ) be an irreducible algebraic subset. The C-rank function is the map Using Theorem 9, we make the following observations.
(1) If X does not detect a closed essential surface, Rk C X (α) = ∞ if and only if α is strongly detected by X. For X an irreducible algebraic component of X(N ) we define X • to be the union of the irreducible components X of X(N ) such that (1) X detects a closed essential surface if and only if X detects a closed essential surface and (2) for every slope α, X detects α if and only X detects α.
First, we assume that X does not detect a closed essential surface. Theorem 9 implies that X • is the union of the irreducible algebraic components X of X(N ) such that, for every slope α, The automorphism φ determines an automorphism on any complex polynomial ring by acting on the coefficients. This automorphism descends to an isomorphism φ : Since φ : C → C restricts to the identity on Q and I α is represented by a polynomial with integer coefficients we have φ( To prove the converse, we apply φ −1 to any integral dependence relation for The case when X does detect a closed essential surface follows from the above argument along with Lemma 5.3 of [3]. is an algebraic set. The rational trace ring T Q (X) is the smallest Q-algebra that contains T (X). Theorem 10. Suppose N is a knot manifold, X is an irreducible algebraic subset of X(N ), and α ∈ S. The following are equivalent.
(1) X does not detect a closed essential surface and α is not detected by X.
(3) As a Q[I α ]-module, T Q (X) is a finitely generated and free.
Proof. The first paragraph of the proof of Theorem 9 shows that (2) and (3) are equivalent. Theorem 9 shows that (2) implies (1).
It remains to show that (1) implies (2). Suppose then that X does not detect a closed essential surface and α is not detected by X. T Q (X) is generated by T (X) as a Q-algebra, so it suffices to show that elements of T (X) are integral over Q[I α ].
Let {X j } n 1 be the set of irreducible algebraic components of X • and take f ∈ T (X). The map C[X • ] → C[X] induced by inclusion is given by restriction and is surjective. Take F ∈ C[X • ] such that F | X = f . Let π : X • → C 2 be the map (F, I α ). Let (x, y) be the coordinates on Im(π) determined by xπ = F and yπ = I α .
Fix j ∈ {1, . . . , n}. By Theorem 8, I α : X j → C is non-constant and π(X j ) is an irreducible plane curve. Let P j ∈ C[x, y], be an irreducible polynomial which defines π(X j ). X • is defined over Q (Proposition 3.7) and π is given by polynomials with Z-coefficients, so we may assume that For each j, Theorem 9 gives that F | Xj is integral over C[I α ]. This means that there is a polynomial in C[x, y] which is zero on π(X j ) and monic as a polynomial in (C[x])[y]. Select one such polynomial p j with minimal degree. Then p j is irreducible and must differ from P j only by multiplication by some α j ∈ C. We have Hence α j ∈ Z and ( α −1 j ) · P gives an integral dependence relation for f over is the rank of T Q (X) as a Q[I α ]-module. It will require more work to show that there are related theorems which work over Z. We begin by establishing some lemmas concerning valuations. Proof. If every a i is zero then x = 0 and so ν(x) = ∞, so we may assume that at least one of the a i 's is non-zero. Let j be such that ν(a j ) + jν(x) is minimal in the set {ν(a i ) + iν(x) | 0 ≤ i < k}. We have Proof. Let R be the valuation ring corresponding to ν and let m be the maximal ideal in R. We know that Z ⊂ R. Note that m ∩ Z is prime (maximal) in Z so ν is p-adic.

Valuations
The following lemma is essentially Lemma 2.2 of [7]. Proof. Since (c k , p) = 1 and h ∈ Z[t] we know that r is an integer and at least zero. Now, for i ∈ {0, . . . , k − 1}, define a i = c i p k−1−i . We have p (k−1)r c i = a i (p r ) i , so Then h * ∈ Z[t] and h * (d) = 0. We have Let ν p be an extension of the p-adic valuation to Q(b). Since (c k , p) = 1, ν p (c k ) = 0. Also, for every i, a i ∈ Z so ν(a i ) ≥ 0. Hence, Lemma 4.1 implies that ν p (d) ≥ 0.
