In other words, if you have a pile of cannonballs in a square pyramid, can you rearrange them as a cube instead, or do you have to shell someone first? Or, instead, does $$n(n+1)(2n+1)=6x^3$$ have any nontrivial integer solutions?
The analogous problem where you want to put the cannonballs in a ...
ideal is closed under subtraction and closed under multiplication so if I have something like http://prntscr.com/dd06sh so does this mean that $a \in R$ and $f(x) \in I$ then for closure under subtraction it's $a-f(x)$ and multiplication is $af(x)$ ??
@robjohn do you know how to prove ideals? I know it's nonempty, we have closure under subtraction, and closure under multiplication but I think what I typed is kinda basic
@robjohn the problem I see, correct me if I'm wrong, is that Pascal's recursion is only defined for integers. And with factorials: we are not there yet to use gammafunction, so (-1)! is not defined yet. So the only way for me (as i see it) is using $\prod$ algebra or the "polynomial of degree n" argument.
$$\binom{x}{n}=\prod_{i=1}^{n} \frac{x+1-i}{i}$$
and this is really painful if I don't know the "magic" trick haha
@robjohn x is of degree one, x(x-1) of 2, [...], x(x-1)...(x-n) of degree n
the problem i have is: with the equality $\binom{x}{n+1}-\binom{x-1}{n}=\binom{x-1}{n+1}$ i have one polynom of degree n+1 and the other two are of degree n. or am i wrong?
$\sqrt{397}=[19;\overline{1,12,3,4,9,1,2,1,2,1,1,2,1,2,1,9,4,3,12,1,38}]$ and if you get a convergent $\frac{p_n}{q_n}=[19;1,12\dotsb 4,3,12,1]$ then $p_n^2-397q_n^2=\pm 1$
I have conjectured that this holds for all numbers
@Sophie Well you know that $\left|\frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}\right|=\frac1{q_nq_{n+1}}$ when $\frac{p_n}{q_n}$ are continued fraction approximations
If a ring R is simple because the only ideals are 0 and R does this mean that we have a zero ideal and an identity? I got this for closure under multiplication $0 \cdot r =0 = r \cdot 0$ $1 \cdot r = r = r \cdot 1$
@arctictern do you know ideal proofs... like for something to be an ideal it must be nonempty, closure under subtraction, and closure under multiplication
If $K$ is a field and $a\in K$, then the splitting field of $x^n-a$ over $K$ will be $K(b,\xi)$ where $b$ is any root of $x^n-a$, and $\xi$ is a primitive $n$th root of unity. To show this, show every root of $x^n-a$ is expressible with $b$ and $\xi$, and conversely that $b$ and $\xi$ are expressible in terms of roots of $x^n-a$.
I'm on this http://prntscr.com/dd0jw9 so we need to have a nonempty subset, closure under subtraction, and closure under multiplication , so I'm getting something weird like $a-f(x)$ and then $af(x)$...$f(x)a$ ... I think the $f(a) = 0$ is throwing me off
@Adeek the galois group must have size 6 because that is the degree of the extension (exercise: prove this), so it suffices to see the galois group is nonabelian, so it suffices to find two symmetries that don't commute
We are trying to show $I$ absorbs ambient multiplication. Suppose $f(x)\in I$, and let's multiply it by something arbitrary like $r(x)\in R[x]$. How do you check if $r(x)f(x)$ is still an element of $I$?
points at the definition of I given in the problem
@arctictern first since we have the roots are combination of $(2)^1/3$ and $\omega$ so our splitting field is just $\mathbb{Q}((2)^1/3,\omega)$ but we can delete the 1/2's so we get the splitting field is $\mathbb{Q}((2)^1/3,i\sqrt{3})$
the condition "$f(a)=0$" in the definition of $I$ tells you how to determine if a polynomial is in $I$: plug in $a$ and see if the result is $0$
So let's start again. Suppose $f(x)\in I$ and we multiply it by an arbitrary $r(x)\in R[x]$. We get $r(x)f(x)$. If we plug $a$ into $r(x)f(x)$ what do we get?
