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QUADRATIC RINGS PETE L. CLARK 1. Quadratic fields and quadratic rings Let D be a squarefree integer not equal to 0 or 1. Then D is irrational, and Q [ D ], the subring of C obtained by adjoining D to Q , is a field. From an abstract algebraic perspective, an explanation for this can be given as follows: since D is irrational, the polynomial t 2 - D is irreducible over Q . Since the ring Q [ t ] is a PID, the irreducible element t 2 - D generates a maximal ideal ( t 2 - D ), so that the quotient Q [ t ] / ( t 2 - D ) is a field. Moreover, the map Q [ D ] Q [ t ] / ( t 2 - D ) which is the identity on Q and sends D t is an isomorphism of rings, so Q [ D ] is also a field. We may write Q [ D ] = { a + b D | a, b Q } , so that a basis for Q [ D ] as a Q -vector space is 1 , D . In particular Q [ D ] is two-dimensional as a Q -vector space: we accordingly say it is a quadratic field . It is also easy to check by hand that the ring Q [ D ] is a field. For this and for many other things to come, the key identity is ( a + b D )( a - b D ) = a 2 - Db 2 . For rational numbers a and b which are not both zero, the rational number a 2 - Db 2 is also nonzero: equivalently there are no solutions to D = a 2 b 2 , because D is irrational. It follows that – again, for a , b not both 0 – we have a + b D · a a 2 - Db 2 - b a 2 - Db 2 D = 1 , which gives a multiplicative inverse for a + b D in Q [ D ]. We wish also to consider quadratic rings , certain integral domains whose frac- tion field is a quadratic field Q ( D ). Eventually we will want a more precise and inclusive definition, but for now we consider Z [ D ] = { a + b D | a, b Z } . 1 2. Fermat’s Two Squares Theorem The rings Z [ D ] occur naturally when we study Diophantine equations. E.g: 1 This equality is a fact which is not difficult to check; it is not the definition of Z [ D ]. By way of comparison, we recommend that the reader check that the ring Z [ D 2 ] is not of the form Z α + Z β for any two fixed elements α, β of Z [ D 2 ]. In fact its additive group is not finitely generated as an abelian group. 1

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2 PETE L. CLARK Question 1. Which prime numbers p can be expressed as a sum of two squares? More precisely, for for which prime numbers p are there integers x and y such that (1) x 2 + y 2 = p ? Remark: Evidently 2 is a sum of squares: 1 2 +1 2 = 2. Henceforth we assume p > 2. At the moment we have exactly one general technique 2 for studying Diophantine equations: congruences. So let’s try to apply it here: if we reduce the equation x 2 + y 2 = p modulo p , we get x 2 + y 2 0 (mod p ). Suppose ( x, y ) Z 2 is a solution to (1). It cannot be that p | x , because then we would also have p | y , and then p 2 | x 2 + y 2 = p , a contradiction. So both x and y are nonzero modulo p (hence invertible in Z /p Z ) and the congruence can be rewritten as x y 2 ≡ - 1 (mod p ) .
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