> Variant pi or "pomega" ($\varpi$ or ϖ) is a glyph variant of lower case pi sometimes used in technical contexts as though it were a lower-case omega with a macron, though historically it is simply a cursive form of pi, with its legs bent inward to meet.
We have the finite extension $E/F$. We have the $F$-monomorphism $\tau :E\hookrightarrow \tau (E)\subseteq E$. Then it follows that $\tau (E)$.
The proof is the following:
Since it is a monomoprhism, we have that $\tau (E)\cong E$. So, $[E:F]=[\tau (E) : F]$. If $e_1, \ldots , e_n$ is a basis of $E/F$ then $\tau (e_1), \ldots , \tau (e_n)$ is a basis of $\tau (E)/F$, so $[e:\tau (E)]=1$. Therefore, $E=\tau (E)$.
Do we have that $\tau (E)\cong E$, because $\tau (E)$ is the image and so the mapping is surjective?
Hello, I'm looking for some vocabulary to google what are probably existing algorithms.
I need an initialization vector of N values 0.0 -> 1.0. Let's say N is 16. I would like to generate that initialization vector from a single value X, 0.0 -> 1.0.
Such that, if X is very close, let's say 0.1 and 0.10000000000000000000000000001, the initialization vectors generated are also close. But when not close, let's say 0.1 and 0.2, the initialization vectors look completely different, random.
Is there a term for what I'm trying to do, so I can look up research papers on it?
Let $\sigma : A\rightarrow A$ be an injective endomorphism, with $|A|<\infty$. Since $\sigma$ is injective we have that for $x_1\neq x_2$ we have $\sigma (x_1)\neq \sigma (x_2)$. How does it imply then that for each $x\in A$ there is a $y\in A$ such that $\sigma (y)=x$ ? I got stuck right now... @Starfall
The regular expression for the complement of the language $L = \{a^nb^m \mid n≥4, m≤3\}$ is:
$(λ + a + aa + aaa)b^* + a^*bbbb^* + (a + b)^*ba(a + b)^*$
$(λ + a + aa + aaa)b^* + a^*bbbbb^* + (a + b)^*ab(a + b)^*$
$(λ + a + aa + aaa) + a^*bbbbb^* + (a + b)^*ab(a + b)^*$
$(λ + a + aa + aaa)b^* + a...
Consider the category the set $\mathbb{Z}^{+}$ of positive integers endowing it with divisibility relation. Thus there exists one morphism $d \mapsto m$ in this category iff d divides m without remainder; There is no morphism between d and m otherwise. Show that this category has products and coproducts. What are their convential names.
It is very cool that we have relationship among gcd lcm and normal set products and disjoint union. I guess it makes sense somewhat but it is very cool. @TedShifrin
Using the residue theorem it can be brought to an expression ${100\choose 50}\frac1{2^{100}}$ rather quickly, but I cannot evaluate that in under 5 mins
> A student who takes much more than five minutes to calculate the mean of $\sin^{100} x$ with 10% accuracy has no mastery of mathematics, even if he has studied non standard analysis, universal algebra, supermanifolds, or embedding theorems.
[Abstract algebra] Theorem(???): For any countable or finite associative semigroup S, given a left (right) multiplicative identity and a pair of inverses $uv=1$ where $v\neq 1$ ($u\neq 1$). If $u1\neq u$ ($1v \neq 1$), then the semigroup is inconsistent since no $vx$ ($xu$) can be defined for all $x$ such that the system is consistent.
Proof: Has 4 short columns, working through all 6 special elements in the semigroup. To be pasted shortly
@TobiasKildetoft Because it is a semigroup, inconsistent here means at least one associative rule will fail. By the way, I am not sure if I had a semigroup where not all pairs of products are defined is still a semigroup, if that is still a semigroup, then this can be the counterexample
@Secret As is so often the case, I am really not sure of your usage of these words. A semigroup is by definition associative, and has the operation defined everywhere
@TedShifrin the residue theorem trick is useful for things like $\frac{\sin}{\cos}$ not polynomials in $\sin$ and $\cos$, that can be seen by direct binomial expansion right^
@Secret I have no idea which parts of those were the assumptions on the semigroup. Was it just that you had a left identity and a pair of inverse elements with respect to that left identity?