Since h is irreducible over Z and Q(b) = Q(d), it follows that h * is irreducible over Z. Hence, neither b nor d is integral over Z. By Corollary 5.2 of [1], we have a valuation ν on Q(b) with ν(d) < 0. The restriction of ν to Q cannot be the p-adic valuation and (again using Lemma 4.1) cannot be trivial. By Lemma 4.2, ν| Q = ν q for some prime q ∈ Z which is different from p. Therefore, ν(p) = 0. Also Throughout this section, N is a knot manifold and Γ is its fundamental group. Next, we introduce some terminology from [20].
Definition 5.1. A representation ρ : Γ → SL 2 C is algebraic non-integral if the image of ρ is in SL 2 F , where F is a number field and χ ρ (γ) is not an algebraic integer for some γ ∈ Γ. If ρ is an algebraic non-integral representation we abbreviate this by saying that ρ is an ANI-representation.
Similar to Theorem 7, the following result again follows from the work of Culler and Shalen. See Lemma 2 of [20].
Theorem 11. For every ANI-representation ρ of Γ there is an associated nonempty essential surface Σ ⊂ N . Suppose that α ∈ S is not a boundary slope. The surface Σ may chosen to have empty boundary if and only if I α (χ ρ ) is an algebraic integer.

Definitions 5.2.
(1) Suppose that ρ is an ANI-representation of Γ and Σ ⊂ N is a non-empty essential surface. Σ is associated to ρ if it is contained in a surface given by ρ and Theorem 11. (2) Suppose that X is an algebraic subset of X(N ). A surface Σ is ANI-detected by X if there is an ANI-representation ρ of Γ whose character lies on X and Σ is associated to ρ.
Proposition 5.3. If X is an irreducible algebraic subset of X(N ) and N contains a closed essential surface which is ANI-detected by X, then N contains a closed essential surface detected by X.
Example 1. In [8] Cooper and Long ask whether there is a one-cusped manifold whose character variety detects a closed essential surface. This was answered by Tillmann in [25] where he presented a calculation which shows that the character variety of the Kinoshita-Terasaka knot (11n42) detects a closed essential surface. Together with data collected by Goodman, Heard, and Hodgson [12], http://www. ms.unimelb.edu.au/~snap/, Proposition 5.3 gives many more examples.
Suppose that N is a one-cusped hyperbolic 3-manifold and let X 0 be an algebraic component of X(N ) which contains the character of a discrete faithful representation ρ 0 . By Theorem 11 and Proposition 5.3, if ρ 0 is an ANI-representation then N contains a closed essential surface detected by X (N ).
For all manifolds in the Callahan-Hildebrand-Weeks census of cusped hyperbolic manifolds with up to 7 tetrahedra [4], and for all complements of hyperbolic knots and links up to 12 crossings, Goodman, Heard, and Hodgson use the computer program Snap to attempt to (among other things) decide whether or not the holonomy representation is an ANI-representation. They find 252 such manifolds, 21 of which are complements of knots in S 3 (this is almost certainly not a complete list). The knots are 9a30, 9a31, 10a89, 10a96, 10a103, 11n97, 12n156, 12n245, 12n246, 12n260, 12n494, 12n508, 12n518, 12n600, 12n602, 12n604, 12n605, 12n694, 12n888, 12a1205, and 12a1288.
Roughly, the following theorem is proven by applying Theorem 10 and arguing as in [7]. Theorem 12. Suppose N is a knot manifold, X is an irreducible algebraic subset of X(N ), and α ∈ S. The following are equivalent.
(1) X does not detect a closed essential surface and α is not detected by X.
