# What is the essence of the constant factor in the standard definitions of the discriminant?

Let $f(x) = x^m+\sum_{j=0}^{m-1}f_{m-j}x^j\in P[x]$ be a monic polynomial over a field $P$ and let $f(x) = (x-\alpha_1)\cdot\ldots\cdot(x-\alpha_m)$ be a factorization of $f$ over an extension field $Q$ of $P$.

Then it is quite natural to consider a value (called discriminant) $$\prod_{1\leq i<j\leq m}(\alpha_i-\alpha_j)^2,$$ whose being zero or not determines whether or not the polynomial has any multiple roots (in any extension field of $P$, not only $Q$).

But, when $f$ is not monic, say $f = f_0 \cdot x^m+\sum_{j=0}^{m-1}f_{m-j}x^j$, a usual definition of the discriminant reads $$f_0^{2m-2}\cdot \prod_{1\leq i<j\leq m}(\alpha_i-\alpha_j)^2.$$

So, my question is:

What is the essence of the factor $f_0^{2m-2}$?

• That is not quite correct: you left out a factor $(-1)^{m(m-1)/2}$. – Robert Israel Sep 6 '17 at 17:36
• Thank you, but I think that I forgot a square =) – Mikhail Goltvanitsa Sep 6 '17 at 18:11
• Worth pointing out: in the literature one sometimes finds authors explicitly terminologically distinguishing the so-called normalized discriminant, which simply has the factor in question left out, from the so-called standard discriminant, which has the factor $f_0^{2m-2}$. – Peter Heinig Sep 6 '17 at 18:40
• @Peter, I don't understand the edits you made. You changed "But, when $f$ is not monic" to "But, when $f$ is monic," which seems to me to be exactly wrong. Also, with the indexing you introduced on the sums, the polynomials now have constant term zero, and leading term $2f_0x^m$. Can you please reconsider? – Gerry Myerson Sep 6 '17 at 23:21
• @GerryMyerson: many thanks for pointing out. The missing 'not' was simply a mistake. The (indeed) wrong indexing of the sums is eery, not matter how it happened, since I do not remember changing the indexes of the sums, I only consciously changed the indexation of the product, which the OP made range over all pairs of distinct indices, which is (fields being free of zero-divisors) not wrong, yet highly nonstandard. Anyway, I will correct it, yet only in two hours from now, since I will now have to go offline. – Peter Heinig Sep 7 '17 at 6:06

As Robert said, if you want everything to work in $\mathbb Z[f_0,\ldots,f_m]$, you need that factor. I'll also mention that your polynomial indexing is messed up, you probably meant the sum to go from $j=0$ to $j=m-1$, not $j=1$ to $j=m$.

In any case, things become clearer if you study the theory of resultants and ask when two polynomials have a common root. If they're not monic, you really want to view them as homogeneous polynomials that may have a common root "at infinity" if their leading coefficients both vanish. The discriminant of $f(x)$ is, essentially, the resultant of $f(x)$ and $f'(x)$.

• thank you very much for yor answer and for correction. I also thought that the primary thing is resultant. But why it is logically to speak about the root at "infinity", when leading coefficients vanish? – Mikhail Goltvanitsa Sep 6 '17 at 18:27
• @MikhailGoltvanitsa Consider two polys $f(x)$ and $g(x)$ of degrees $d$ and $e$ respectively. If we view their homogenizations $F(x,y)=y^df(x/y)$ and $G(x,y)=y^eg(x/y)$, then the roots of $F$ and $G$ are points in projective space $\mathbb P^1$. The "point at infinity" is $[1,0]$, and $F$ and $G$ have a common root at that point if and only if their leading coefficients both vanish. – Joe Silverman Sep 6 '17 at 18:31
• @Peter Heinig: The discriminant (as defined in the question) is equal to the resultant only up to a factor $f_0$, cf my comment to Robert Israel's answer. – js21 Sep 7 '17 at 9:20
• @js21: thanks for pointing out. Yes, my comment was hastily written. I will delete it. – Peter Heinig Sep 7 '17 at 10:06

The factor $f_0^{2m-2}$ makes the discriminant a polynomial in the coefficients $f_0, \ldots, f_m$.

• Thank you. And, why the exponent is exactly $2m-2$? – Mikhail Goltvanitsa Sep 6 '17 at 18:18
• The resultant polynomial $R(f,f')$ is equal to $\pm f_0^{2m-1} \Delta^2$. However $R(f,f')$ is divisible by $f_0$, since the first row of the Sylvester matrix is. Hence $f_0^{-1} R(f,f') =\pm f_0^{2m-2} \Delta^2$ is a polynomial. – js21 Sep 7 '17 at 9:16
• Moreover the exponent $2m-2$ is optimal since the discriminant of $f_0 x^m + f_1 x^{m-1} + f_m$ is equal to $\pm m^m f_0^{m-1} f_m^{m-1} \pm (m-1)^{m-1} f_{1}^{m} f_m^{m-2}$ which is not divisible by $f_0$. – js21 Sep 7 '17 at 9:43

