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Recent questions in Polynomial arithmetic
Algebra IIAnswered question
Demarion Ortega Demarion Ortega 2022-11-19

Why is the triangle inequality equivalent to a 4 + b 4 + c 4 2 ( a 2 b 2 + b 2 c 2 + c 2 a 2 )
Consider the existential problem of a triangle with side lengths a , b , c 0. Such a triangle exists if and only if the three triangle inequalities
(0) a + b c , b + c a and c + a b
are all satisfied.
Alternatively, if 1 2 3 are the values of a,b and c ordered in ascending order, then the triangle exists iff 1 + 2 3 .
Interestingly, the three triangle inequalities can be recast into a single quartic polynomial inequality. Let 0 , x , y R 2 be the three vertices of the triangle, with x = a , y = b and x y = c. Then c 2 = x y 2 = x 2 2 x , y + y 2 = a 2 + b 2 2 x , y
Therefore x T y = x , y = 1 2 ( a 2 + b 2 c 2 ) and
(1) ( x T y T ) ( x y ) = 1 2 ( 2 a 2 a 2 + b 2 c 2 a 2 + b 2 c 2 2 b 2 ) .
The RHS of (1) must be positive semidefinite because the LHS is a Gram matrix. Conversely, if the RHS is indeed PSD, it can be expressed as a Gram matrix. Hence we obtain x and y and the triangle exists.
As 2 a 2 and 2 b 2 are already nonnegative, the RHS of (1) is positive semidefinite if and only if ( 2 a 2 ) ( 2 b 2 ) ( a 2 + b 2 c 2 ) 2 0, by Sylvester's criterion. That is, the triangle exists if and only if
(2) ( a 4 + b 4 + c 4 ) + 2 ( a 2 b 2 + b 2 c 2 + c 2 a 2 ) 0.
This polynomial inequality can be derived by more elementary means. See circle-circle intersection on Wolfram MathWorld. The geometric explanation for the necessity of (2) is given by Heron's formula, which states that the square root of the LHS is four times the area of the triangle.
Since both (0) and (2) are necessary and sufficient conditions for the existence of the required triangle, the two sets of conditions must be equivalent to each other. Here are my questions. Is there any simple way to see why (0) and (2) are equivalent? Can we derive one from the other by some basic algebraic/arithmetic manipulations?

Algebra IIAnswered question
vedentst9i vedentst9i 2022-11-19

Prove that given projective curve has genus 1.
Show that
X = { ( ( x 0 : x 1 ) , ( y 0 : y 1 ) ) : ( x 0 2 + x 1 2 ) ( y 0 2 + y 1 2 ) = x 0 x 1 y 0 y 1 ) } P 1 × P 1
is a smooth curve of genus 1.
I can prove it with the following reasoning.
Using the Segre Embedding, the curve consists on all the elements ( x : y : z : w ) P 3 satisfying the equations
x 2 + y 2 + z 2 + w 2 x w = 0 , x w y z = 0 ,
i.e. the vanishing set of I= I = ( x 2 + y 2 + z 2 + w 2 x w , x w y z ). I can use the Jacobian criterion to prove it's smooth; no issue.
To find the genus, since I'm better at it, I decided to compute the arithmetic genus. To do that, I proved that
{ x 2 + y 2 + z 2 + w 2 x w , x w y z }
is in fact a Gröbner basis (Using GRevLex ordering), so
L T ( I ) = ( x 2 , y z ) .
Then every minimal free resolution of the quotient over a monomial ideal (I think I can remove the hypothesis of it being monomial; I'm not completely sure, but in such case I don't need to compute LT(I) or even prove that I have a Gröbner basis) generated by two elements has the form 0 S S 2 S 0, in particular, in this case it has the form
0 S ( 4 ) S ( 2 ) 2 S 0
which allows me to compute the Hilbert polynomial of X, by the method in the first section of The Geometry of Syzygies from Eisenbud (I remember it's also used in Cox's), and with the Hilbert polynomial I also have that the arithmetic genus is 1.
But can I prove it without computing the Hilbert polynomial?

Algebra IIAnswered question
Mark Rosales Mark Rosales 2022-11-15

Evaluation of polynomial modulo in GF(2)
I'm new to this forum and I have a question about about modular arithmetics. My background is in computational chemistry and I work now as a software developer. For my work I need to learn about cryptography and I'm reading the book "Cryptography Made Simple" by Nigel P. Smart.
In paragraph 1.2 on Finite Fields, the author gives an example of polynomial modulo in F 2 ( x ) / f ( x ) F 2 ( x ) as such:
( 1 + x + x 2 ) ( x + x 3 ) ( mod x 4 + 1 ) = x 2 + 1
However, when I do the evaluation myself I get a different result, which is confirmed by using the PolynomialMod function in Mathematica. I will show the steps I took below. First of all I expand the multiplication and evaluate modulo 2:
x 5 + x 4 + 2 x 3 + x 2 + x ( mod 2 ) = x 5 + x 4 + x 2 + x
Then:
x 5 + x 4 + x 2 + x ( mod x 4 + 1 )
x ( x 4 + 1 ) + x 4 + x 2 ( mod x 4 + 1 )
x ( x 4 + 1 ) + ( x 4 + 1 ) + x 2 1 ( mod x 4 + 1 )
Which finally results in:
( x + 1 ) ( x 4 + 1 ) + x 2 1 ( mod x 4 + 1 ) = x 2 1
I put in attachment a screenshot 1 from Mathematica to confirm it matches my result. Initially I thought it was an error in the book, but in a following section the author gives another example:
( x 3 + 1 ) ( x 4 + 1 ) ( mod x 7 + x + 1 ) = x 4 + x 3 + x
Again, evaluating the expression manually I obtain:
( x 3 + 1 ) ( x 4 + 1 ) ( mod x 7 + x + 1 ) = x 4 + x 3 x
Which is confirmed once more by Mathematica, see screenshot 2.
Does anyone know if there is a reason behind the difference in the results? Any help would be much appreciated.

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