ON PROPERTY D(2) AND COMMON SPLITTING

FIELD OF TWO BIQUATERNION ALGEBRAS

A.S. Sivatski

Department of Mathematics St.Petersburg Electrotechnical University 197376, St.Petersburg, Russia

Abstract. Let F be a field of characteristic ≠ 2. We say that F has property D(2) if for any quadratic extension L/F and any two binary quadratic forms over F having a common nonzero value over L this value can be chosen in F. We show that if k is a field of characteristic ≠ 2 having at least two distinct quadratic extensions, then for the field k(x) property D(2) does not hold. Using this we construct two biquaternion algebras over a field K = k(x)((t))((u)) such that their sum is a quaternion algebra, but they do not have a common biquadratic (i.e. a field of the kind K(\sqrt{a}, \sqrt{b}), where a, b ∈ K^{∗}) splitting field.

Let F be a field of characteristic different from 2. By definition, F has property D(2) if for any quadratic extension L/F and any two binary quadratic forms q_{1}, q_{2} over F the existence of a common value of the forms q_{1}_{L}, q_{2}_{L} implies the existence of a common value of the forms q_{1}_{L}, q_{2}_{L}, which lies in F. Examples of fields of characteristic 0 not satisfying this property has been given in [5]. Later in [1] starting from such a field, it has been shown that the answers to the following questions are negative in general:

1) Let (a_{1}, b_{1}) and (a_{2}, b_{2}) be quaternion algebras over a field K. Suppose c ∈ K^{∗} is such that (a_{1}, b_{1}) ⊗ (a_{2}, b_{2})_{K(\sqrt{c})} is not a division algebra. Is it true that there exists d ∈ K^{∗} such that

(a_{1}, b_{1})_{K(\sqrt{c},\sqrt{d})} = (a_{2}, b_{2})_{K(\sqrt{c},\sqrt{d})} = 0 ?

2) Let φ be an 8-dimensional form from I^{2}(K) whose Clifford algebra has index 4. Is it true that φ is a direct sum of two forms similar to 2-fold Pfister forms ?

3) Let φ be a 14- dimensional form from I^{3}(K). Is it true that φ is similar to the difference of the pure parts of two 3-fold Pfister forms ?

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 13C12, 13F20.

Key words and phrases. Quadratic form, Biquaternion algebra, Second residue map.

The work under this publication was partially supported by Royal society Joint Project”Quadratic forms and central simple algebras under field extensions”

By means of quite a different and very nonelementary technique the negative answers to the same questions have been given in [3].

The above questions stipulate our interest to property D(2). As far as we know, examples of fields of positive characteristic not satisfying property D(2) are not known. In this note independently of characteristic of the field we give simple counterexamples to property D(2) such that the 2-cohomological dimension of the ground field F equals 2. Using these examples we show that the following question has the negative answer in general:

Suppose D_{1} and D_{2} are two biquaternion algebras over the field K such that ind(D_{1} + D_{2}) = 2. Does there exist a common splitting field of D_{1} and D_{2} of the type K(\sqrt{a}, \sqrt{b}), where a, b ∈ K^{∗} ?

Our notation is standard. All the fields in the sequel are of characteristic different from 2. By form we always mean a quadratic form over a field. Slightly abusing notation we often identify a form over F with the corresponding class in the Witt group W (F). By φ_{an} we denote the anisotropic part of the form φ. If L/F is a field extension, and φ is a form over F, then D_{L}(φ) is the set of nonzero values of φ_{L}, and G_{L}(φ) is the group of multipliers of φ_{L}, i.e. G_{L}(φ) = {a ∈ L^{∗} : φ ≃ aφ}. By the n-fold Pfister form ⟨⟨a_{1}, …, a_{n}⟩⟩ we mean the form ⟨1, −a_{1}⟩ ⊗ … ⊗ ⟨1, −a_{n}⟩. By (a, b) we denote the quaternion algebra with generators 1, i, j, k and relations i^{2} = a, j^{2} = b, ij = −ji. If k is a field, t is an indeterminate and p ∈ k[t] is a monic irreducible polynomial, then ∂_{p} : W (k(t)) → W (k[t]/p) is the second residue map, i.e. the group homomorphism determined by the following rule: for a squarefree polynomial f ∈ k[t]

∂_{p}(⟨f⟩) = \{ \begin{matrix}0 if p does not divide f, \\ ⟨\frac{f}{p} ⟩ if p divides f \end{matrix} .

