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 2018-12-03, 16:15 #1 Dr Sardonicus     Feb 2017 Nowhere 2×29×53 Posts Basic stuff about discriminants A primer on discriminants Definitions and basic terminology The most common formulas for the polynomial discriminant are given on the Wolfram Mathworld page here. The resultant is then described here, and the Sylvester matrix is here. $f = \sum_{i=1}^{n}c_{i}x^i$ $\Delta(f)\;=\;c_n^{2n-2}\prod_{i We will assume polynomials are monic (leading coefficient is 1), which, besides simplifying the formulas by getting rid of the variable cn, also excludes the zero polynomial from consideration. Also, algebraic integers have defining polynomials that are monic in Z[x]. We will also assume the coefficients are in a field k, the ground field. This will usually be either the rational numbers, or a finite field. A field extension K/k is an inclusion (as sets) of fields, $k \subset K$ in which the arithmetic of k is also included as a subset of the arithmetic of K. In particular, K is a vector space over k. If K is a finite-dimensional k-vector space, the dimension is called the degree of the extension. A field extension K/k is called separable if every polynomial which is irreducible in k[x] and splits into linear factors in K[x] has no repeated roots. An irreducible polynomial in k[x] with repeated factors in K[x] is called inseparable, and the corresponding field extension is called inseparable. Inseparable extensions arise from fields k of characteristic p with elements r having no p-th root in k. The polynomial x^p - r then defines an inseparable extension of k. If k has characteristic p, and every element of k has p-th roots in k, k is called perfect. All finite extensions of a perfect field are separable. All field extensions of a field k of characteristic zero (i.e. k contains the rational field Q) are separable. Also, finite fields of characteristic p are "perfect," so all finite extensions of finite fields are separable. An extension being separable has a certain significance with regard to the discriminant. The discriminant and permutations The discriminant of a monic polynomial has a square root consisting of the product $\prod_{i This product can be used to distinguish whether a permutation of the x's is even or odd. A single transposition changes the sign of exactly one factor, hence changes the sign of the product. It can be shown that any permutation of the x's can be expressed as a product of transpositions. Because the effect of a given permutation on the above product is independent of such an expression, it follows that the number of permutations in a product expressing a given permutation is either always even (the product is unchanged), or is always odd (changes the sign of the product). In the context of a finite normal and separable extension with Galois group G, the this means that the discriminant is a square in k precisely when G consists entirely of even permutations. Sign of the discriminant There is one very simple application of this permutation idea which does not require any knowledge of Galois groups. Let f = f(x) be a monic polynomial with real coefficients and no repeated factors. The only result we need is that any non-real roots of f(x) = 0 occur in complex-conjugate pairs. This is an immediate consequence of the factor theorem: f(z) = 0 when x - z is a factor of f(x). Writing $f(x) \; = \; (x - z)q(x)$ and taking complex conjugates, we have, since f(x) has real coefficients so is invariant under complex conjugation, $f(x) \; = \; (x - \bar{z})\bar{q}(x)$ so if z is a non-real complex root, so is its complex conjugate. Now, the effect of complex conjugation on the roots of f(x) = 0 is clear: it transposes all pairs of complex-conjugate roots of f(x) = 0, and leaves all real roots invariant. Therefore, complex conjugation is an even permutation if there are evenly many complex-conjugate pairs of roots, and an odd permutation if there are oddly many such pairs. We conclude that, if f(x) = 0 has evenly many complex-conjugate pairs of roots, then the discriminant has a square root which is invariant under complex conjugation, i.e. a real square root. The discriminant is therefore positive. If the number of complex-conjugate pairs is odd, then the discriminant has a square root whose sign is changed by complex conjugation; that is, a pure-imaginary square root. The discriminant is therefore