Proof. As with Theorem 10, we need only argue that (1) implies (2). Suppose that (1) is true and take γ ∈ Γ. Our goal is to show that I γ is integral over Z[I α ], so we may assume that I γ is not everywhere zero on X. By Theorem 8, dim(X) = 1 and by Theorem 9, I α must be non-constant on X. By Theorem 10, there is a polynomial f (y, z) ∈ Z[y, z] of the form f (y, z) = cy r + r−1 j=0 q j (z)y j with q j ∈ Z[z] such that c = 0, f (I γ , I α ) is the zero function on X, and the greatest common divisor of the coefficients of f is one. Choose f among all such polynomials to have smallest possible total degree. Then, since I α and I γ are not everywhere zero on X, neither y nor z divides f . Take k ∈ N large enough such that f (y, xy −k ) has no two terms with the same degree in y. Take l ∈ Z as small as possible such that g(x, y) = y l f (y, xy −k ) is a polynomial in Z[x, y]. Then g(x, y) is of the form g(x, y) = cy s + s−1 j=0 a j x tj y j where the t j 's are distinct integers, the greatest common divisor of the a j 's is one, g(I α I k γ , I γ ) = 0 on X, and neither x nor y divides g. We know that I α is non-constant on X so there must be a pointx ∈ X with I α (x) = 0. If I α I k γ is constant on X then I α = KI −k γ for some K ∈ C. Hence, x is an ideal point of X. By Theorem 2.2.1 and Proposition 2.3.1 of [10], either X detects a closed essential surface or an essential surface with boundary slope α. Both are contradictions so I α I k γ must be non-constant on X. Therefore, we may choose a prime p such that gcd(c, p) = 1 and for every root b of g(p, y) there is a character χ ρ ∈ X with χ ρ (α) · χ ρ (γ) k = p and χ ρ (γ) = b. Take g(p, y) = η · u j=0 h j (y) a complete factorization of g(p, y) over Z. So η ∈ Z and each h j is irreducible over Z. We claim that c divides η. The leading coefficient of g(p, y) is c. If we denote the the leading coefficient of h j as c j , we have c = ηc 0 · · · c u . If c doesn't divide η then there is some c i with gcd(α, c i ) > 1. Write c i = p t d where gcd(p, d) = 1. Since gcd(c, p) = 1 we must have gcd(c, d) > 1, in particular d = ±1. Let b be a root of h i . By our choice of p, there is χ ρ ∈ X which corresponds to the solution g(p, b). Using Hilbert's Nullstellensatz, we may assume that the entries of the matrices in ρ(π 1 N ) lie in a number field F . By Lemma 4.3, we have a valuation ν on Q(b) with ν(p) = 0 and ν(b) < 0. Take an extension of ν to F . We have ν(I γ (χ ρ )) < 0 and Since k ≥ 0 we must have ν(I α (χ ρ )) ≥ 0. Hence, N contains a closed essential surface which is ANI-detected by X. Proposition 5.3 shows that this is a contradiction, so we must have that c divides η. The constant η divides every coefficient of g(p, y) and c divides η, so c divides every coefficient of g(p, y). But the coefficient for z j is a j p tj and gcd(c, p) = 1. Hence c divides each a j . The only integers that divide every a j are ±1. Therefore f (I γ , I α ) is an integral dependence relation for I γ over Z[I α ]. Corollary 5.5. Suppose N is a knot manifold, X is an irreducible algebraic subset of X(N ) which is defined over Q. The following are equivalent.
(1) X does not detect a closed essential surface and α ∈ S is not detected by X.
(2) Every f ∈ T (X) satisfies a monic irreducible polynomial with coefficients in Proof. By Theorem 9, need only argue that (1) implies (2). Assume that X does not detect a closed essential surface and let f be a non-zero element of T (X). By Theorem 8, X is a curve. Since X does not detect a closed essential surface, I α is non-constant on X. Let V be the Zariski closure of the image of X under the map (f, I α ). V must be an irreducible affine curve defined over Q. Let p ∈ Q[x, y] be a defining equation for V . By multiplying by an integer we may assume p ∈ Z[x, y]. The proof of Theorem 12, shows that p is an integral dependence relation of f over Definition 5.6. A knot manifold is called small if it does not contain a closed essential surface.