@DHMO Let $W_n = \int_{0}^{\pi} \sin^n$. We have by integration by parts that $W_n = W_{n-2} + {1\over n-1}W_n$ ie $W_n = {n-1\over n} W_{n-2}$ What follows is that $W_{2n} = {(2n)!\over n! 4^n}W_0 = {(2n)!\over n!4^n}\pi$
O wait, I can see it now: the two lines implies av=a1=1. Multiplying by v on the right give avv=a1v=1v => avv=av=v, which is a contradiction since $v \neq 1$
Now consider what x,y can satisfy $$vx=y$$ it is easy to check for contradiction for x=0 and y=u,1,v. For y=a, it reduces to the case uu=u. So we only need to check for b
If vu=b va=v1a=vuva=bva=u1a=b vb=vua=ba a=1a=uva=ub=u contradiction
@DHMO Especially in Vienna (which contains $\approx 25\%$ of the population), it's way more common for people to know some Hungarian or Slovak, since those countries are way closer.
Finally, to show that even if $S$ has countable number of elements (possibly uncountable amount of elements as well???) we can show that a contradiction will result hence concluding the proof as any element c that is not the ones already considered will be expressed as a word consists of the previous elements a,b,u,v,1, thus one cannot bypass the contradiction by picking a new element
Hungerford's Algebra poses the question: Is it true that a semigroup $G$ that has a left identity element and in which every element has a right inverse is a group?
Now, If both the identity and the inverse are of the same side, this is simple. For, instead of the above, say every element has a ...
Ok for $a=u$, this means there is no b (because uu=b, which at that point a,b are arbitrary $\neq u$ thus by relabelling we can equally say it is uu=a, which is shown to cause a contradiction earlier
Actually, I just realised, my proof can as well stop at uu=a since a is arbitrary, thus it account for all case. This means there is no consistent way to define uu
or the mirror inverted version of these properties (mirror inverted as mentioned by DHMO). Then it can be shown that $uu=x$ is inconsistent for any $x \in S$
you only need to show that this is true for polynomials of degree $2$
since every polynomial over $\mathbb R$ factors in a product of irreducible polynomials of degree $1$ and $2$ and the complex roots can't come from the linear factors
@DHMO if $K$ is a finite degree extension of $F$ (that is $F\subseteq K$ and $|K:F|<\infty$) then all of the elements of $K$ are algebraic over $F$ ($K$ is an algebraic extension)
an element $\alpha$ of $K$ can't have minimal polynomial of degree higher than $|K:F|$ (let's call this number $n$) because otherwise $K[\alpha]$ would be a vector subspace of $K$ (over $F$) of dimension bigger than $\dim K$, which is absurd
So elements of $\mathbb{C}$ have minimal polynomials of degree at most $2$ (over $\mathbb R$)
Now suppose you have a polynomial $f\in\mathbb R[x]$ of degree $\ge 3$, we want to show that it's reducible
let's call $\beta\in\mathbb{C}$ a root of $f$ (which exists since $\mathbb{C}$ is algebraically closed), let $g$ be its minimal polynomial over $\mathbb R$, so we have $g|f$ because it's the minimal polynomial and that is a proper factor since it's of degree strictly bigger than $0$ and strictly smaller than that of $f$
The regular expression for the complement of the language $L = \{a^nb^m \mid n≥4, m≤3\}$ is:
$(λ + a + aa + aaa)b^* + a^*bbbb^* + (a + b)^*ba(a + b)^*$
$(λ + a + aa + aaa)b^* + a^*bbbbb^* + (a + b)^*ab(a + b)^*$
$(λ + a + aa + aaa) + a^*bbbbb^* + (a + b)^*ab(a + b)^*$
$(λ + a + aa + aaa)b^* + a...