The tensor product of central simple K-algebras is always considered over the field K.

We turn to the construction of the examples in question. Let k be a field, a, d_{1}, d_{2} ∈ k^{∗} such that d_{1}^{2}−4a, d_{2}^{2}−4a, (d_{1}^{2}−4a)(d_{2}^{2}−4a) ∉ k^{∗}^{2}, x an indeterminate, F = k(x), L = F (\sqrt{x^{2} − 4a}). Set

q_{1} ≃ (x − d_{1})⟨⟨d_{1}^{2} − 4a⟩⟩ ∈ W (F),
q_{2} ≃ (x − d_{2})⟨⟨d_{2}^{2} − 4a⟩⟩ ∈ W (F).

Proposition 1. The field F does not satisfy property D(2), and the forms q_{1}, q_{2} and the quadratic extension L/F provide a counterexample.

Proof. It is trivial to check that

q_{1}((x − d_{1} − \sqrt{x^{2} − 4a})(x − d_{1})^{−1}, (x − d_{1})^{−1})
= q_{2}((x − d_{2} − \sqrt{x^{2} − 4a})(x − d_{2})^{−1}, (x − d_{2})^{−1})
= 2(x − \sqrt{x^{2} − 4a}) ∈ L.

This shows that D_{L}(q_{1}) ∩ D_{L}(q_{2}) ≠ ∅.

Now let us prove that D_{L}(q_{1}) ∩ D_{L}(q_{2}) ∩ F = ∅. We will follow an idea in [6]. Let y_{1}, y_{2} be indeterminates, k̂ a maximal odd degree extension of k(y_{1}, y_{2}). Since d_{1}^{2} − 4a and d_{2}^{2} − 4a are distinct nontrivial elements in k^{∗}/k^{∗}^{2}, it is easy to see that (d_{1}^{2} − 4a, y_{1}) ⊗k̂ (d_{2}^{2} − 4a, y_{2}) is a division algebra. Put

φ = ⟨⟨d_{1}^{2} − 4a, y_{1}⟩⟩ − ⟨⟨d_{2}^{2} − 4a, y_{2}⟩⟩.

In particular, φ ≠ 0 over any quadratic extension of k̂. Put y = \sqrt{x^{2} − 4a}, x − y = 2t. Then x + y = \frac{2a}{t}, hence x = t + \frac{a}{t}. So we have L = F (\sqrt{x^{2} − 4a}) = k(t), F = k(t + \frac{a}{t}). Suppose that u ∈ D_{k̂(t)}(q_{1}) ∩ D_{k̂(t)}(q_{2}) ∩ k̂(x). Obviously, we may

assume that u ∈ k̂[x] and u = u(x) is squarefree. Then

(x − d_{1})u ∈ D_{k̂(t)}(⟨⟨d_{1}^{2} − 4a⟩⟩) = G_{k̂(t)}(⟨⟨d_{1}^{2} − 4a⟩⟩) = G_{k̂(t)}(q_{1}),
(x − d_{2})u ∈ D_{k̂(t)}(⟨⟨d_{2}^{2} − 4a⟩⟩) = G_{k̂(t)}(⟨⟨d_{2}^{2} − 4a⟩⟩) = G_{k̂(t)}(q_{2}).

Therefore, we have

(1)  \begin{matrix}(q_{1}⟨⟨y_{1}⟩⟩ − q_{2}⟨⟨y_{2}⟩⟩)_{k̂(t)} = ((x − d_{1})uq_{1}⟨⟨y_{1}⟩⟩ − (x − d_{2})uq_{2}⟨⟨y_{2}⟩⟩)_{k̂(t)} \\ = u(⟨⟨d_{1}^{2} − 4a, y_{1}⟩⟩ − ⟨⟨d_{2}^{2} − 4a, y_{2}⟩⟩)_{k̂(t)} = uφ_{k̂(t)}. \end{matrix}

Similarly, since 4t = 2(x − \sqrt{x^{2} − 4a}) ∈ D_{L}(q_{1}) ∩ D_{L}(q_{2}), we have

(2)  (q_{1}⟨⟨y_{1}⟩⟩ − q_{2}⟨⟨y_{2}⟩⟩)_{k̂(t)} = tφ_{k̂(t)}.

