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Prove that the discriminant of the polynomial $X^{n}+pX+q$ is $$D=(-1)^{n(n-1)/2}n^{n}q^{n-1}+(-1)^{(n-1)(n-2)/2}(n-1)^{n-1}p^{n}$$

I found the same exercise here: (From Artin)Prove the general formula of discriminant of $x^n + px + q$ for $n ≥ 2$

But my teacher told me to solve it from the Sylvester matrix (sorry if you consider this a duplicate).

Let $F(X)=X^{n}+pX+q$, then $F'(X)=nX^{n-1}+p$. We compute the discriminant as the determinant of the Sylvester matrix of $F$ and $F'$.

$$D(F)=R_{F,F'}=\det\left(\begin{array}{ccccc|cccccc} 1 & 0 & \cdots & 0 & 0 & p & q & 0 & \cdots & 0 & 0\\ 0 & 1 & \cdots & 0 & 0 & 0 & p & q & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & 1 & 0 & 0 & 0 & 0 & \cdots & q & 0\\ 0 & 0 & \cdots & 0 & 1 & 0 & 0 & 0 & \cdots & p & q\\ \hline n & 0 & \cdots & 0 & 0 & p & 0 & 0 & \cdots & 0 & 0\\ 0 & n & \cdots & 0 & 0 & 0 & p & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & 0 & n & 0 & 0 & 0 & \cdots & p & 0\\ 0 & 0 & \cdots & 0 & 0 & n & 0 & 0 & \cdots & 0 & p \end{array}\right)$$

This matrix has size $ (2n-1)\times(2n-1)$.

My teacher told me that there is a special row from which I can start computing the determinant, but I don't know.

Edit2: Using the last row: $$D(F)=(-1)^{n}n\det\left(\begin{array}{ccccc|ccccc} 1 & 0 & \cdots & 0 & 0 & q & 0 & \cdots & 0 & 0\\ 0 & 1 & \cdots & 0 & 0 & p & q & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & 1 & 0 & 0 & 0 & \cdots & q & 0\\ 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & p & q\\ \hline n & 0 & \cdots & 0 & 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & n & \cdots & 0 & 0 & p & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & 0 & n & 0 & 0 & \cdots & p & 0 \end{array}\right)+$$ $$+(-1)^{2n-2}p\det\left(\begin{array}{ccccc|ccccc} 1 & 0 & \cdots & 0 & 0 & p & q & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0 & 0 & 0 & p & q & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 & 0 & 0 & 0 & 0 & \cdots & q\\ 0 & 0 & \cdots & 0 & 1 & 0 & 0 & 0 & \cdots & p\\ \hline n & 0 & \cdots & 0 & 0 & p & 0 & 0 & \cdots & 0\\ 0 & n & \cdots & 0 & 0 & 0 & p & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 0 & n & 0 & 0 & 0 & \cdots & p \end{array}\right)$$

Where every block is a square of size $n-1$. Then we can use this formula for square blocks: $\det\begin{pmatrix}A & B\\ C & D \end{pmatrix}=\det(AD-BC)$

Then: $$D(F)=(-1)^{n}n\det\left(\begin{array}{ccccc} -nq & 0 & \cdots & 0 & 0\\ p-np & -nq & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & -nq & 0\\ 0 & 0 & \cdots & p-np & -nq \end{array}\right)+$$ $$+(-1)^{2n-2}p\det\left(\begin{array}{ccccc} p-np & -nq & \cdots & 0 & 0\\ 0 & p-np & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & p-np & -nq\\ 0 & 0 & \cdots & 0 & p \end{array}\right)$$

$$=(-1)^{n}(-1)^{n-1}nn^{n-1}q^{n-1}+(-1)^{2n-2}pp^{n-1}(1-n)^{n-1}=$$ $$=(-1)^{2n-1}n^{n}q^{n-1}+(-1)^{3n-1}(n-1)^{n-1}p^{n}$$ But the result must be $$(-1)^{\frac{n(n-1)}{2}}n^{n}q^{n-1}+(-1)^{\frac{(n-1)(n-2)}{2}}(n-1)^{n-1}p^{n}$$

I did something wrong with the exponents of the $(-1)$ terms in all this process.

Miguel Mars
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  • Did you ask your teacher? :-| – Mariano Suárez-Álvarez Jan 15 '18 at 20:12
  • In any case, you have lots of 1's. Have you tried doing row and column operations to simplify the matrix? Since you have no idea what your teacher meant, you can try other things... – Mariano Suárez-Álvarez Jan 15 '18 at 20:13
  • @MarianoSuárez-Álvarez I computed the determinant, but didn't get the expected result. – Miguel Mars Jan 15 '18 at 22:11
  • If $A$ and $D$ are square blocks of size $n-1$, then the whole matrix has to have size $2n-2$. It doesn't. – Mariano Suárez-Álvarez Jan 15 '18 at 22:18
  • @MarianoSuárez-Álvarez Right, $D$ is a block of size $n$. Thank you. – Miguel Mars Jan 15 '18 at 22:52
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    It looks like the ideal row might be the last row. Each row gives a lot of zeros. But the two minorants you get with the last row, seem to involve some fairly simple diagonal matrix or a sum of a diagonal and off-diagonal matrix. –  Jan 15 '18 at 23:49
  • @Robert Thank you, I almost have it now. Now I only have trouble with the exponents of $(-1)$. – Miguel Mars Jan 16 '18 at 00:28
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    According to this paper: http://www2.math.uu.se/~svante/papers/sjN5.pdf the discriminant of your polynomial isn't given precisely by the determinant of this Sylvester matrix. Rather, there is a factor of $(-1)^{n(n-1)/2}$ that should be in front. Also, when you do row expansion the exponent on the $(-1)$ should be $i+j$ where $(i,j)$ is the coordinate of the coefficient of the minorant. This should clear up any sign issues. –  Jan 16 '18 at 16:34

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