If my field had four elements and they were $0,1,\omega,\omega+1$ then I could have all the elements defined, and I get $0+0=0,1+1=0,\omega+\omega=\omega,\omega+1+\omega+1=\omega$ and that is fine, but apparently I should obtain $x+x=0$ which implies that $\omega=0$ and now I have only two elements again, e.g I am working with $Z_2$
@beginner I just got back online. Your keyboard problem sounds like a driver issue. Next time you plug that keyboard in, check your device manager to see that it properly recognized the device and has a current driver for it
A quick mention @beginner about the diagram I linked, arrows are pointed upwards in each case, and there are also directed edges to neighbors of neighbors which aren't drawn. For example, there is an undrawn edge from 3 to 30 since $3|30$
How come it is fine to say all fields of size $p$ prime are isomorphic? I feel this is true, and I am using the word closer to the meaning of graph isomorphism? I can see that they certainly have a bijective mapping, but I don't know how I would prove the homomorphism
I think I saw something about generators, it is a single element that reaches all elements of the field, e.g. $1$ for addition, and something else for multiplication. Yeah @JMora I am always lagging out hehe
I can see it is true that for $\Bbb F_p$ where $p$ is an odd prime I have exactly half of the non-zero elements as squares for the first few $p$ odd prime, but I am unsure how to prove this. Has this also got something to do with generators?
Oh I got this mistletoe hat without trying haha
I thought the hats thing was going to be a purchasable hat that was MSE oriented that you really bought and wore
The answer to $x+3^x<4$ is $x<1$ by plotting the graph of $y=x+3^x$ and $y =4$. Is there a way to get to the solution algebraically?
Updated: is there a way to get to the solution of $x+a^x<b$ in general, algebraically?
@BalarkaSen The cofinite topology is always $T_1$, but $\Bbb C[X]$ has a generic point: since $(0)$ is prime, we have $\overline{(0)}=V(0)={\rm Spec}(\Bbb C[x])$. So $(0)$ is not closed.
Hmm, OK. Define the map f : Spec C[x] - (0) to C. Equip C with the topology whose open sets are C minus a finite number of points.
Then inverse image of some point a in C goes to (x-a)
An open set U around a in C is complement of a finite number of points along with a.
So taking inverse image maps that to the subset of Spec C[x] - (0) which consists of the whole space minus a finite number of points (ideals) along with (x - a)
So that's an open set around (x - a).
Hence f is continuous, and since it is bijective, it is a homeomorphism.
Given a point $\mathfrak{p} \in \text{Spec}(A)$, the open neighborhood $X_\mathfrak{p}$ around $\mathfrak{p}$ in $\text{Spec}(A)$ by definition of Zariski topology doesn't contain $\mathfrak{p}$
@GustavoMontano Nothing much. Started algebraic topology, done Munkres and now beginning Hatcher. Meaning to do commutative algebra a bit along with it too.
Wha..., @Vrouvrou? You were asking questions about algebraic topology 6 months ago and then doing general topology (homeomorphisms and stuff) only a week ago and now you're onto fundamental inequalities?
Galois theory is very easy to visualize, on the other hand, @Pedro. Think about larger fields as their positions in the lattice and think about Galois groups are the edges joining the nodes (nodes corresponds to fields) in the lattice ;)
@beginner If you have a polynomial over a field F, splitting field of the polynomial is the minimal extension of the field F which contains all the roots of the polynomial.
Where I can find a reference that the pullback of the metric of the exponential map can be expressed by Riemannian curvature tensor? Also why must this be true?
The answer to $x+3^x<4$ is $x<1$ by plotting the graph of $y=x+3^x$ and $y =4$. Is there a way to get to the solution algebraically?
Updated: is there a way to get to the solution of $x+a^x<b$ in general, algebraically?