Corollary 5.7. Let X be an irreducible algebraic subset of X(N ), where N is a small knot manifold. A slope α is detected by X if and only if Rk Z X (α) = ∞.

PSL 2 C-character varieties
The theorems from Sections 3 and 5, also apply in the PSL 2 C setting after making appropriate definitions and using virtually identical arguments. See [2] for details in what follows.
As usual, let N be a compact irreducible orientable 3-manifold with torus boundary and let Γ = π 1 N . Denote the set of PSL 2 C-representations of Γ as R(N ). PSL 2 C acts on R(N ) by conjugation. Lett : R(N ) → Y(N ) denote the corresponding algebro-geometric quotient. R(N ) and Y(N ) are affine algebraic sets and the mapt is a surjective regular map. Y(N ) is called the PSL 2 C-character variety for N and a point in Y(N ) is called a PSL 2 C-character. As with the SL 2 C-character variety, every irreducible algebraic component of Y(N ) has dimension at least one [6].
We would like to relate Y(N ) to SL 2 C-character varieties and define T (Y) where Y is an irreducible algebraic subset of Y(N ). First we claim that there is a welldefined equivalence class of Z 2 -central extensions and a finite regular map from the SL 2 C-character variety X(Γ) ofΓ to Y(N ) which contains Y in its image. First note that ifρ 1 ,ρ 2 ∈ R(N ) have the same image undert then the representations are either both reducible or are both irreducible. Hence we may refer to the points of Y as being either reducible or irreducible.
Suppose that every element of Y is reducible and chooseχ ∈ Y smooth in Y(N ). There is a diagonal representationρ ∈t −1 (χ). This representation determines an extension (2) and a liftρ :Γ → SL 2 C ofρ. Moreover, the isomorphism class ofΓ is independent of our choices ofχ andρ. The natural epimorphisms SL 2 C → PSL 2 C andΓ → Γ together induce a regular map φ : X(Γ) → Y(N ). In fact, the image of φ is the quotient of X(Γ) under the natural action of H 1 (Γ; Z 2 ). Moreover, Y ⊆ Im(φ).
Otherwise Y contains an irreducible character which is a smooth point of Y(N ). Letχ be such a point andρ ∈t −1 . Exactly as before, we obtain an extension (2) and a liftρ ofρ toΓ. Again, the isomorphism class ofΓ is independent of our choices, we obtain a finite regular map φ : X(Γ) → Y(N ), the image of φ is the quotient of X(Γ) by H 1 (Γ; Z 2 ), and Y ⊆ Im(φ).
In either case, the induced map φ * : C[Im(φ)] → C[X(Γ)] is injective. We define the trace ring T (Im(φ)) to be to be the image of T (Im(φ)) under this epimorphism.
Note that if γ ∈ Γ andγ 1 ,γ 2 ∈Γ are the preimages of γ then I 2 γ1 = I 2 γ2 . It follows that there is a well-defined squared trace function I 2 γ ∈ T (Y). To obtain results in the PSL 2 C-setting corresponding to those in Sections 3 and 5, we use the PSL 2 C-definitions rather than SL 2 C-definitions and replace I α with the squared trace function I 2 α whenever I α is not in the PSL 2 C-trace ring.
Remark 3. The equivalence classes of extensions (2) are in bijective correspondence with the elements of H 2 (Γ; Z 2 ).

Ranks and examples
Throughout this section we assume that N is a knot manifold and Γ = π 1 N . For an irreducible algebraic set X ⊆ X(N ), observe that Among other things, this section gives examples which show that these inequalities need not be equalities. If R is a ring, M is an R-module, and S ⊂ M then we write S R to denote the R-module generated by S.

