I found multifarious duplicates that I listed at https://math.stackexchange.com/a/631364/53934.
I edged the purple part because my answer proves it more efficiently.
I remember that any function $f: A -> f(A)$ is always onto but I misread. Sorry. Hence I removed original question (1.).
(2....
Currently, I have a homomorphism mapping $f$ from a monoid $A$ to a group $B$. I found the kernel of $f$ and I want to have group structure on the space $A/ker(f)$.
By first isomorphism theorem, we know that $A/ker(f)$ is isomorphic to the group $B$.
I forgot to mention that $f$ is onto.
Is it possible somehow to find the group structure on $A/ker(f)$=
I'm preparing to learn galois theory; does it make sense to skip dummit and foote chapter on polynomial rings? I ask because I'm not gonna be using dummit and foote for galois theory (so won't be using studying part iii of the book, and polynomial rings will introduced in my galois theory course/book, I assume).
Define $T^k(\Omega)$, $\Omega$ an open subset of $\mathbb{R}^m$ (with a smooth boundary), as a space of function equivalance classes, with the norm defined as $$ \|f\|_{T^k(\Omega)}^2 = \|f\|_{L^2(\Omega)}^2 + \|(\sum\limits_{i=1}^m(\frac{\partial^{k}f}{\partial x_i^{k}})^2)^{\frac{1}{2}}\|_{...
@Ryan if you're sure you'll learn the content later, then there's no harm, but it may be the case that a self-contained Galois theory book will take polynomial rings as given, and since Galois theory uses polynomial rings... It's definitely something to learn, by some means or other
@Daminark In fact, looking a bit more it's not covered (some of it is, but it's largely assumed/skipped over). So I'll have to learn it from D&F. Thanks.
(Standard) Distributivity is an additive map: $\phi (a+b) = \phi (a) + \phi (b)$ for any element $a,b$
Multiplication by anything is a homomorphism as additive identities are preserved
Division by zero, however is not a homomorphism
The most important property of any division by zero algebra is that all elements are idemponent under addition. This can be easily observed by considering the division by zero map $\phi (0) = 1, \phi(1) = 0$
$0+0=0$ (additive identity)
$\phi (0+0) = \phi (0)$ (applying $\phi$)
$\phi(0)+\phi(0)=\phi(0)$ ($\phi$ is additive)
$1+1=1$ (Definition of $\phi$)
Now, observe that $1$ is a multiplicative identity by definition, the following is immediately concluded:
$x+x=x$ for any $x$
hence all elements in a division by zero algebra are idempotent under +
While it is possible to have $0+0 \neq 0$ (in which case one obtains a "shifted" Rock Paper Scissors (RPS) magma (see commutative magma for details) where the winning rules are altered), the same relation still results due to how additive identities get mapped to multiplicative identities
More generally, given an additive identity $e$, multiplicative identity $f$, a constant $a$ and the division by zero map $\phi$, and for all elements $b$, we have the following:
$e+e=a$
$\phi (e) + \phi (e) = \phi(a)$
$f+f = \phi (a)$
$bf+bf = b \phi(a)$
$b+b = b \phi (a)$
Thus unlike many algebraic structures, division by zero algebras are often highly constrained by their axioms
(and here I have not assumed any associativity, thus this is a very general result)
and therefore, the closest thing that is related to division by zero algebras (assuming we keep the usual distributivity rule so that we don't end up with things like wheels, meadows etc.) are RPS magmas, which means they generalise rock paper scissors games in some systematic fashsion
now as for whether generalised rock paper scissors games are useful, we will leave this question to the game theorists...
(also typo remove the words "$\phi (1) = 0$")
Conjecture: The meaning of division by zero (as defined above) is the imposing of a partial ordering to pairs of elements in an algebraic structure such that they become generalised RPS magmas
More details will be discussed later in here when technicalities are finalised:
All discussions on the ongoing project of algebraic structures...
To get an idea on how constrained division by zero algebras and division by zero magmas are, cyclic groups are one of the few examples which are "more constrained" than them (recall a group is cyclic if its generator consists of just one element <a>)
@GaurangTandon The problem is that, for example, you could have $7,65$ already in the list, so you want to be sure that the new number $0.b_1b_2\dots$ you're building is not $7,64\overline{9}$, the easiest way to make sure that won't happen is to avoid numbers with repeating nines altogether
@AlessandroCodenotti yep, i was getting to the same btut now i am wondering if theere are other digits like $8$ which on repetition could give another number
@AlessandroCodenotti given a repeating number $z$, such that $$\begin{align}z&=x+0.\bar{9}\\10z&=10x+9.\bar{9}\end{align}$$ substracting we get $$z=x+1$$ wow, nice! the same does not occur with any repeating digit
ja vielleicht werden Kenner die Hände über den Kopf zusammenschlagen, dass ich diese fast schon Klischee-Autoren vorschlage, aber es sind halt Klassiker und das zu Recht.
