A doubt about the multiplicity of polynomials in two variables

[V9999]

TL;DR

Here, I present a question about multiplicity and the number of solutions of polynomials in two variables that I think might interest those that subscribe to Physics Forums.

Let $P(x,y)$ be a multivariable polynomial equation given by
$$P(x,y)=52+50x^{2}-20x(1+12y)+8y(31+61y)+(1+2y)(-120+124+488y)=0,$$
which is zero at $q=\left(-1, -\frac{1}{2}\right)$. That is to say,
$$ P(q)=P\left(-1, -\frac{1}{2}\right)=0.$$

My doubts relie on the multiplicity of this point $q=\left(-1, -\frac{1}{2}\right)$.

It is generally well-known that the number of times a given factor appears in the factored form of a polynomial equation is called multiplicity. Bearing this in mind and going through straightforward algebraic manipulations, one may verify that the polynomial equation above may be expressed as follows
$$P(x,y)=(1+x)^2+\pmb{\Bigg[}\frac{1-60x+122y}{25}\pmb{\Bigg]}(1+2y)=0,$$

whence we recognize, not surprisingly, that $P(x,y)=0$ at $q=\left(-1,-1/2\right)$. That is, $P(x,y)$ is zero when
$$(1+2y)=0 \implies y=-\frac{1}{2}$$,
and
$$(1+x)^2=0 \implies x_{1}=-1\quad \text{e} \quad x_{2}=-1.$$

Based on the previous equations right above, I ask:
1. Is it possible to say that $x_{1,2}=-1$ is of multiplicity 2?
2. One may write that $P(x,y)=0$ in two equal points? That is to say, is it correct to establish that
$$P(x_{1}, y)=P(x_{2},y)=0,$$
on account of the multiplicity of $x$?

Thank you very much!

 

[andrewkirk]

What you have written above is a sum, not a factorisation. Neither $(1+x)$ nor $(1+2y)$ is a factor of the polynomial. So there is no reason to suppose any multiplicity.

For $(1+x)$ to be a factor would require that $P(-1,y)=0$ for all $y$ and that is not the case. Similarly for $(1+2y)$.

 

[V9999]

What you have written above is a sum, not a factorisation. Neither $(1+x)$ nor $(1+2y)$ is a factor of the polynomial. So there is no reason to suppose any multiplicity.

For $(1+x)$ to be a factor would require that $P(-1,y)=0$ for all $y$ and that is not the case. Similarly for $(1+2y)$.

Hi, andrewkirk. Thanks for commenting. I see your point. However, I did not understand your last statement. Correct me, if I am not mistaken, but by substituting $y=-1/2$ into $P(x,y)$ we will have $P(x,-1/2) \equiv \tilde{P}(x)=50x^2+100x+50$, or, simply,$\tilde{P}(x)= 50(x+1)^2$. Based on this latter, would the definition of multiplicity be applicable? Thanks again!

 

[andrewkirk]

Hi, andrewkirk. Thanks for commenting. I see your point. However, I did not understand your last statement. Correct me, if I am not mistaken, but by substituting $y=-1/2$ into $P(x,y)$ we will have $P(x,-1/2) \equiv \tilde{P}(x)=50x^2+100x+50$, or, simply,$\tilde{P}(x)= 50(x+1)^2$. Based on this latter, would the definition of multiplicity be applicable? Thanks again!

No it would not give multiplicity, because that factorisation only applies when you have a specific value of y. (y = -1/2). Multiplicity is based on factorisations that can be done without setting any of the variable values.

 

[mathwonk]

The concept of multiplicity of a root is a bit more complicated than what you said, in 2 variables. If a polynomial in two variables vanishes at a point, and its partial derivatives do also, then the point is considered as a root of multiplicity at least 2, (a generalization of the fact that a polynomial in one variable has a double root if the derivative also vanishes at the root). That is the case here. I have not made any further calculations, but since your polynomial has degree 2, that fact alone implies that your polynomial should factor into 2 linear factors, both passing through that point, assuming I did not make a mistake in evaluating ∂P/∂x and ∂P/∂y. I have not computed those factors. (I.e. it is only in degree 2 that a double point forces a factorization.)

