If $f:A \to B$ is a split monomorphism of rings, does it follow that the preimage of a maximal ideal is a maximal ideal?
Let $\mathfrak b$ be a maximal ideal of $B$ such that $f^{-1}(\mathfrak b)$ is not a maximal ideal of $A$, i.e. $f^{-1}(\mathfrak b) \subsetneq \mathfrak a$ for some ideal $\mathfrak a \subsetneq A$. Then, $\mathfrak b \subsetneq \mathfrak b + \langle f(\mathfrak a) \rangle \subsetneq B$. amirite?
The dot product of vectors $\mathbf{a}$ and $\mathbf{b}$ is defined as:
$$\mathbf{a} \cdot \mathbf{b} =\sum_{i=1}^{n}a_{i}b_{i}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}$$
What about the quantity?
$$\mathbf{a} \star \mathbf{b} = \prod_{i=1}^{n}a_{i} + b_{i}=(a_{1} +b_{1})\,(a_{2}+b_{2})\cdots \,...
Sanity check: Let $A_0 = \Bbb Q$. Let $A_{n+1} = A_n[X] / \langle X^2 \rangle$. Let $f_n : A_n \to A_{n+1}$ be the inclusion that sends $a \in A_n$ to $a + \langle X^2 \rangle \in A_{n+1}$. Then, let $A = \varinjlim A_n$. Then, for every $n \in \Bbb N$ there is $x \in A$ such that $x^{2^n} = 0$ but $x^m \ne 0$ where $m < 2^n$.
so the idea is that you can relate $\chi(M)$ to the behavior trace of the heat operator on a riemannian manifold via the hodge decomposition of the de rham cohomology groups
Suppose you have a metal sphere, and you put a net charge on it. Since like charges repel, that net charge will gather itself on the surface of the sphere
Moreover, from Gauss's law and spherical symmetry, the electric field outside the sphere is identical to what you'd have if all the charge was concentrated at the center
which amounts to E = kq/r^2 where q is the net charge and r is your distance from the sphere (k is Coulomb's constant)
in particular, if you look at the electric field right near the surface of the sphere, you have $E = k q/a^2$ where $a$ is the radius of the sphere.
which if you write $k=1/4\pi \epsilon_0$ (because that's how it's defined) is $E=\frac{1}{\epsilon_0} \frac{q}{4\pi a^2}=\frac{\sigma}{\epsilon_0}$
where $\sigma$ is the surface charge density (total charge / surface area of sphere)
You can moreover show that this is a quite generic result: The electric field at the surface of a charged conductor (not necessarily spherical) is given by $E=\sigma/\epsilon_0$ where $\sigma$ is the surface charge density at that point of the surface
Can someone check my answer here? He is saying his TA has told him that the answer is "weaker" instead of "not comparable". I want to make sure I'm not leading him in the wrong direction:
I agree with "for every $n \in \Bbb N$ there is $x \in A$ such that $x^{2^n} = 0$ but $x^m \ne 0$ where $m < 2^n$", I don't see why it holds that "in $A$ there is $0=x_0, x_1, x_2, \cdots$ such that $x_{n+1}^2 = x_n$ for every $n \in \Bbb N$"
@LeakyNun sure. Easy example $(0) \subset (p)$ in $\Bbb Z$
Are you sure? I think you mean a different ring than you actually wrote down. What you wrote down is $\Bbb Q[X_1, X_2, \dots]/(X_1^2, X_2^2, \dots)$. I think you wanted to write down $\Bbb Q[X_1, X_2, \dots](X_1^2,X_1-X_2^2, X_2-X_3^3, \dots)$
Sanity check: Let $A_0 = \Bbb Q$. Let $A_{n+1} = A_n[X] / \langle X^2 \rangle$. Let $f_n : A_n \to A_{n+1}$ be the inclusion that sends $a \in A_n$ to $a + \langle X^2 \rangle \in A_{n+1}$. Then, let $A = \varinjlim A_n$. Then, for every $n \in \Bbb N$ there is $x \in A$ such that $x^{2^n} = 0$ but $x^m \ne 0$ where $m < 2^n$.
@LeakyNun in each step, you're just adjoining an extra variable and modding out by the square, you don't adjoin any square root. Note that the colimit is just a union in this case
@LeakyNun yes. Good obversation, it's actually an exercise in Atiyah-Macdonald to come up with a power series all whose coefficients are nilpotent, but which is not nilpotent itself
@LeakyNun I think if you want to have increasing orders of nilpotency, it's easier to come up with $\Bbb Q[X_1, X_2, X_3, \dots]/(X_1,X_2^2,X_3^3, \dots)$ at least that's what I did
