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### 14.4 Functions and Variables for algebraic extensions

Function: algfac (f, p) ¶

  Returns the factorization of f in the field \(K[a]\). Does the same as `factor(f, p)` which in fact calls `algfac`. One can also specify the variable a as in `algfac(f, p, a)`.

  Examples:

      (%i1) algfac(x^4 + 1, a^2 - 2);
                                 2              2
      (%o1)                    (x  - a x + 1) (x  + a x + 1)
      (%i2) algfac(x^4 - t*x^2 + 1, a^2 - t - 2, a);
                                 2              2
      (%o2)                    (x  - a x + 1) (x  + a x + 1)

  In the second example note that \(a = sqrt(2 + t)\).

Function: algnorm (f, p, a) ¶

  Returns the norm of the polynomial \(f(a)\) in the extension obtained by a root a of polynomial p. The coefficients of f may depend on other variables.

  Examples:

      (%i1) algnorm(x*a^2 + y*a + z,a^2 - 2, a);
                                  2              2      2
      (%o1)/R/                   z  + 4 x z - 2 y  + 4 x

  The norm is also the resultant of polynomials f and p, and the product of the differences of the roots of f and p.

Function: algtrace (f, p, a) ¶

  Returns the trace of the polynomial \(f(a)\) in the extension obtained by a root a of polynomial p. The coefficients of f may depend on other variables which remain “inert”.

  Example:

      (%i1) algtrace(x*a^5 + y*a^3 + z + 1, a^2 + a + 1, a);
      (%o1)/R/                       2 z + 2 y - x + 2

Function: bdiscr (args) ¶

  Computes the discriminant of a basis \(x_i\) in \(K[a]\) as the determinant of the matrix of elements \(trace(x_i*x_j)\). The args are the elements of the basis followed by the minimal polynomial.

  Example:

      (%i1) bdiscr(1, x, x^2, x^3 - 2);
      (%o1)/R/                             - 108
      (%i2) poly_discriminant(x^3 - 2, x);
      (%o2)                                - 108

  A standard base in an extension of degree n is \(1, x, ..., x^{n - 1}\). In this case it is known that the discriminant of this base is the discriminant of the minimal polynomial. This is checked in (%o2) above.

Function: primelmt (f_b, p_a, c) ¶

  Computes a prime element for the extension of \(K[a]\) by a root b of a polynomial \(f_b(b)\) whose coefficients may depend on a. One assumes that f_b is square free. The function returns an irreducible polynomial, a root of which generates \(K[a, b]\), and the expression of this primitive element in terms of a and b.

  Examples:

      (%i1) primelmt(b^2 - a*b - 1, a^2 - 2, c);
                                    4       2
      (%o1)                       [c  - 12 c  + 9, b + a]
      (%i2) solve(b^2 - sqrt(2)*b - 1)[1];
                                        sqrt(6) - sqrt(2)
      (%o2)                       b = - -----------------
                                                2
      (%i3) primelmt(b^2 - 3, a^2 - 2, c);
                                    4       2
      (%o3)                       [c  - 10 c  + 1, b + a]
      (%i4) factor(c^4 - 12*c^2 + 9, a^4 - 10*a^2 + 1);
                       3    2                       3    2
      (%o4) ((4 c - 3 a  - a  + 27 a + 5) (4 c - 3 a  + a  + 27 a - 5)
                                 3    2                       3    2
                       (4 c + 3 a  - a  - 27 a + 5) (4 c + 3 a  + a  - 27 a - 5))/256
      (%i5) primelmt(b^3 - 3, a^2 - 2, c);
                         6      4      3       2
      (%o5)            [c  - 6 c  - 6 c  + 12 c  - 36 c + 1, b + a]
      (%i6) factor(b^3 - 3, %[1]);
                  5       4        3        2
      (%o6) ((48 c  + 27 c  - 320 c  - 468 c  + 124 c + 755 b - 1092)
                 5        5         4       4          3        3          2        2
       ((- 48 b c ) - 54 c  - 27 b c  + 64 c  + 320 b c  + 360 c  + 468 b c  + 149 c
                                 2
       - 124 b c - 1272 c + 755 b  + 1092 b + 1606))/570025

  In (%o1), f_b depends on `a`. Using `solve`, the solution depends on sqrt(2) and sqrt(3). In (%o3), \(K[sqrt(2), sqrt(3)]\) is computed, and we see that the the primitive polynomial in (%o1) factorizes completely here. In (%i5), we compute \(K[sqrt(2), 3^{1/3}]\), and we see that `b^3 - 3` gets one factor in this extension. If we assume this extension is real, the two other factors are complex.

Function: splitfield (p, x) ¶

  Computes the splitting field of the polynomial \(p(x)\). In the generic case it is of degree \(n!\) in terms of the degree \(n\) of p, but may be of lower order if the Galois group of p is a strict subgroup of the group of permutations of \(n\) elements. The function returns a primitive polynomial for this extension and the expressions of the roots of p as polynomials of a root of this primitive polynomial. The polynomial f may be irreducible or factorizable.

  Examples:

      (%i1) splitfield(x^3 + x + 1, x);
                                                    4         2
                    6         4         2       alg1  + 5 alg1  - 9 alg1 + 4
      (%o1)/R/ [alg1  + 6 alg1  + 9 alg1  + 31, ----------------------------,
                                                             18
                                       4         2          4         2
                                   alg1  + 5 alg1  + 4  alg1  + 5 alg1  + 9 alg1 + 4
                                 - -------------------, ----------------------------]
                                            9                        18
      (%i2) splitfield(x^4 + 10*x^2 - 96*x - 71, x)[1];
                   8           6           5            4             3
      (%o2)/R/ alg2  + 148 alg2  - 576 alg2  + 9814 alg2  - 42624 alg2
                                                          2
                                             + 502260 alg2  + 1109952 alg2 + 18860337

  In the first case we have the primitive polynomial of degree 6 and the 3 roots of the third degree equations in terms of a variable `alg1` produced by the system. In the second case the primitive polynomial is of degree 8 instead of 24, because the Galois group of the equation is reduced to D8 since there are relations between the roots.

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