TITLE: Rings with zero divisors in which $x^2=x$ implies $x\in\{0,1\}$
QUESTION [3 upvotes]: Any idempotent element (other than 0 or 1) must be a zero-divisor, so in particular there are no nontrivial idempotents in domains.

Are there examples of rings with zero divisors without nontrivial idempotents? (If so, is there anything interesting to say about such rings, e.g., characterizations?)

REPLY [3 votes]: Idempotents of $R$ correspond to clopen subsets of $X = \text{Spec } R$. When $R$ is noetherian, a clopen subset is necessarily a finite disjoint union of connected components.
On the other hand, zero divisors of $R$ correspond to closed subsets of $X$ whose complement is contained in a proper closed subset of $R$. When $R$ is noetherian, such a closed subset is a finite, nonempty union of irreducible components.
Thus, we can obtain a wide variety of examples from geometry. Whenever $X$ is connected but not irreducible, $R$ has nontrivial zero divisors but no nontrival idempotents. For instance, the variety consisting of two intersecting lines is connected but not irreducible, so the ring $\mathbf C[x,y]/(xy)$ has nontrivial zero divisors (eg. $x$ and $y$!) but no nontrivial idempotents.


TITLE: Why should it be hard to generalize Dvir's proof of the finite field Kakeya conjecture to the Euclidean case?
QUESTION [4 upvotes]: Let $q$ be prime and let
$q\delta \sim 1.$ Let $K$ be any set of $Cn\delta$-separated tubes in $B(0,2)$, where $C_n$ is some constant depending on $n$. Let us consider a grid of $q^n$ points scaled with $\delta^n$. We embed this evenly spaced orthogonal grid in $\mathbf{R}^n$ so that the origins of the two sets meet. Let $K' \subset K$ so that each tube in $K$ contains a line from $K'$. The lines  $a + vt \in K'$ are such that if $v \in tu$ then $a + tu = a + vt$. So that we certainly have $q^{n−1}$ lines in $K$ and the directions are $\delta$-separated. By Dvir's theorem we have $\lvert K'\rvert \ge \frac{q^n}{2^n}$. So we have at least that many points $x \in K' \subset K$ that are $\delta$-separated. Multiplying above by $\delta^n$ we have $\lvert K\rvert  \ge \lvert K'\rvert\delta^n \ge q^n\delta^n \frac{1}{2^n} \gtrsim \frac{1}{2^n}$ , because the $C_n\delta$-tubes contain the $\delta$−tubes. Isn't this a straightforward generalization?

REPLY [8 votes]: I don't entirely follow your proposed approach, but let me make a few remarks which might help you think about this:
(1) Clearly this can't work, since this argument purports to show that a Kakeya set must have positive Lebesgue measure which, in contrast to the finite field setting, we know is false in Euclidean space.
(2) Note that the discretization of a tube is not necessary an algebraic line or even approximately an algebraic line. Indeed consider R^2 and a tube oriented along the y axis. Now adjust the angle slightly in either direction. What does a discretization of this tube look like? Well it depends on the scales involved but, roughly in the case of a $\delta$ tube rotated by a $~\delta$ degree angle, the discretization will be the union of long segments of vertical lines (rectangles/blocks) stacked on top of each other slowly drifting in the direction of the rotation. This is not an algebraic line.
(3) In the discrete (finite field) setting two lines intersect at a single point. However, for reasons similar to the discussion in (2), two tubes can have substantial overlap. This is sometimes refereed to as "small angle issues" in the literature.


TITLE: does an angle of 0 degrees between 2 vectors mean they are linearly dependent
QUESTION [0 upvotes]: If the angle between 2 vectors is 0 degrees or 180 degrees does that imply the 2 vectors are linear dependent?

REPLY [0 votes]: Yes.
Use the equality condition for Cauchy-Schwarz Inequality.


TITLE: Solving difficult Lagrangian
QUESTION [0 upvotes]: \begin{equation}
\mathcal{L} = x - \lambda_1 ( - \frac{1}{2}x^2 + 2x-y) - \lambda_2 ( \frac{1}{2}x^2 -2x + 4 -y)
\end{equation}
This gives the following FOC's
\begin{equation}
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial x} & = 1 + \lambda_1x -2\lambda_1  - \lambda_2x + 2\lambda_2 = 0 \\
\frac{\partial \mathcal{L}}{\partial y} & = \lambda_1  + \lambda_2 = 0 \\
y &= -\frac{1}{2}x^2 + 2x\\
 y &= \frac{1}{2}x^2 -2x + 4
\end{aligned}
\end{equation}
Here we have $-\lambda_1=\lambda_2$ substituting this into the first equation gives us
\begin{equation}
x = 2\lambda_1 - \frac{1}{2}
\end{equation}
Substituting this result into first constraint gives:
\begin{equation}
y = -2\lambda_1^ 2 + \frac{1}{8}
\end{equation}
substituting into the next constraint gives:
\begin{equation}
0 = 4\lambda_1^2 - 7/4 \lambda_1 + 71/16
\end{equation}
the last equation has no real roots so I cant solve the problem yet I know there should be a solution. Also I am not one 100% sure if I understand how to check for the second order conditions in this case. Could someone please help? Also, this Lagrangian should be also solvable using level curve method. Could someone explain me how that would work?

REPLY [1 votes]: If the problem is
$$
(\min,\max) x \ \ \mbox{s.t.}\ \ \left\{\begin{array}{rcl}g_1(x,y) & = & -\frac 12 x^2 + 2x - y=0\\ g_2(x,y) & = & \frac 12 x^2-2x-y+4=0\end{array}\right.
$$
we have that $g_1(x,y)\cap g_2(x,y) = (2,2)$ so the feasible solution is at $x = 2$