Abelian curves.
Suppose that H 1 (N ; Z) ∼ = Z. There is a curve X A ⊆ X(N ) consisting of all characters of representations which factor through H 1 (N ; Z). Let α ∈ Γ be a peripheral element whose image in H 1 (N ; Z) generates H 1 (N ; Z).
Every character in X A is the character of a diagonal representation and the image subgroup in SL 2 C is generated by the image of α. It follows that every trace function I α ∈ T (X A ) is represented by a polynomial in Z[I α ]. Using the single coordinate I α on X A , we have X A = C. In particular, we have the following well-known proposition.  For example, if N is hyperbolic then there is a hyperbolic curve X 0 ⊆ X(N ). By Proposition 3.1.1 of [10], X 0 is a norm curve component. Proposition 7.3. Suppose that N is a knot manifold and X is an irreducible algebraic subset of X(N ). Let α be a slope. If X is a norm curve component then Suppose that X is a norm component and that Rk C X (α) < 2. By Theorem 8, dim(X) = 1. Hence, Rk C X (α) = 1 and we have f In particular, f has an inverse f −1 ∈ C[I α ]. We have Therefore, X has exactly one ideal point. On the other hand, [3,Proposition 4.5] and Theorem 7 show that X strongly detects at least two boundary slopes, so X must have more than one ideal point. 7.3. Two-bridge knots. Suppose that K is a two bridge knot. Define N = S 3 −K and Γ = π 1 N . Let µ, β | ωβ = µω be the standard presentation for Γ as given in Section 4.5 of [18]. The element µ ∈ Γ is a meridian for the knot.
Suppose that X is an algebraic component of X(N ) which is defined over Q. Let x = I µ and y = I µβ in T (X). It is shown in [14] that N is small and the slope represented by µ is not a boundary slope. By Corollary 5.5, we have a set of polynomials {p j } n j=0 in Z[x] where X is given as the zero set in C 2 of P (x, y) = y n − n j=0 p j (x)y j and P is irreducible in C[x, y]. Let B = {y j } n−1 j=0 ⊂ T (X). Since P is zero in T (X), B generates T (X) as a Z[x]-module, T Q (X) as a Q[x]-module, and C[X] as a C[x]-module. In particular, (4) Rk Z X (µ) ≤ n. Proof. First, we induct on k to show that M k = 1, y, . . . , y k C[x] is free of rank k + 1 whenever k ≤ n − 1. The base case is immediate since But k < n and P is irreducible, so each p j must be zero. Therefore, B is a free basis for C[X] as a C[x]-module and using the inequalities (3) and (4) we have To see that B is a free basis for T (X) as a Z[x]-module, it remains only to show that this module is free. To do this, we show that 1, y, . . . , y k−1 which we have already seen is trivial.
Example 2. Let K be a trefoil knot. A calculation shows that X(N ) = X A ∪ X 0 , where X 0 is an irreducible curve. The polynomial P which defines X 0 is y − 1. Therefore Rk C X0 (µ) = Rk Q X0 (µ) = Rk Z X0 (µ) = 1. Proposition 7.3 reaffirms the well-known fact that X 0 is not a norm curve.