so $f(a) = a^2-3a+2 = 0$ $f(a) = (a-2)(a-1)=0$ $a = 1,2, $ and plugging those values should be 0 so if we plug in a for r(x)f(x) don't we get r(a)f(a)?
hmm do we still have the $f(a) =0$ in play? I'm thinking that $f_{1}(a) \in I$, $f_{2}(a) \in I $ so $ f_{1}(a)+f_{2}(a) \in I $ but that would mean that $f_{1}(a) = 0$ and $f_{2}(a) =0$ so both will be 0 when added together
@robjohn i think i was actually on the right track while i used $\prod$. it just gets really fast really confusing^^ (since descending factorials are the same)
@arctictern what do you think of this reasoning so $[Q((2)^1/3), Q] = 3$ as $x^3 - 2$ is irreducible over Q by Eisenstein. Since $x^2 - 3$ is irreducible polynomial with $\sqrt(-3)$ as a root, so the extension $[Q((2)^1/3,\sqrt(-3)), Q(2^{1/3})]$ has at most degree 2.
@robjohn finding a closed form is always possible for linear recursives. But what would be an example that is not linear and can't be expressed in close form?
When people do optimization with Lagrangian Multipliers. Do you take the gradient of the entire Lagrangian? or do you just take the sum of partials of with respect to the original variables (just $\vec x$, for example)?
Assume each multiplier is only associated with an equality constraint, to make things easier.
just b on this one http://prntscr.com/dd1cz8 wouldn't the matrix be closed under subtraction and closed under multiplication for it to be an ideal? $\begin{bmatrix}0&b\\0&0\end{bmatrix}$ $\begin{bmatrix}0&a\\0&0\end{bmatrix}-\begin{bmatrix}0&b\\0&0\end{bmatrix}=\begin{bmatrix}0&a-b\\0&0\end{bmatrix}$
okay. let's check if I is closed under ambient multiplication. pick an arbitrary element of I. then pick an arbitrary element of S. multiply them out and see if the result is still an element of I,
so $$\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \begin{pmatrix} 0 & x \\ 0 & 0 \end{pmatrix} $$
so that's closure under subtraction and closure under multiplication $\begin{bmatrix}0&0\\0&r\end{bmatrix}$ $\begin{bmatrix}0&0\\0&r\end{bmatrix}-\begin{bmatrix}0&0\\0&s\end{bmatrix}=\begin{bmatrix}0&0\\0&r-s\end{bmatrix}$
I have some problem @arctictern so I defined my map as before I don't think it is $S_3$ because there is some elements which commute with each other but I haven't verified it.
Let's check if $J$ is closed under outside multiplication. That means given any element of $J$, we can multiply it by an arbitrary element of the ring $M_2(\Bbb R)$ and it will still be an element of $J$.
You are not multiplying two elements of $J$ together. You are multiplying an element of $J$ by an element of the ring. The ring $M_2(\Bbb R)$ is bigger than the subset $I$.
no, matrix multiplication
which is the multiplication in the ring $M_2(\Bbb R)$
@Adeek and how do you extend that to a map on a Q-basis for Q(2^1/3,theta)/Q?
@Adeek say the three roots of x^3-2 are a,b,c. the automorphism which sends a real root to a complex root cannot possibly commute with complex conjugation.
so it's required that we need those $\begin{bmatrix}0&0\\0&r\end{bmatrix}$ but then row 2 column 2 can either be 0 or a letter as long as it's similar to this $\begin{bmatrix}0&0\\0&r\end{bmatrix}$ so b needs to be 0 otherwise we'll have $\begin{bmatrix}0&br\\0&dr\end{bmatrix}$
for example if we consider $\sigma_1$ given by $(2)^{1/3} \mapsto (2)^{1/3}\theta$ and $i\sqrt{3} \mapsto i\sqrt{3}$. Consider $\sigma_2$ defined same way on $(2)^{1/3}$ and sends $i*\sqrt{3} \mapsto -i\sqrt(3)$ then we find that they don't commute @arctictern
How about this. Let $r$ be a real root of $x^3-2$ and let $s_1,s_2$ be the complex conjugate roots. Let $\alpha$ be a symmetry that sends $r$ to $s_1$, and let $\beta$ be complex conjugation. Then check $(\alpha\circ\beta)(r)=s_1$ but $(\beta\circ\alpha)(r)=s_2$.