Perhaps you will find the following helpful, which expands a bit on Robert Israel's answer: Write $f(x)=f_0x^m+\ldots+f_m=f_0\cdot(x-\alpha_1)\cdots(x-\alpha_m).$ Multiplying out the RHS we get $f(x)=f_0\cdot \sum_{j=0}^m(-1)^j\sigma_j(\alpha_1,\ldots,\alpha_m)\cdot x^{m-j}$ where $\sigma_j=\sigma_j(\alpha_1,\ldots,\alpha_m)$ is the $j^{th}$ elementary symmetric polynomial in $\alpha_1,\ldots,\alpha_m$. Then matching coefficients we get that $$\frac{f_j}{f_0}=(-1)^{j}\cdot\sigma_j.$$ Now in the polynomial ring ${\mathbb{C}}[\alpha_1,\ldots,\alpha_m]$, the symmetric group $W=\mathfrak{S}_m$ acts by permuting variables, and a well known fact from invariant theory says that the invariant subring is generated by the elementary symmetric polynomials, i.e. $${\mathbb{C}}[\alpha_1,\ldots,\alpha_m]^W={\mathbb{C}}[\sigma_1,\ldots,\sigma_m]={\mathbb{C}}\left[\frac{f_1}{f_0},\ldots,\frac{f_{m-1}}{f_0}\right].$$ Now the discriminant $$\Delta^2=\prod_{1\leq i<j\leq m}(\alpha_i-\alpha_j)^2$$ is also invariant under $W$, hence we must have that $$\Delta^2=P\left(\frac{f_1}{f_{0}},\ldots,\frac{f_{m-1}}{f_0}\right)$$ for some polynomial $P\in{\mathbb{C}}[y_1,\ldots,y_m]$. Multiplying $\Delta^2$ by an appropriate power of $f_0$ will clear denominators to give you an honest polynomial in the $f_0,\ldots,f_m$. Certainly the exponent $m(m-1)$ will work, but it is not quite clear to me why $2(m-1)$ will, as you claim in your original statement...and perhaps this is the point of your question...?

• Thank you, for clarification. This is one of the subquestions of my question. But as I understand there is another algebraic essence of this factor as Peter Heinig pointed out. – Mikhail Goltvanitsa Sep 8 '17 at 11:19

A trivial observation, which was not pointed out so far: the non-normalized discriminant is 'better' in that it comes 'closer' to being a 'homomorphism' $P[x]\rightarrow P$, though it 'still' is not such a homomorphism, it is merely 'multiplicative up to a multiplicative computable polynomial-in-the-coefficients constant'.

I think one could characterize the factor $([x^{\mathrm{deg}(f)}](f))^{2\deg(f)-2}$, that the OP is asking the 'essence' of, as the unique factor which gives the 'most efficient' multiplicativity-identity, though I have never seen this and don't have time to try to make this observation rigorous. I expect this to be widely known. It seems some sort of answer to the OP's question for the 'essence' of the normalizing factor, and perhaps even for the OP's question about the '$2m-2$'.

Here is the trivial observation.

Let $\mathrm{disc}$ denote the (standard) discriminant, in the sense of the (corrected) OP.

Let $\mathrm{d}f$ denote the degree of a polynomial $f$.

Let $\mathrm{ndisc}(f) := \frac{1}{a_{\mathrm{d}f}^{2\mathrm{d}f-2}}\cdot \mathrm{disc}(f)$ denote the normalized discriminant.1 .

( $\mathrm{ndisc}$ is defined since a leading coefficient by definition is non-zero )

Then, for all polynomials $f_0,f_1$ of degree $\geq 1$ we have the useful and normalization-factor-free identity

$\mathrm{disc}(f_0\cdot f_1) = \mathrm{resultant}(f_0,f_1)\cdot \mathrm{disc}(f_0)\cdot \mathrm{disc}(f_1)$. ${}\qquad$ (multiplicative.up.to.a.polynomial.factor)

For the 'normalized discriminant' $\mathrm{ndisc}$, i.e. the one without the factor that the OP is asking about, we however find a less efficient identity.

Using that $\mathrm{d}( f_0\cdot f_1) = \mathrm{d}f_0 + \mathrm{d}f_1$ and that the leading coefficient of $f_0\cdot f_1$ is $a_{\mathrm{d}f_0} \cdot a_{\mathrm{d}f_1}$, we find the less efficient identity

$\small\text{$(a_{\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1})^{2\cdot(\mathrm{d}f_0+\mathrm{d}f_1)-2}\cdot \mathrm{ndisc}(f_0\cdot f_1) = \mathrm{resultant}(f_0,f_1)\cdot a_{\mathrm{d}f_0}^{2\mathrm{d}f_0-2}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1-2}\cdot\mathrm{ndisc}(f_0)\cdot \mathrm{ndisc}(f_1)$}$

equivalently

$\small\text{$\mathrm{ndisc}(f_0\cdot f_1) = \frac{\mathrm{resultant}(f_0,f_1)}{a_{\mathrm{d}f_0}^{2\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1}} \cdot \mathrm{ndisc}(f_0)\cdot \mathrm{ndisc}(f_1)$}$ ${}\qquad$ (multiplicative.up.to.a.nonpolynomial.factor).

So from this point of view, the 'normalized discriminant' suddenly looks less normalized than the OP's discriminant.

The only mathematical difference between (multiplicative.up.to.a.polynomial.factor) and (multiplicative.up.to.a.nonpolynomial.factor) I can think of is:

• if $a_{\mathrm{d}f_0}^{2\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1}$ does not divide resultant($f_0,f_1$), where 'divides' refers to the ring generated by the coefficients, then the 'multiplicative constant' in (multiplicative.up.to.a.nonpolynomial.factor) is not a polynomial in the coefficients, while in (multiplicative.up.to.a.polynomial.factor) it is.

This perhaps might make it possible to characterize the factor the OP is asking about by an algebraic property.

1 Which one could argue is the 'simplest' $\mathrm{auxiliarypolynomial}(f)$ that can be used to 'encode' the logical statement 'there exists a splitting field of $f$ in which there exists a multiple root of $f$ ' in terms of *an equation over the signature of field theory', in the form '$\mathrm{auxiliarypolynomial}(f)=0$'. Though of course 'simplicity-criteria' like 'smallest number of steps to write down the auxiliar polynomial' tend to have little mathematical meaning, and the discriminant is an easy example of this. Slightly more complicated expressions are often simpler to work with.