Example 3. Let K be the figure-eight knot. A calculation shows that X(N ) = X A ∪ X 0 , where X 0 is a hyperbolic curve. The polynomial P which defines X 0 is y). It follows that T (Y 0 ) can be identified with the subring (with 1) of T (X 0 ) generated by x 2 and y. Moreover, Y 0 can be identified with the zeros of y 2 +(−1−ξ)y +(−1+2ξ) where ξ = x 2 . We have . Then λ commutes with µ and represents the boundary slope determined by a Seifert surface. Since σ(λ) is the identity, l = I λ ∈ T (Y 0 ). If we take (l, ξ, y) as coordinates on Y 0 then the ideal determined by Y 0 is generated by the following polynomials in C[l, ξ, y]. p 1 = (−1 − 4l) + (−11 + 4l)y + (16 − l)y 2 − 7y 3 + y 4 Let M = 1, y, y 2 , y 3 C[l] . The first polynomial describes the image of Y 0 under the map χ → (L(χ), y(χ)). Hence, p 1 is irreducible and M is free of rank 4. The last polynomial shows that ξ i y j ∈ M for every i, j ∈ Z + . It follows that Let s = I µ 2 λ . Then s ∈ T (Y 0 ). Similar calculations show that {1, y, y 2 , y 3 , ξ} generates C[Y 0 ] as a C[s]-module and that

Other examples.
Example 4. Let N be the Snappea census manifold M 003 from [4]. N is a finite volume hyperbolic 3-manifold with a single cusp. γ, η | γηγ −2 ηγη 3 is a presentation for Γ = π 1 N . The elements µ = (η 2 γηγ) −1 and λ = (γηγ) −1 ηγη together generate a peripheral subgroup. Define Let X 0 ⊂ X(N ) be a hyperbolic curve and let Y 0 be the image of X 0 in the PSL 2 Ccharacter variety. We can take (m, x, y, z) as coordinates on X 0 . The ideal in C[m, x, y, z] determined by X 0 is generated by the following three polynomials The first polynomial is irreducible and gives that 1, z, z 2 , z 3 C[m] is free of rank 4. The last two polynomials show that x, y ∈ 1, z, z 2 , z 3 C[m] . It follows that It is straightforward to check that m ∈ T (Y 0 ) and that (m, ζ) can be taken as coordinates on Y 0 . As such, Y 0 is the zero locus of the polynomial ζ 2 + (−m − 1)ζ + 1. It follows that Rk C Y0 (µ) = Rk Q Y0 (µ) = Rk Z Y0 (µ) = 2. We see that these ranks achieve the smallest possible value for the rank of a norm curve. In contrast to Example 3, Rk C Y0 (µ) < Rk C X0 (µ). Example 5. Let N be the exterior of the knot 8 20 . The knot 8 20 is hyperbolic. Let X 0 ⊂ X(N ) be a hyperbolic curve. As outlined in the introduction to this paper, Γ is generated by a pair of elements µ and γ where µ is a meridian. Define x = I µ , y = I γ , and z = I γµ . The functions x, y, z give an embedding of X 0 into C 3 .
We use the symbol ≡ to indicate congruence modulo (G) in C[x, y, z] and we write LM (f ) for the leading monomial of f ∈ C[x, y, z] with respect to our chosen monomial order. For 1 ≤ j ≤ 5, let m j = LM (g j ). Then (m j ) 5 1 = (yx 4 , y 3 z, y 2 xz, x 2 z, z 2 ) and m j < m j+1 .
Oertel's work in [19] shows that N is a small knot manifold, so Corollary 5.5 implies that there are integral dependencies for y and z over Z[x] given by irreducible polynomials. The polynomial g 1 is the polynomial for y and and a quick calculation produces the polynomial for z. These polynomials also show that the set B = y i z j | 0 ≤ i, j ≤ 4 is a generating set for C[X 0 ] as a C[x]-module, T Q (X 0 ) as a Q[x]-module, and T (X 0 ) as a Z[x]-module. However, B is to large to be a basis for any of these modules. Lemma 7.6. If k ≥ 1 then there is a polynomial P ∈ C[x, y] such that Proof. Induct on k and use the polynomial g 4 .
Proof. Induct on k and use the polynomial g 2 .
Lemma 7.8. If k ≥ 1 then there is a polynomial P ∈ C[x, y] and d 0 , . . . , d 4 ∈ C such that Proof. The lemma is trivial if k = 1. If k is even, a straightforward application of Lemmas 7.6 and 7.7 gives numbers d 0 , d 1 , and d 2 such that x k z is equivalent to a polynomial in If k ≥ 3 is odd, we apply the result in the even case to obtain c 0 , c 1 , c 2 ∈ C where x k z is equivalent to a polynomial in To finish the proof, notice that g 3 shows that xy 2 z is represented by a polynomial in C[x, y].
