By Hans Sterk
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Additional info for Algebra 3: algorithms in algebra [Lecture notes]
So a0 (X, Y ) + a1 (X, Y )f (X, Y ) + · · · + am (X, Y )f (X, Y )m ∈ (p(X), q(Y )). Altogether we find that F (X, Y, Z) = (Z − f (X, Y ))h(X, Y, Z) + a0 (X, Y )+ a1 (X, Y )f (X, Y ) + · · · + am (X, Y )f (X, Y )m ∈ ((p(X), q(Y ), Z − f (X, Y )). Now the kernel of φ is the kernel of the composition ψ◦j, where j denotes the inclusion map. , g(Z) ∈ I (note that we identify Q [Z] with its image under j). √The minimal polynomial of 2 is X 2 − 2 and the minimal √ √polynomial of 3 2 is X 3 − 2. To find the minimal polynomial of 2 + 3 2 we compute a Gr¨obner basis for the ideal (X 2 − 2, Y 3 − 2, Z − X − Y ) in Q [X, Y, Z] with respect to the lex order, where X > Y > Z.
From the analyst’s point of view, the arctan may be the best way of writing the integral, but for an algebraist the arctan obscures the structure of the integral in the following sense. The identity x2 i 1 1 1 =− ( − ) +1 2 x+i x−i enables us to rewrite the integral with the help of logarithms as x2 i 1 dx = − (log(x + i) − log(x − i)), +1 2 at the expense of extending the coefficients to Q (i). In the exercises you will show that more integrals, which are usually not expressed in terms of logarithms, can be expressed in terms of logarithms.
For example, in characteristic 2, we have X 8 + X 4 + 1 = (X 2 )4 + X 4 + 14 = (X 2 + X + 1)4 . Now let’s suppose we are given g n and we wish to find g. The first approach is to differentiate g n , this yields (g n ) = ng g n−1 , and divide g n by this derivative. ) This goes well if p does not divide n and if g = 0. If g = 0, then all exponents in g are divisible by p. By using the above equality repeatedly, we can absorb all these factors p and rewrite g n as k g(X)n = h(X p )t , with p relatively prime with t and with at least one of the exponents of h(X).
Algebra 3: algorithms in algebra [Lecture notes] by Hans Sterk