Combining (1) and (2) we conclude that

(3)  uφ_{k̂(t)} = tφ_{k̂(t)}

Substituting x for t + \frac{a}{t} in the left part of (3) and comparing the residues at t of the both parts of this equality we see that the degree of the polynomial u in x is odd. Since the field k̂ has no proper extensions of odd degree, we conclude that there is c ∈ k̂ such that x − c divides u. Notice that x − c = \frac{p(t)}{t}, where p(t) = t^{2} − ct + a. Comparing the residues at p(t) of the both parts of the equality

(3) we get a contradiction, since ∂_{p}(tφ) = 0, and ∂_{p}(uφ) = φ_{k̂(\sqrt{c^{2}−4a})} ≠ 0. The proposition is proved.

Corollary 2. Let k be a field, b_{1}, b_{2} ∈ k^{∗}. Then the following conditions are equivalent.

1) The elements 1, b_{1}, b_{2} ∈ k^{∗}/k^{∗}^{2} are pairwise distinct.

2) The field k(x) has not property D(2) and there exist a ∈ k^{∗}, s_{1}, s_{2} ∈ k(x)^{∗} such that the extension k(x, \sqrt{x^{2} − 4a})/k(x) and the forms s_{1}⟨⟨b_{1}⟩⟩ and s_{2}⟨⟨b_{2}⟩⟩ provide a counterexample.

Proof. 1) ⇒ 2). By Proposition 1 it suffices to find a, d_{1}, d_{2} ∈ k^{∗} such that b_{1} = d_{1}^{2} − 4a, b_{2} = d_{2}^{2} − 4a. One can put, for instance, d_{1} = \frac{b_{1}−b_{2}+1}{2}, d_{2} = \frac{b_{2}−b_{1}+1}{2}, a = \frac{1}{4}(d_{1}^{2} − b_{1}).

2) ⇒ 1). If b_{1} is a square, then D_{L}(⟨⟨b_{1}⟩⟩) = L^{∗} for any extension L/k(x), hence for any s_{1}, s_{2} ∈ k(x)^{∗} the forms s_{1}⟨⟨b_{1}⟩⟩ and s_{2}⟨⟨b_{2}⟩⟩ do not provide a counterexample to property D(2). Assume that b_{1}b_{2} is a square, i.e. b_{1} ≡ b_{2}( mod k^{∗}^{2}). Let L/k(x) be an arbitrary field extension. Assume that c ∈ D_{L}(s_{1}⟨⟨b_{1}⟩⟩) ∩ D_{L}(s_{2}⟨⟨b_{2}⟩⟩). Then ⟨⟨b_{1}, cs_{1}⟩⟩ = ⟨⟨b_{1}, cs_{2}⟩⟩ = 0, hence ⟨⟨b_{1}, s_{1}s_{2}⟩⟩ = 0, and so

s_{1} ∈ D_{L}(s_{1}⟨⟨b_{1}⟩⟩) ∩ D_{L}(s_{2}⟨⟨b_{2}⟩⟩) ∩ k(x).

Remark. If k_{0} is an algebraically closed field, and k is the rational function field over k_{0}, then cd_{2}(k) = 1 and cd_{2}(k(x)) = 2. By Corollary 2 the field k(x) does not have property D(2). On the other hand, if I^{2}(K) = 0, then the field K has property D(2), since 1 is a value of any binary form over K.