negative. Thus, if f(x) is a monic polynomial with real coefficients and no repeated factors, its discriminant is positive or negative, according to whether the number of pairs of complex-conjugate roots is even or odd. Another definition of "discriminant" If K/k is a field extension of degree n, there is another definition of "discriminant," namely the discriminant of a basis of K as a vector space over k. This is possible because, besides addition in K, and "scalar multiplication" by elements of k, there is also multiplication by other elements of K. Multiplication by a element of K defines a k-linear transformation of K, as a vector space over k: $T_{\alpha}:\;x\;\rightarrow \alpha x, \; x\;\in\;K$ This transformation is invertible, except for multiplication by 0. The trace (from K to k) of an element of K is the trace of the linear transformation defined by multiplication. The value of the trace is an element of k. Now, suppose K/k is a separable field extension of degree n, and $[x_1,\;\dots\;x_n]$ is a k-basis of K as a vector space. Then the discriminant of this basis is defined as $\Delta(x_1,\;\dots\;x_n)\;=\;\text{Det}(M)\text{, where }M\text{ is the nxn matrix with }M_{ij}\;=\;trace(x_{i}x_{j})$. Since the elements of this matrix are in k, the discriminant is an element of k. We note that the matrix is symmetric. To describe the discriminant more clearly, we note that every element of K is algebraic over k. We imagine k and K to be imbedded in an algebraically closed field. This allows us to use another formulation of the trace, as the sum of the "algebraic conjugates" of x. This gives us the decomposition of M as a product of transpose matrices: $M\; = \; NN^{T}:\; N_{ij}=\sigma_{j}(x_{i}),\;N_{ij}^{T} = \sigma_{i}(x_{j})$ We can glean one important fact directly from this formulation: Changing the basis of K as a vector space over k, changes the discriminant by a factor which is a non-zero square in k -- namely, the square of the determinant of the nxn matrix defining the change of basis. This notion of discriminant is related to the polynomial discriminant as follows: Assume there is a "primitive element" for the extension K/k, i.e. an element x of K such that f(x) = 0 for a (monic) irreducible polynomial of degree n in k[x]. Then $[x_i = x^{i-1},\; i\; = \; 1,\;\dots\;n]$ is a k-basis of K. For this "power basis," the matrix N has the form $N_{ij}=\sigma_{j}(x^{i-1})$ so N a Vandermonde matrix. The usual formula for the determinant of a Vandermonde matrix shows that the discriminant of the power basis equals the polynomial discriminant of f. We are assuming that K/k is separable, so that the roots of f(x) = 0 are all distinct, so the polynomial discriminant is non-zero. Therefore, the discriminant of any k-basis of K is non-zero. Now, in the usual situation where K is a number field and k = Q, the rationals, we generally have a defining polynomial which is a monic irreducible polynomial in Z[x]. The field K may then be viewed as the "residue field" Q[x]/(f(x)) in which the arithmetic is defined modulo f(x). The discriminant of any Q-basis of K is a rational number. It is easily shown that any basis may be transformed into a basis consisting entirely of algebraic integers (roots of monic polynomials). The discriminant of a basis consisting of algebraic integers is a non-zero rational integer. There is obviously a minimum absolute value to such discriminants. The discriminant having this minimum absolute value is called the field discriminant of the extension K/Q. A basis of algebraic integers of K having this discriminant is a Z-basis of the ring R = OK of algebraic integers of K. A tantalizing question is, whether the ring R of algebraic integers of K has a power basis, i.e. R = Z[x] (modulo f(x)) for some defining polynomial f(x). In this case, the ring R is sometimes called "monogenic," meaning it is generated as a ring over Z by a single element. If K/Q is quadratic, there are always polynomials for which R = Z[x] as is well known. When K/Q is of degree 3 or more, though, there are examples where there is no such f(x). There are also results showing that, in many cases, Abelian extensions of Q do not have power bases for their rings of integers. One case in which the ring of integers does have a power basis, is cyclotomic fields of n-th roots of unity. Last fiddled with by Dr Sardonicus on 2018-12-03 at 17:02
2018-12-05, 10:43   #2
Nick