Proof. The proof is by induction on deg(g).
Since G is a Groebner basis for (G), some m j must divide the leading term of the polynomial on the right. This forces f = p.
Let k ≥ 1 and assume that the result holds for polynomials with degree less than k. Suppose that g = k 0 c j x j . By Lemma 7.8, we have q ∈ C[x, y] and r 0 , . . . , r 4 ∈ C such that c k x k z ≡ q + r 0 z + r 1 yz + r 2 y 2 z + r 3 xz + r 4 xyz. Therefore, By the inductive hypothesis, we have As before, we must have f = p since G is a Groebner basis for (G). Proof. Take h ∈ 1, y, . . . , y k . Then there are polynomials p ∈ C[x, y] and g ∈ C[x] such that p represents h in C[X 0 ] and p ≡ −gz. Define f = p + gz and notice that f ∈ (G). By Lemma 7.9, p ∈ (G) and h = 0.
is a free C[x]-module with rank 6.
Proof. Lemmas 7.5 and 7.10, reduce the problem to showing that the intersection is trivial. Since g 3 ∈ G, we have that 1, . . . , y 3 , xy 2 z C[x] = 1, . . . , y 4 C[x] . Together with Lemmas 7.5 and 7.10, this shows that 1, . . . , y 3 , xy 2 z, z C[x] is free with rank six. Now suppose that h is an element of the intersection (5). Then xh is an element of both 1, . . . , y 3 , z C[x] and xy 2 z C[x] . But we know that the intersection of this module is trivial. Hence, hx = 0 and so h = 0. Now to prove Theorem 13, we need only show that B spans the modules in question. This is straightforward using the polynomials G.
The elements τ and λ = [α, β] form a basis for a peripheral subgroup of N . The functions give an embedding of X(N ) into C 8 . X(N ) contains exactly two hyperbolic curves, we label them X where ∈ {0, 1}. It is straightforward to verify that X strongly detects the slopes represented by t and lt and does not detect a closed essential surface. This is done in [11]. It is also easy to use Theorem 9 to confirm these facts.
We prove Theorem 14 as a series of lemmas. Suppose k is a field, let R = k[x 0 , . . . , x n ], and I ⊆ R and ideal. Take < to be the monomial order generated by x i < x j iff i < j and let G = {g j } r 1 be a Groebner basis for I with respect to <. Write m j = LT (g j ). For f ∈ R, writef to denote its image in R/I. is free of rank k − 1, (2) n i < n j for every i < j, and (3) n k is not divisible by any m j then the k[x 0 ]-submodule generated by {n j } k 1 is free of rank k.
Then there is a set of polynomials In particular g := k j=1 p j n j ∈ I.
If p k = 0 thenf = 0 and we are done. Otherwise, let cx s 0 be its leading term. Then, by (2), LT (g) = cx s 0 n k . Since g ∈ I and G is a Groebner basis for I, we must have that LT (g) is divisible by some m j . This contradicts (3), so we must have p k = 0.
Remark 4. Recall that Groebner bases can be used to solve the ideal membership problem.
This shows that the projection given by (l, t, u, v, w, x, y, z) → (l, t, y, z), restricts to an isomorphism (defined over Q) from X 0 ∪ X 1 onto its image. Hence, we have natural identifications of C[X ] and C[X 0 ∪ X 1 ] with quotients of C[l, t, y, z]. Define and G = {g 1 , g 2 , g 3 }. Let < be the pure lexicographic monomial order for C[l, t, y, z] induced by the relationship l < z < y < t. It can be easily verified that G is a Groebner basis for (G) and (G) is the kernel of the natural map C[l, t, y, z] → C[X ]. Similarly, {g 1 , g 2 , h} is a Groebner basis for the ideal it generates and this ideal is the kernel of the natural map C[l, t, y, z] → C[X 0 ∪ X 1 ].
