Let $n > 0$ be some integer and let $G$ be a splitting field of the set of all polynomials of degree at most $n$ over a field $F$. I need to prove that $G$ is not an algebraic closure of $F$ in the following two cases:
$F = \mathbb{Q}$
$F = \mathbb{Z}_{p}$ for prime $p$ (NOT the $p$-adic number...
Actually, I think one of them does. The Greek word "logos" where the English word "logic" comes from is how Jesus is described in the beginning of the Gospel according to John, which was originally written in Greek.
It could be translated as "In the beginning, there was logic."
But, personally, I don't think that would really change anybody's mind about it.
If I wanted to prove that $G$ that I mention above is not an algebraic closure of $\mathbb{Q}$, somebody said I should find a polynomial of degree $>n$ that does not split over $G$. I'm still having so much trouble wit hthis, though! How do I come up with such a polynomial?
How do I do that when I don't even know what $G$ is?
if F is infinite, there should be infinitely many things of degree <=n, which means X is infinite and so m (the number of things taken out of X to make a particular rational expression) can be arbitrarily large
One of the people who commented told me to find a polynomial of degree $>n$ over $\mathbb{Q}$, at least for part 1. of the question that is irreducible in $\mathbb{Q}$.
I was trying to do that. I need a polynomial $a_{p}x^{p} + a_{p-1}x^{p-1}+ \cdots a_{1}x + a_{0}$ where $p|a_{p-1}$, $\cdots\, p|a_{1}, \, p|a_{0}$ but $p$ does not divide $a_{p}$ and $p^{2}$ does not divide $a_{0}$.
sure. it doesn't even have to be the same prime, could be $x^q-\ell$ (since the prime $\ell$ is irrelevant and $q>n$ can be different from $p$ in the $\Bbb F_p$ part of the problem)
@Secret Write $f(n)=\sum_{k=1}^n k^2$. Denote $(\Delta g)(n):=g(n+1)-g(n)$. Then $\Delta f(n)=(n+1)^2$, so $\Delta^2 f(n)=2n+3$. Using $\Delta \binom{n}{k}=\binom{n}{k-1}$ (when $k$ is fixed), you can "discretely integrate" the equation $\Delta^2f(n)=2\binom{n}{1}+3\binom{n}{0}$ twice (adding constants).
@Secret $\binom{n+1}{k}-\binom{n}{k}$ counts the $k$-subsets of $\{1,\cdots,n+1\}$ that include $n+1$, which are in bijection with the $(k-1)$-subsets of $\{1,\cdots,n\}$
Here's a thought for a construction: let $E(G)$ be the free group on the underlying set of a group $G$ mod the relations $[g]^k=[g^k]$, for example I think in the case of the quaternion group $Q_8$ we get $C_4\ast_{C_2}C_4\ast_{C_2} C_4$.
Idea came about when considering braid groups, symmetry groups of polytopes and orientation entanglement.
I guess really the construction could be done on any $\Bbb Z$-set $G$
well, no, the orbits would need distinguished points.
Ah now I know why there's that extra 1 in the summation by parts: The finite difference product rule has a cross term $\Delta f\Delta g$ and unlike derivatives, it does not vanish
I don't know why some people come on here, basically post questions where they're asking people to do their homework for them, nobody touches them. I post something where I've at least done SOME work, although I sound totally clueless, and I get downvoted.
But I kind of understand what that other guy Sean was saying.
I also have no idea why it's important that $x^{2}+x+1 \neq x^{2} + x$ in the context of this problem. Apparently, that's the connection between Sean's commenets and Lubin's answer, and I don't really see where it's coming from.
@ALannister He's saying that $z^2 + z$ is not surjective; every value it takes is 0. So you can just translate it by $1$ to get $z^2 + z + 1$ which has no roots over $\Bbb F_2$, say.
Let $x$ be some divergent sequence. Construct the tuple $(x,s)$ which consists of a divergent series and its distribution of brackets given as a string of relative positions $L, R$ and $|$. The <have not figured out the name yet> is defined as:
@ALannister This is why Lubin's last comment is relevant. $z + z^2$ has two elements in the kernel as a map from the field to itself, so the image of this is going to have only half the elements of the field.
And then you do the shift argument (Martin beat me to it)
For example, let: $s =\sum_{n=0}^{\infty} (-1)^n$. Then $[s,\{\}]$... O cr**, the LR | notations is insufficient to denote brackets of the form: x+(x+x)+(x+x)+... vs (x+x)+(x+x)+(x+x)+...
[Random] The bracket operator: Let $\mathop{B}$ be the bracket operator, which is defined as follows: Given some string $s$ that describe how brackets are distributed, and a formula $f$, then $B(f)$ applies brackets onto $f$ in that order. For example:
$$\mathop{B}(\cdot ((\cdot \cdot) \cdot)(\cdot \cdot)),1+2+3+4+5) = 1+((2+3)+4)+(5+6)$$ More simply, we can treat the string of brackets as one operator: $$(\cdot ((\cdot \cdot) \cdot)(\cdot \cdot))\circ (1+2+3+4+5) = 1+((2+3)+4)+(5+6)$$
Need to came up more systematic ways to wrote that string of brackets...
That also @Balarka but the reason why everybody hated B.G. was because Windows 98 included Internet Explorer, and as they say, "video killed the radio star".
Let $n > 0$ be some integer and let $G$ be a splitting field of the set of all polynomials of degree at most $n$ over a field $F$. I need to prove that $G$ is not an algebraic closure of $F$ in the following two cases:
$F = \mathbb{Q}$
$F = \mathbb{Z}_{p}$ for prime $p$ (NOT the $p$-adic number...