It has been established in [1] that if the field F has not property D(2), then the field F_{1} = F ((t)) has not property CS. Recall that property CS for the field F_{1} means that for any quaternion algebras Q_{1}, Q_{2} over F_{1} and c ∈ F_{1}^{∗} such that (Q_{1} ⊗ Q_{2})_{F_{1}(\sqrt{c})} is not a division algebra there exists d ∈ F_{1}^{∗} such that Q_{1}_{F_{1}(\sqrt{c},\sqrt{d})} = 0, Q_{2}_{F_{1}(\sqrt{c},\sqrt{d})} = 0. In fact, if the binary forms s_{1}⟨⟨b_{1}⟩⟩, s_{2}⟨⟨b_{2}⟩⟩ and the quadratic extension F (\sqrt{c})/F provide a counterexample to property D(2) for the field F, then the quaternion algebras Q_{1} ≃ (b_{1}, s_{1}t), Q_{2} ≃ (b_{2}, s_{2}t) give a counterexample to property CS for the field F ((t)).

Let now F be an arbitrary field, for which property CS does not hold. Assume that quaternion algebras Q_{1}, Q_{2} and a quadratic extension F (\sqrt{c})/F provide a counterexample. In particular, (Q_{1} ⊗ Q_{2})_{F (\sqrt{c})} is not a division algebra, which implies that Q_{1} ⊗ Q_{2} ≃ (c, d) ⊗ (e, f) for some d, e, f ∈ F^{∗}. Notice that (e, f) ≠ 0, for otherwise Q_{1}and Q_{2} would have a common quadratic subfield. Consider the biquaternion algebras D_{1} ≃ Q_{1} ⊗ (c, u) and D_{2} ≃ Q_{2} ⊗ (c, du) over the field F ((u)).

Proposition 3. 1) ind(D_{1} ⊗ D_{2}) = 2.

2) The algebras D_{1} and D_{2} do not have a common biquadratic splitting extension. In other words, for any p, q ∈ F ((u))^{∗} either D_{1}_{F ((u))(\sqrt{p},\sqrt{q})} ≠ 0, or D_{2}_{F ((u))(\sqrt{p},\sqrt{q})} ≠ 0.

Proof.

1) D_{1} + D_{2} = Q_{1} + Q_{2} + (c, u) + (c, du) = Q_{1} + Q_{2} + (c, d) = (e, f), which proves the first part of the proposition.

2) Assume the contrary, i.e. that D_{1}_{F ((u))(\sqrt{p},\sqrt{q})} = 0, D_{2}_{F ((u))(\sqrt{p},\sqrt{q})} = 0. Since

F ((u))^{∗}/F ((u))^{∗}^{2} = F^{∗}/F^{∗}^{2} ⊕ ℤ/2ℤ, we may assume that p ∈ F^{∗}. If q ∈ F^{∗}, it is easy to see that F (\sqrt{c}) ⊂ F (\sqrt{p}, \sqrt{q}), i.e. F (\sqrt{p}, \sqrt{q}) = F (\sqrt{c}, \sqrt{c′}) for some c′ ∈ F^{∗}. Hence

Q_{1}_{F (\sqrt{c},\sqrt{c′})} = Q_{2}_{F (\sqrt{c},\sqrt{c′})} = 0,

which is impossible, since the algebras Q_{1}, Q_{2} and the extension F (\sqrt{c})/F provide a counterexample to property CS. Therefore, we may suppose that q = au, where a ∈ F^{∗}. Then, obviously,

(Q_{1} + (c, a))_{F ((u))(\sqrt{p},\sqrt{au})} = (Q_{1} + (c, u))_{F ((u))(\sqrt{p},\sqrt{q})} = 0,
(Q_{2} + (c, ad))_{F ((u))(\sqrt{p},\sqrt{au})} = (Q_{2} + (c, du))_{F ((u))(\sqrt{p},\sqrt{q})} = 0.

Hence,

(Q_{1} + (c, a))_{F (\sqrt{p})} = 0, (Q_{2} + (c, ad))_{F (\sqrt{p})} = 0,

which implies that

Q_{1}_{F (\sqrt{c},\sqrt{p})} = Q_{2}_{F (\sqrt{c},\sqrt{p})} = 0,

a contradiction to the fact that the algebras Q_{1}, Q_{2} and the extension F (\sqrt{c})/F provide a counterexample to property D(2). The proposition is proved.

