## June 3rd, 2017

### On the adjugate

Originally published at 狗和留美者不得入内. You can comment here or there.

I learned that the adjugate is the transpose of the matrix with the minors with the appropriate sign, that as we all know, alternates along rows and columns, corresponding to each element of the matrix on which the adjugate is taken. The matrix, multiplied with its adjugate, in fact, yields the determinant of that matrix, times the identity of course, to matrix it. Note that the diagonal elements of its result is exactly what one gets from applying the minors algorithm for calculating the determinant along each row. The other terms vanish. There are $n(n-1)$ of them, where $n$ is the number of rows (and columns) of the (square) matrix. They are, for each column of the adjugate and each column of it not equal to the current column, the sum of each entry in the column times the minor (with sign applied) determined by the removal of the other selected column (constant throughout the sum) and the row of the current entry. In the permutation expansion of this summation, each element has a (unique) sister element, with the sisterhood relation symmetric, determined by taking the entry of the adjugate matrix in the same column as the non minor element to which the permutation belongs and retrieving in the permutation expansion of the element times minor for that element the permutation, the product representing which contains the exact same terms of the matrix. Note that shift in position of the swapped element in the minor matrix is one less than that in the adjugate matrix. Thus, the signs of the permutations cancel. From this, we arrive at that the entire sum of entry times corresponding minor across the column is zero.

A corollary of this is that $\mathrm{adj}(\mathbf{AB}) = \mathrm{adj}(\mathbf{B})\mathrm{adj}(\mathbf{A})$.

### Galois group of x^10+x^5+1

Originally published at 狗和留美者不得入内. You can comment here or there.

This was a problem from an old qualifying exam, that I solved today, with a few pointers. First of all, is it reducible? It actually is. Note that $x^{15} - 1 = (x^5-1)(x^{10}+x^5+1) = (x^3-1)(x^{12} + x^9 + x^6 + x^3 + 1)$. $1 + x + x^2$, as a prime element of $\mathbb{Q}[x]$ that divides not $x^5-1$ must divide the polynomial, the Galois group of which we are looking for. The other factor of it corresponds to the multiplicative group of $\mathbb{F}_{15}$, which has $8$ elements. Seeing that it has $3$ elements of order $2$ and $4$ elements of order $4$ and is abelian, it must be $C_2 \times C_4$. Thus, the answer is $C_2 \times C_2 \times C_4$.