complex conjugation is obviously an automorphism, and there's a theorem of galois theory that the galois group acts transitively on the roots of a splitting polynomial, so there is a symmetry $\alpha$ that sends $r$ to $s_1$
could you check this? I did matrix multiplication (two entries above me) and some elements aren't absorbed for left so that makes it not a right ideal prntscr.com/dd1sx3
RA is a left ideal - it absorbs multiplication from the left
you're checking if it's a right ideal - if it absorbs multiplication from the right
compute $$\begin{bmatrix} 0 & a \\ 0 & c\end{bmatrix}\begin{bmatrix} w & x \\ y & z\end{bmatrix}$$ and see if it's still in the form $\begin{bmatrix} 0 & \ast \\ 0 & \ast \end{bmatrix}$.
prntscr.com/dd1x68 I wish Ra is I = Stuff -____-!!! otherwise I know how it works. closure under subtraction, closure under multiplication and nonempty. Could I let Ra be I = whatever it is instead? the notation is messing me up
@MikeMiller I'm not sure if it exists. I was told someone at Stony Brook had some things to say about errors in 4-manifold theory, but that looks similar the Zinger thing and I think the person maybe was just conflating that with something else.
any polynomial of degree 3 has atleast one root root ?
I am proving the following if the galois group of irreducible polynomial f(x) of degree 3 is $\mathbb{Z}_3$ then f must have real roots.
Suppose $\mathbb{K}$ is the splitting field of f(x), then $[\mathbb{K} : \mathbb{Q}] = [\mathbb{K} : \mathbb{K(x)}] [K(x) : \mathbb{Q}]$. If the galois group is $Z_3$ then we have $[\mathbb{K} : \mathbb{K(x)}] = 1$ for any root x for the polynomial f(x).
@Adeek any odd degree real polynomial has a real root. either look at asymptotic behavior of the graph in the real cartesian plane or consider complex roots come in conjugate pairs.
even without considering the full galois group, if complex roots come in conjugate pairs and the degree = # of roots counted with multiplicity, then odd degree implies real root. this is college algebra.
Suppose $f(x) = c_nx^n + \cdots + c_1x + c_0 \in \mathbb{R}[x]$ has a complex root $\alpha$. This means $c_n\alpha^n + \cdots + c_1 \alpha + c_0 = 0$. Now take the complex conjugate of both sides of this equation, and on the LHS apply the fact that complex conjugation is both additive and multiplicative.
That's the easiest way to see it with the least machinery IMO
@MikeMiller The first email is already ridiculous. Look at that cc. That's the initial contact with the authors.
user228700
Can anybody please help me to understand one-one and onto functions in the specific case of composition of two functions? I have been trying to figure this out for a few days and have found myself no closer to understanding it than before .__.
What conditions must the domains and codomains of two functions $f(x)$ and $g(x)$ obey for the composition of these two functions, $g[f(x)]$ to be one-one/onto, given that both, $f(x)$ and $g(x)$ are one-one and onto?
I know of a much more recent example in this field (one of the collaborators is active on MO) and the other is near the top, but the authors retracted everything very near when the error was found.
@Kaumudi Suppose both g(x) and f(x) are one-to-one. then g(f(x))=g(f(x')) implies f(x)=f(x') implies x=x', so g(f(x)) is one-to-one as well. Suppose both g(x) and f(x) are onto. Then for all u there exists a v such that g(v)=u, and there exists a w such that f(w)=v, in which case g(f(w))=u. So g(f(x)) is also onto.
user228700
But regarding bijective functions, this is what my book says:
> "The composite of two bijections is a bijection iff $f$ and $g$ are two bijections such that $g \circ f$ is defined, $g \circ f$ is also a bijection only when codomain of $f$ is equal to the domain of $g$"