Remark. To give a concrete example of biquaternion algebras D_{1} and D_{2} satisfying the conditions of Proposition 3 we start from the binary forms q_{1} ≃ (x − d_{1})⟨⟨d_{1}^{2} − 4a⟩⟩ and q_{2} ≃ (x − d_{2})⟨⟨d_{2}^{2} − 4a⟩⟩ over the field k(x) from Proposition

1. Then Q_{1} ≃ (d_{1}^{2} − 4a, (x − d_{1})t), Q_{2} ≃ (d_{2}^{2} − 4a, (x − d_{2})t). It is easy to check the following equality comparing the residues on its both sides:

\begin{matrix}Q_{1} + Q_{2} = (x^{2} − 4a, −(x − d_{1})(x − d_{2})((d_{1} + d_{2})x − 4a − d_{1}d_{2}))+ \\ ((d_{1}^{2} − 4a)(d_{2}^{2} − 4a), ((d_{1} + d_{2})x − 4a − d_{1}d_{2})(d_{1} + d_{2})^{−1}t). \end{matrix}

Tracing back the construction of the algebras D_{1}, D_{2} from the algebras Q_{1}, Q_{2}, we get over the field k(x)((t))((u))

D_{1} ≃ (d_{1}^{2} − 4a, (x − d_{1})t) ⊗ (x^{2} − 4a, u),

D_{2} ≃ (d_{2}^{2} − 4a, (x − d_{2})t) ⊗ (x^{2} − 4a, −(x − d_{1})(x − d_{2})((d_{1} + d_{2})x − 4a − d_{1}d_{2})u).

Notice that if cd_{2}k = 1, K = k(x)((t))((u)), then cd_{2}K = 4, so one can construct a counterexample with a field K of 2-cohomological dimension 4. On the other hand, there is no such a counterexample for a field of cohomological dimension 2. More precisely we have the following

Proposition 4. Let K be a field such that I^{3}(K) = 0. Let further D_{1}, D_{2} be biquaternion algebras such that ind(D_{1} + D_{2}) = 2. Then there exist p, q ∈ K^{∗} such that

D_{1}_{K(\sqrt{p},\sqrt{q})} = D_{2}_{K(\sqrt{p},\sqrt{q})} = 0.

Proof. By ([2], Lemma 14.2) we have I^{3}(K(\sqrt{a})) = 0 for any a ∈ K^{∗}. Let φ_{1}, φ_{2} be Albert forms corresponding to D_{1}, D_{2}. Then by ([4], Ch.2, Th.14.4) we get

(φ_{1} − φ_{2})_{K(\sqrt{b})} ∈ I^{3}(K(\sqrt{b})) = 0

for any b ∈ K^{∗} such that (D_{1} + D_{2})_{K(\sqrt{b})} = 0. Hence φ_{1} ⊥ −φ_{2} ≃ ⟨⟨b⟩⟩ ⊗ ψ, where dim ψ = 6. Let ψ_{1} ∈ I^{2}(K) be such a form that ψ = ψ_{1} + τ, where dim τ = 2. Then we have

φ_{1} − φ_{2} = ⟨⟨b⟩⟩ ⊗ ψ_{1} + ⟨⟨b⟩⟩ ⊗ τ = ⟨⟨b⟩⟩τ,

since ⟨⟨b⟩⟩ ⊗ ψ_{1} ∈ I^{3}(K) = 0. Therefore, dim(φ_{1} ⊥ −φ_{2})_{an} ≤ 4. This means that φ_{1} and φ_{2} have a common 4-dimensional subform, say

φ_{1} ≃ ⟨a, b, c, d, e_{1}, −abcde_{1}⟩,

6 A.S. SIVATSKI

φ_{2} ≃ ⟨a, b, c, d, e_{2}, −abcde_{2}⟩.

Then, obviously,

φ_{1}_{K(\sqrt{−ab},\sqrt{−cd})} = φ_{2}_{K(\sqrt{−ab},\sqrt{−cd})} = 0,

which is equivalent to

D_{1}_{K(\sqrt{−ab},\sqrt{−cd})} = D_{2}_{K(\sqrt{−ab},\sqrt{−cd})} = 0,

so we can put p = −ab, q = −cd. The proposition is proved.