Lemma 7.13. The set B C = {1, z, z 2 , z 3 , y, zy, z 2 y, z 3 y} is a basis for C[X ] as a C[l]-module.
Proof. Since the natural map C[l, t, y, z]/(G) → C[X ] is an isomorphism, we need only show that B C is a C[l]-module basis for C[l, t, y, z]/(G).
Let M ⊆ C[l, t, y, z]/(G) be the C[l]-submodule generated by B C . Lemma 7.12 shows that M is free with rank 8.
The polynomial g 3 shows that t may be expressed in terms of z. So, to finish the proof, it is enough to show that y j z k ∈ M for every pair of integers j, k ≥ 0. By definition of B C and g 1 , z j ∈ M, for every integer j ≥ 0. Using g 2 , we also have that y 2 z j ∈ M, for every integer j ≥ 0. Now, using this fact and the definition of B C , we conclude that if f ∈ M then yf ∈ M. In particular, y j z k ∈ M for every pair of integers j, k ≥ 0. Therefore, M = C[l, y, z]/(G).
Lemma 7.14. The set Proof. The natural map C[l, t, y, z] → C[X ] restricts to a surjection Q[l, t, y, z] → T Q (X ). The kernel of this map is (G) ∩ Q[l, t, y, z]. We claim that {g 1 , g 2 , h} is a Groebner basis for this ideal with respect to the monomial order <. It is immediate to verify that this is a Groebner basis for (g 1 , g 2 , h), so we need only show that (g 1 , g 2 , h) = (g 1 , g 2 , g 3 ) ∩ Q[l, t, y, z]. The inclusion (g 1 , g 2 , h) ⊆ (g 1 , g 2 , g 3 ) ∩ Q[l, t, y, z] is clear. To prove the opposite, suppose f ∈ (G) ∩ Q[l, t, y, z]. Then f (χ) = 0 for every χ ∈ X . Observe that complex conjugation induces a bijection X 0 → X 1 . So f (χ) is also zero. In particular, f is zero on X 0 ∪ X 1 . Therefore f ∈ (g 1 , g 2 , h). Let M ⊂ T Q (X ) be the Q[l]-submodule generated by B Q . Since {g 1 , g 2 , h} is a Groebner basis for the kernel of the map Q[l, t, y, z] → T Q (X ), Lemma 7.12 shows that M is free with rank 16. The remainder of the proof follows an argument similar to the last part of the proof of Lemma 7.13.
Notice that the constant terms of p 1 and p 3 must be zero and the constant term of p 2 is −1. Apply ψ to the equation 2vy + lux = (6 + l)t to see that the equation 2p 3 + lp 2 = (6 + l)p 1 must hold in Z[l]. For a contradiction, observe that the the degree one coefficient of the left hand side of this equation is odd and on the right hand side the coefficient is even.
Lemma 7.19. The Z[l]-module T (X ) is torsion free but not free and Rk Z X (λ) = 17. Proof. C[X ] is an integral domain, so T (X ) is torsion free as a Z[l]-module. Direct summands of free modules are projective. Therefore, by Lemmas 7.17 and 7.18, T (X ) is not a free Z[l]-module. By Lemma 7.16, Rk Z X (λ) ≤ 17. Suppose that {a j } 16 1 generates T (X ) as a Z[l]module. Then every element of B Z can be written as a Z[l]-linear combination of the a j 's. By Lemma 7.14, {a j } 16 1 is a free Q[l]-basis for T Q (X ). It follows that the Z[l]-module generated by {a j } 16 1 is also free. This is a contradiction because we know that T (X ) is not a free Z[l]-module. Therefore, Rk Z X (λ) = 17.