Here is a completely different kind of answer to this question.
A perfectoid space is a term of type perfectoid_space in the Lean theorem prover.
Here's a quote from the source code:
class perfectoid_ring (R : Type*) extends Tate_ring R :=
(complete : is_complete R)
(uniform : is_uniform R)
(...
Wenn ein gemäsigter Schotte losbrabbelt behaupt ich dann versteht der Durchschnittdeutsche mehr als wenn ein Südengländer loslegt (behaupte ich jetzt mal).
Ich find den österreichischen Akzent leichter zu verstehen als den Deutschen, wahrscheinlich weil ich dort gewohnt hab, aber ich find's irgendwie "klarer"
@AlessandroCodenotti given a repeating number $z$, such that $$\begin{align}z&=x+0.\bar{9}\\10z&=10x+9.\bar{9}\end{align}$$ substracting we get $$z=x+1$$ wow, nice! the same does not occur with any repeating digit
When we have a linear map $f:\mathbb{R}^3\rightarrow \mathbb{R}^3$ with $f(x,y,z)=\begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}$, how can we find f-invariant subspaces $V_1, V_2$ of $\mathbb{R}^3$ with $\mathbb{R}^3=V_1\oplus V_2$ ?
@LeakyNun That means that I calculate the eigenvalues and the corresponding eigenvectors and then $V_1$ is the set of the zero vector and the eigenvector of the one eigenvalue and $V_2$ is the set of the zero vector and the eigenvector of the other eigenvalue?
Are the eigenspaces per definition f-invariant? As for the direct sum, the two eigenvectors $\begin{pmatrix}1 \\ -2 \\ 1\end{pmatrix}$ and $\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}$ are linearly independent and they span $\mathbb{R}^3$, that's why we can consider them, right? @LeakyNun
Analogous to zero divisors, zero addends do not have additive inverses, for if they do then the following will happen:
$x+b=u$
$(x+b)-b=u-b$ $x+(b-b)=u$ $x+0=u$ $x=u$
which is a contradiction since $x\neq u$ by definition of zero addends
Compared to zero divisors, zero addends may potentially have extra properties that their multiplicative counterpart does not have, due to extra opportunity to interact with the distributive law. Consider an additive domain (a semiring lacking zero addends) $A$ which has the following expression:
$xy+xz=u$
Then the zero summand theorem (the additive analogue of the zero product theorem) allow us to do:
$xy=u \lor xz=u$
Meanwhile, observe that $xy+xz=u \implies x(y+z)=u$
Thus we have:
$xy=x(y+z) \lor xz=x(y+z)$
which gives:
$xy=xy+xz \lor xz=xy+xz$
If these elements have additive inverses, then we have:
$0=xz \lor 0=xy$
This suggests a connection between zero divisors and additive absorbers in additive domains
Meanwhile there is no such analogue in multiplicative domains:
$(a+b)(c+d)=0 \implies a+b=0 \lor c+d=0$
Here we have additive inverses tied to multiplicative absorbers instead
Similar to division by zero, it is possible to do "subtraction by u" and the properties of the associators are identical to those in division by zero magmas.
$x+u=u$ $x+u+v=u+v$ $x+0=0$
However, as you saw above, zero will then become an additive absorber. Similar to division by zero, there is some degree of constraint imposed on the multiplication structure due to the distributive rule, but if zero addends exists, then the constraint is less than in division by zero. This is because the distributive law is "unidirectional", in that it is multiplication distribute over addition, but not vise versa
We consider the surface $S$ of the space $\mathbb{R}^3$ that is defined by the equation $2(x^2+y^2+z^2-xy-xz-yz)+3\sqrt{2}(x-z)=1$. How can we find (using symmetric matrices) an appropriate orthonormal system of coordinates $(x_1, y_1, z_1)$ for which the above equation has the form $ax_1^2+by_1^2+cz_1^2=d$, for some $a,b,c,d\in \mathbb{R}$ ?
I found multifarious duplicates that I listed at https://math.stackexchange.com/a/631364/53934.
I edged the purple part because my answer proves it more efficiently.
I remember that any function $f: A -> f(A)$ is always onto but I misread. Sorry. Hence I removed original question (1.).
(2....
Define $T^k(\Omega)$, $\Omega$ an open subset of $\mathbb{R}^m$ (with a smooth boundary), as a space of function equivalance classes, with the norm defined as $$ \|f\|_{T^k(\Omega)}^2 = \|f\|_{L^2(\Omega)}^2 + \|(\sum\limits_{i=1}^m(\frac{\partial^{k}f}{\partial x_i^{k}})^2)^{\frac{1}{2}}\|_{...
Here is a completely different kind of answer to this question.
A perfectoid space is a term of type perfectoid_space in the Lean theorem prover.
Here's a quote from the source code:
class perfectoid_ring (R : Type*) extends Tate_ring R :=
(complete : is_complete R)
(uniform : is_uniform R)
(...
