Math Challenge - March 2020

[suremarc]

4. Let $k$ be a field that is not algebraically closed. Let $n\geq 1$. Show that there exists a polynomial $p\in k[x_1,\ldots,x_n]$ that vanishes only at the origin $(0,\ldots,0)\in k^n$. (IR)

Spoiler: Solution

Since $k$ is not algebraically closed, we can find a polynomial $p\in k[x]$ with no zeroes.

Now we construct a new polynomial $p_h(x,y)$ by homogenizing: $p_h(x,y)=y^dp(\frac{x}{y})$, where $d$ is the degree of $p(x)$. One can show that this is a well-defined polynomial, and in addition it vanishes only at the origin: suppose $p_h(x_0,y_0)=0$. Clearly this implies that $y=0$: we have $p_h(x_0,y_0)=y_0^dp(\frac{x_0}{y_0})=0\Rightarrow y_0=0\lor p(\frac{x_0}{y_0})=0$. Since $p$ has no zeroes, it then must be true that $y_0=0$. But $p_h(x_0,0)=cx_0^d$ (the leading term) which only vanishes when $x_0=0$. Hence we have shown that $x_0=y_0=0$ and we are done.

Now we use $p_h$ to inductively prove the existence of polynomials in arbitrarily many variables which vanish only at 0. Suppose $Q(x_1,\ldots,x_n)$ vanishes only at 0. Then $Q(x_1,\ldots,p_h(x_n,x_{n+1}))$ is a polynomial in n+1 variables that vanishes only at 0, because $Q(x_1,\ldots,p_h(x_n,x_{n+1}))=0\iff x_1=\ldots=p_h(x_n,x_{n+1})=0$. But $p_h(x_n,x_{n+1})=0\iff x_n=x_{n+1}=0$.

We have shown that, if there exists a polynomial in n variables that vanishes only at 0, there also exists a polynomial in n+1 variables that vanishes only at 0. Taking the monomial $x$ for $n=1$ as the base case, this proves that there exist polynomials in any number of variables that vanish only at 0. $\square$

 

[Infrared]

@suremarc Looks perfectly correct to me!

I'll just remark that the converse is basically obvious: If $n>1$ and $p\in k[x_1,\ldots,x_n]$ vanishes only at a single point, then $k$ must not be algebraically closed. So this gives an equivalent condition for a field (not) being algebraically closed.

 

[QuantumQuest]

For problem 5,

Spoiler: Is this the answer?

$\text{Area}=(\frac{m}{a_1b_2-a_2b_1})^2\sqrt{(A_1^2+B_1^2)(A_2^2+B_2^2)-(A_1A_2+B_1B_2)^2}$
where $A_1=a_1+a_2$ and $A_2=a_2-a_1$ and likewise for B in terms of b

Sorry I didn’t simplify all the way or explain because it was tedious and I think I might’ve made an error.

You can see the correct answer in the spoiler of post #65 by Not anonymous. In any case I don't see how you reached your answer.

 

[Infrared]

Just to point out another way to do 65 since it's already been solved: Let $P$ be the region we want the area of. The transformation $(x,y)\mapsto (u,v)=(a_1x+b_1y+c_1,a_2x+b_2y+c_2)$ is diffeomorphism from $P$ to the region bounded to the square $|u|+|v|= m$. The Jacobian determinant $a_1b_2-a_2b_1$. The area of the square $|u|+|v|\leq m$ is $2m^2$, so by the change of variables formula,

$2m^2=\int_{|u|+|v|\leq m} du dv=\int_P |a_1b_2-a_2b_1| dx dy$

Dividing by the Jacobian gives the desired area of $P$. This is really the same as @Not anonymous's answer since computing the Jacobian is equivalent to taking a cross product, but maybe it's helpful to see anyway and shows the idea of the change of variables formula: applying a linear map to a region scales the area/volume by the determinant (and to the first order, applying a function is the same as multiplying by its Jacobian matrix, up to translations).

 

[Adesh]

@archaic You have an algebra mistakes. You should have $3x_1^2=\frac{8q^2-p^2}{q^2}$. The approach is definitely fine though.

Sir, I think it was my solution . It was a minor typo I think. Is everything all right except that typo?

 

[Infrared]

I still don't think it's right. It looks like to me that you're claiming that since $3$ divides $8q^2-p^2$, it must divide both $8q^2$ and $p^2.$ I don't see why this should be true without further argument. For example, $3$ divides $1+5$ but not $1$ or $5$.

Also, in the future, please write your solutions in your post- I'd rather not have to follow links.

 

[PeroK]

Generalisation:

Spoiler: Question 13

Let's look at $f(x) = x^3 - px$, where $p$ is prime.

We'll look for $f(x) = f(y)$, where $x = \frac a b, \ y = \frac c d$ are in the lowest form.
$$f(\frac a b) = f(\frac c d) \ \Rightarrow \ \frac{a(a^2 - pb^2)}{b^3} = \frac{c(c^2 - pd^2)}{d^3}$$

Note that $\frac{a(a^2 - pb^2)}{b^3}$ is also in its lowest form, so we have $b = d$. And:
$$a(a^2 - pb^2) = c(c^2 - pb^2) \ \Rightarrow \ a^2 + ac + c^2 = pb^2 \ \Rightarrow \ (2a+c)^2 = 4pb^2 - 3c^2$$
So, we need solutions to:
$$4pb^2 - n^2 = 3c^2$$
Where $n = 2a + c$.

We have solutions for $p = 3$. Namely, $x = -2, y = 1$ and $x = -1, y = 2$.

Otherwise, note that neither of $b, n$ can be divisible by $3$ without $b, c$ having a common factor of $3$. Hence $b^2 = n^2 = 1 \ (mod \ 3)$ and we have:
$$p = 1 \ (mod \ 3)$$
For example, we have no solutions for $p = 2, 5$, but we have a triple solution for $p = 7$:
$$x= -3, y = 1, z = 2\ \ \text{with} \ f(x) = f(y) = f(z) = -6$$

 

[fresh_42]

We have opened a second part:
https://www.physicsforums.com/threads/math-challenge-march-2020-part-ii.985311/