Chapter 3: Polynomial and Rational Functions
... squared can ever be negative. But Italians trying to solve more complicated equations found they needed to use solutions to equations like our example, in passing, on their way to real number solutions of the equations that interested them. So imagine, they said, that quadratics like the above have ...
... squared can ever be negative. But Italians trying to solve more complicated equations found they needed to use solutions to equations like our example, in passing, on their way to real number solutions of the equations that interested them. So imagine, they said, that quadratics like the above have ...
Algebra Notes
... intersecting them, we construct the point (x, y). Now imagine you’re trying to find all the constructible numbers. You already know that every rational number is constructible, so by the above claim, every point whose coordinates are rational. To find more constructible points, you try to find a way ...
... intersecting them, we construct the point (x, y). Now imagine you’re trying to find all the constructible numbers. You already know that every rational number is constructible, so by the above claim, every point whose coordinates are rational. To find more constructible points, you try to find a way ...
IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION
... for some i ∈ f1, 2, 3g. Now, if (x0 , y0 ), (x00 , y00 ) 6= (0, 0) are 2 solutions to (†) then (x00 , y00 ) = (λx0 , λy0 ) for some non-zero constant λ: We can deduce this since we had (ui , vi ) 6= (0, 0). Consequently, these 2 solutions to (†) (for a fixed i) give us the same intersection point (x ...
... for some i ∈ f1, 2, 3g. Now, if (x0 , y0 ), (x00 , y00 ) 6= (0, 0) are 2 solutions to (†) then (x00 , y00 ) = (λx0 , λy0 ) for some non-zero constant λ: We can deduce this since we had (ui , vi ) 6= (0, 0). Consequently, these 2 solutions to (†) (for a fixed i) give us the same intersection point (x ...
Key Recovery on Hidden Monomial Multivariate Schemes
... monomial of the form P (x) = x1+q over an extension F of degree n of the base finite field K is hidden by two linear bijective mappings S and T . The public key is P = T ◦ P ◦ S and if some polynomials of the public key are removed, we get a SFLASH public key. In[5], the authors consider the case wh ...
... monomial of the form P (x) = x1+q over an extension F of degree n of the base finite field K is hidden by two linear bijective mappings S and T . The public key is P = T ◦ P ◦ S and if some polynomials of the public key are removed, we get a SFLASH public key. In[5], the authors consider the case wh ...
SOLVING QUADRATIC EQUATIONS OVER POLYNOMIAL RINGS
... (1) with n ≥ 1 can be reduced to solving in B a finite sequence of at most 1 + n + · · · + nd polynomial equations in one variable over B, each of them of degree ≤ n. Proof: Without loss of generality, we can assume that an 6= 0. We proceed by induction on d. When d = −∞, i.e., t = 0, we do not need ...
... (1) with n ≥ 1 can be reduced to solving in B a finite sequence of at most 1 + n + · · · + nd polynomial equations in one variable over B, each of them of degree ≤ n. Proof: Without loss of generality, we can assume that an 6= 0. We proceed by induction on d. When d = −∞, i.e., t = 0, we do not need ...
Math 248A. Norm and trace An interesting application of Galois
... Note that this takes care of characteristic 0. But of course what is more interesting is that even in positive characteristic, such as for finite fields, the trace is non-vanishing for separable extensions. Proving this (even uniformly across all characteristics at once) requires a better technique. ...
... Note that this takes care of characteristic 0. But of course what is more interesting is that even in positive characteristic, such as for finite fields, the trace is non-vanishing for separable extensions. Proving this (even uniformly across all characteristics at once) requires a better technique. ...