Indeed your calculation in post # 3 agrees with that conclusion, and does imply that either the root has multiplicity 2, or else the root has multiplicity one and the line y = -1/2 is tangent to the curve at that point. I.e. you have calculated the intersection number of the given curve with the line y = -1/2 and found it to be 2. If the point has itself multiplicity one, then only the tangent line has intersection number 2, but if the point has multiplicity 2, which is what I computed, then almost every line has intersection number 2 at that point.

In fact setting x =-1 also gives a double root at y = -1/2, so we have two perpendicular lines both having intersection number 2 with this curve at (-1, -1/2), hence both cannot be tangent if it is a point of multiplicity one. Thus the point is a "double point" of the curve, or a point of multiplicity 2. It follows that the polynomial must factor into two lines, at least over the complex numbers, but I have not found the factors.

Actually setting y = 0, seems to give two complex intersections, so conceivably this curve has only one real point, the double point you found, but I am not entirely sure. Still itseems that if there were two real linear factors, then I should not have found only complex points with y=0.

Maybe you can explore a bit further. a simpler example is x^2 + y^2 = 0, which has only one real point, namely (0,0), which is a double point. Probably you can compute an invariant in your case to see whether your example is equivalent to this one, or to x^2-y^2, or to -x^2 -y^2. Presumably you only have to find a determinant of a symmetric quadratic form. Indeed that determinant seems to be positive so you should be in the x^2 + y^2 case (or minus that, no difference here). I am not an algebra expert however so I might be making some errors here. But my conjecture is this polynomial factors into two linear complex factors, and has only one real point, a point of multiplicity 2.

 

[andrewkirk]

Let $P(x,y)$ be a multivariable polynomial equation given by
$$P(x,y)=52+50x^{2}-20x(1+12y)+8y(31+61y)+(1+2y)(-120+124+488y)=0,$$
....
one may verify that the polynomial equation above may be expressed as follows
$$P(x,y)=(1+x)^2+\pmb{\Bigg[}\frac{1-60x+122y}{25}\pmb{\Bigg]}(1+2y)=0,$$

I just realised this appears to be wrong.
The coefficient of $x^2$ in the first expression is 50, while in the second it is 1. So the two polynomials cannot be the same.
$q=(-1, -0.5)$ is a double root of the second polynomial in the sense described by @mathwonk, but not of the first polynomial.
It looks like we should forget about the first polynomial and focus on the second.

Regarding @mathwonk's simpler example $x^2+y^2=0$. I note that factorises to $(x-iy)(x+iy)=0$ whose solution is all points of form $(c, ic)$ and $(c, -ic)$ for complex $c$. Taking $c=0$ gives us our only real root, and it's a double root because it is the intersection of the $(c, ic)$ and $(c, -ic)$ branches.

Maybe Wolfram can factorise the second poly in the OP, if anybody wants to type it in.

 

[mathwonk]

ok, i'm not sure about this, but as andrewkirk said, it seems the original polynomial was written wrong, and perhaps the last parenthesis should contain (-120x + 124 + 488y), i.e. an "x" was apparently omitted after -120. at least this seems to make the polynomial have a double root at (-1,-1/2). moreover it is possible that then the original polynomial, after this modification, equals 50 times the second one, accounting for the different x^2 coefficients, but I have not checked that.

 

[DaveE]

one may verify that the polynomial equation above may be expressed as follows

My computer doesn't agree. Although I suppose that doesn't change the basic question about multiplicity of roots in multi-dimensions.

it is possible that then the original polynomial equals 50 times the second one, accounting for the different x^2 coefficients, but I have not checked that.

It's a mess, none of them agree, except for the root at (-1,-1/2). I find it annoying when polynomials aren't presented in a standard form. Maybe the weird format makes sense to the OP, but I don't understand why.

 

[andrewkirk]

I plugged the second polynomial ( Q(x,y) in the notation introduced by @DaveE ) into Wolfram here.
That gave the factorisation as:
$$Q(x,y) = \frac1{25} (5 x - (12 + 10 i) y - (1 + 5 i)) (5 x - (12 - 10 i) y - (1 - 5 i))$$
So $y = (5x - 1 - 5i) / (12 + 10 i)$ and $y = (5x - 1 + 5i) / (12 - 10 i)$ are the two branches of the solution.
Putting x = -1 gives y = -1/2 for both branches. So the point (-1, -1/2) is a double root because it occurs in both branches.

Further it is the only double root because a double root must satisfy the equations of both branches and, since they are two linear equations in two variables, they can have at most one solution.