Homepage › Solution manuals › Ivan Niven › An Introduction to the Theory of Numbers › Exercise 3.6.10 ($f(x,y)$ and $f_0(x,y)=x^2 + y^2$ are properly equivalent if $f_0 = f\cdot M,\ M \in \mathrm{GL}(2,\mathbb{Z})$)

An Introduction to the Theory of Numbers

Exercise 3.6.10 ($f(x,y)$ and $f_0(x,y)=x^2 + y^2$ are properly equivalent if $f_0 = f\cdot M,\ M \in \mathrm{GL}(2,\mathbb{Z})$)

Suppose that a matrix M with integral elements and determinant − 1 takes a form f ( x , y ) = a x 2 + bxy + c y 2 to x 2 + y 2 . Prove that f and x 2 + y 2 are equivalent by showing that there is another matrix M 1 with integral elements and determinant + 1 , that also takes f to x 2 + y 2 .

Answers

Proof. By hypothesis, the forms f ( x , y ) = a x 2 + bxy + c y 2 and f 0 ( x , y ) = x 2 + y 2 satisfy f 0 = f ⋅ M , where M = ( α β γ δ ) ∈ GL 2 ( ℤ ) and det ⁡ ( M ) = − 1 , so that

x 2 + y 2 = a ( αx + βy ) 2 + b ( αx + βy ) ( γx + δy ) + c ( γx + δy ) 2 .

Then, using the substitution ( x , y ) ← ( x , − y ) ,

x 2 + y 2 = x 2 + ( − y ) 2 = a ( αx + β ( − y ) ) 2 + b ( αx + β ( − y ) ) ( γx + δ ( − y ) ) + c ( γx + δ ( − y ) ) 2 = a ( αx − βy ) 2 + b ( αx − βy ) ( γx − δy ) + c ( γx − δy ) 2 .

Therefore f 0 = f ⋅ M 1 , where

M 1 = ( α − β γ − δ ) = ( α β γ δ ) ( 1 0 0 − 1 )

satisfies det ⁡ ( M 1 ) = − det ⁡ ( M ) = 1 .

Then M 1 ∈ Γ and M 1 takes f to x 2 + y 2 , so f and x 2 + y 2 are properly equivalent. □

Note: we used the matrix A = ( 1 0 0 − 1 ) , which is an automorphism of f 0 in GL 2 ( ℤ ) , i.e. f 0 ⋅ A = f 0 . Then M 1 = MA satisfies

f ⋅ M 1 = ( f ⋅ M ) ⋅ A = f 0 ⋅ A = f 0 .

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2024-11-29 11:21
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