The following theorem [7] is the main motivation of the present note, and for the sake of completeness we give here its proof. Theorem 5. Let D_{1} and D_{2} be biquaternion algebras over a field k such that ind(D_{1} + D_{2}) = 2. Then there exists a field extension l/k of degree 4 such that D_{1}_{l} = D_{2}_{l} = 0. Proof. Let D_{1} + D_{2} = Q = (a, b). Then by the dimension count D_{1} ⊗ Q ≃ M_{2}(D_{2}). In particular, Q ↪ M_{2}(D_{2}). It follows that there exist I, J ∈ M_{2}(D_{2}) such that

I^{2} = ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ), J^{2} = ( \begin{matrix}b & 0 \\ 0 & b \end{matrix} ), and IJ = −JI.

Let I = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ). We may assume that α^{2} ≠ a, and D_{2} is a division algebra (in the opposite case the theorem is simple and left to the reader).

Lemma 6. There is a matrix S ∈ GL_{2}(D_{2}) such that SIS^{−1} = ( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ).

Proof. Since α^{2} ≠ a and I^{2} = ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ), we have β ≠ 0. Then for any x ∈ D_{2}

( \begin{matrix}1 & 0 \\ x & 1 \end{matrix} ) I( \begin{matrix}1 & 0 \\ x & 1 \end{matrix} )^{−1} = ( \begin{matrix}1 & 0 \\ x & 1 \end{matrix} ) ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ( \begin{matrix}1 & 0 \\ −x & 1 \end{matrix} ) = ( \begin{matrix}α − βx & ∗ \\ ∗ & ∗ \end{matrix} ) .

Hence setting x = β^{−1}α we may assume that α = 0. Since ( \begin{matrix}α & β \\ γ & δ \end{matrix} )^{2} = ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ), we get βγ = a, βδ = 0. Since D_{2} is a division algebra, we conclude that δ = 0, i.e.

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) = ( \begin{matrix}0 & aγ^{−1} \\ γ & 0 \end{matrix} ). Put C = ( \begin{matrix}1 & 0 \\ 0 & aγ^{−1} \end{matrix} ). Then

C ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) C^{−1} = ( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ) .

The lemma is proved.

In view of the above lemma, conjugating by a suitable matrix we may assume that I = ( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ). Let J = ( \begin{matrix}x & y \\ z & t \end{matrix} ). Since

( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ) ( \begin{matrix}x & y \\ z & t \end{matrix} ) = − ( \begin{matrix}x & y \\ z & t \end{matrix} ) ( \begin{matrix}0 & 1 \\ a & 0 \end{matrix} ) ,

we get that z = −ay, t = −x. Therefore, ( \begin{matrix}x & y \\ −ay & −x \end{matrix} )^{2} = ( \begin{matrix}b & 0 \\ 0 & b \end{matrix} ), which means

that x^{2} − ay^{2} = b and xy = yx. Hence the subalgebra l_{1} = k[x, y] is a subfield of D_{2}. Moreover, Q_{l_{1}} = 0. If l is a maximal subfield of D_{2} containing l_{1}, then Q_{l} = (D_{2})_{l} = 0, which implies (D_{1})_{l} = 0, and proves Theorem 5.

REFERENCES

[1] Hoffmann D.W., Tignol J.-P., On 14-dimensional quadratic forms in I^{3}, 8-dimensional forms in I^{2}, and the common value property, Doc. Math. (1998), 189-214.

[2] Izhboldin O.T., Fields of u-invariant 9, Ann. Math. 154 (2001), 529-587.

[3] Izhboldin O.T., Karpenko N.A., Some new examples in the theory of quadratic forms, Math. Z. 234 (2000), 647-695.

[4] Scharlau W., Quadratic and Hermitian forms, Springer, Berlin Heidelberg New York (1985).

[5] Shapiro D.B., Tignol J.-P., Wardsworth A.R., Witt rings and Brauer groups under multiquadratic extensions, II, J. Algebra 78 (1982), 58-90.

[6] Sivatski A.S., Nonexcellence of the function field of the product of two conics, K-theory 34 (2005), 209-218.

[7] Risman L.J., Zero divisors in tensor products of division algebras, Proceedings of the AMS 51 (1975), 35-36.

e-mail: sivatsky@AS3476.spb.edu