Dec 2012
The Netherlands

101010001002 Posts

Quote:
 Originally Posted by Dr Sardonicus A tantalizing question is, whether the ring R of algebraic integers of K has a power basis
Determining the ring of integers at all in cases where the discriminant is too large to factorize over ℤ is already a hard problem!

2018-12-05, 14:00   #3
Dr Sardonicus

Feb 2017
Nowhere

60028 Posts

Quote:
 Originally Posted by Nick Determining the ring of integers at all in cases where the discriminant is too large to factorize over ℤ is already a hard problem!
If you can't factor the discriminant, you're sort of up a stump -- even with a quadratic polynomial, where the ring of integers is known to be monogenic.

Using the formulation of the discriminant as a product of squares of differences of roots, it is clear that a change of variable consisting of a translation by an element of the ground field (usually, a rational integer) does not affect either the discriminant or the extension(s) defined by the polynomial.

The following discussion covers the basics pretty well, including the fact that, if there is a power basis, the number of monic irreducible polynomials in Z[x] defining such a basis is (up to integer translates of the variable) finite.

Which number fields are monogenic? and related questions

There are some general results giving classes of number fields in which answer to the "power basis" question is in the negative. I've read the first of the following references (which is mentioned in the above link) and (I think) the second. If the second reference is the one I'm thinking of, the result applies to all Abelian extensions of degree relatively prime to 6, and greater than some modest bound.

M.-N. Gras, Non monogeneite de l'anneau des entiers des extensions cycliques de Q de
degre premier l >= 5, J. Number Theory 23 ( 1986 ), 347-353.

M.-N. Gras, Non monogeneite de l'anneau des entiers de certaines extensions abeliennes
de Q, Publ. Math. Sci. Besancon, Theor. Nombres, 1983-1984.

M.-N. Gras. Condition necessaire de monogeneite de l’anneau des entiers d’une extension abelienne de Q.
In Seminaire de theorie des nombres, Paris 1984–85, volume 63 of Progr. Math., pages 97–107. Birkhauser Boston, Boston, MA, 1986.

 2018-12-05, 22:34 #4 Dr Sardonicus     Feb 2017 Nowhere 2·29·53 Posts More fun and games with discriminants More discriminant formulas Assume f = f(x) is a monic polynomial of degree n with real coefficients. Then f(x) may be expressed as a product of linear factors $\prod_{i=1}^n (x - r_{i})$ where some of the ri may be complex. Using the product formula for the derivative, we find that f'(x) is the sum of the products of n-1 terms of this product. Substituting a given root of f(x) = 0 into this formula causes all but one of these terms to have a factor of 0, giving, for each i from 1 to n, $f'(r_{i}) = \prod_{j\ne i}(r_{i}-r_{j})$ Multiplying these products together, and noting that the pairs (i, j) and (j, i) appear in equal and opposite factors, we obtain $\text{disc}(f)\; = \; (-1)^{\frac{n(n-1)}{2}}\prod_{i=1}^n f'(r_{i})$ This formula gives (with some algebraic manipulation) an evaluation of the discriminants of trinomials of the form x^n + a*x + b. A generalization to trinomials x^n + a*x^m + b with 1 < m < n is not quite as amenable to this formulation. However, using the factored form of the derivative, $f'(x) = n\prod_{j=1}^{n-1}(x - \xi_{j})$ and regrouping, we find $\text{disc}(f) = n^{n}(-1)^{\frac{n(n-1)}{2}} \prod_{j=1}^{n-1}f(\xi_{j})$ This expression may be a bit easier to evaluate for trinomial f(x). There is a special kind of polynomial, called a self-reciprocal polynomial, $f(x) = x^{n}f(1/x)$ for which the discriminant can be rewritten in a non-trivial way. Such a polynomial is "palindromic" in that the coefficient of x^k is equal to the coefficient of x^(n-k). We will assume that f(1) and f(-1) are both non-zero; that is, neither x-1 nor x+1 is a factor of f, and that f is monic. Under these assumption, the degree of f(x) is even, n = 2k,and the roots of f(x) = 0 occur in reciprocal pairs. Thus, and we may write $f(x) = x^{\frac{n}{2}}\sum_{i=1}^{\frac{n}{2}}(c_{i}(x^{i}+x^{-i}))$ which can be rewritten as $f(x) = x^{k}g(x+\frac{1}{x})$ where g is a polynomial of degree k = n/2. Under our assumption that f is monic, g is also monic. We note that if f(r) = 0, then g(r+1/r) = 0. We then find, if f(r) = 0 (and remembering the chain rule!) $f'(r) = r^{k}g'(r+\frac{1}{r})(1 - \frac{1}{r^{2})$ In the formula for the discriminant as a product over values of f'(r), the power of -1 here is (-1)^k(2k - 1) which is equal to (-1)^k. As r runs through the roots of f(x) = 0 once, r + 1/r runs through the roots of g(x) = 0 twice. Also, since the r's occur in reciprocal pairs, the product of all the r^k is 1. We then have $\text{disc}(f) = (-1)^{k}\prod_{i=1}^{2k}f'(r_{i}) = (-1)^{k}(\text{disc}(g))^{2}f(1)f(-1)$

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