For the field $Q(\sqrt[3]{2})$, the elements in the field are in the form of $a + b\sqrt[3]{2}+c\sqrt[3]{4}$, where a,b,c, are elements of Q. Is the automorphism group that fixes Q  $C_{3}$? multiplying all elements in $Q(\sqrt[3]{2})$ by $\sqrt[3]{2}$, will result a cyclic cycle of period 3, but what about mapping $\sqrt[3]{2}$  to  $-\sqrt[3]{2}$? For extension fields like $Q(\sqrt{2},\sqrt{3})$, the mapping towards a radical's negative counterpart is part of the automorphism group. Should this also be part of the automorphism group that fixes Q?
I am having difficulty with this proof. I have previously proved that for an automorphism on the extension field $K$ of $F$ that when restricted to $F$ is the identity automorphism must take roots of $f(x)\in F[x]$ in $E$ to roots of $f(x)$ in $E$. I was thinking of using the minimal polynomial $x^{3}-2\in\mathbb{Q}[x]$ and the result of previous problem I solved, but I suspect there are automorphisms that don't fix the subfield, so the result can't be used or I need to prove that any automorphism of an extension field must fix the subfield that is being extended or always sends roots in $E$ to roots in $E$.  I am not asking for a solution to this, but maybe a hint or pointer in the right direction. This problem is introduced in the text prior to Galois theory, so please keep that in consideration.
to infinitive as direct object?! (as in : I promised to get there in time.) I'm a passionate Korean English-learner, and I would like to ask you a question about grammar.   It was 3 years ago when I first started to study the to-infinitive as studying grammar with Grammar In Use series and English Grammar series by Longman, having been exposed to English for 3 years. There wasn't any mention of the author saying that the usage of the to-infinitive is as an (direct) object. And for me also, the times I have encountered sentences in which the usage of the to-infinitive, so to speak, as direct objects were used, I've read them as some kind of obligatory adverbial that shows what the verbs are about, not as objects, which means they are noun phases! 100% of Korean textbooks say the usage is as a direct object, but I doubt it. I think it's some kind of obligatory adverbial that shows what the verbs are about.    I thought it was better for me to ask Americans themselves. Do you read this usage of to-infinitive as a noun phrase or an adverbial? I do not understand how dummy variables work in math.   Suppose we have:  $$I_1 = \int_{0}^{\infty} e^{-x^2} dx$$  How is this equivalent to:  $$I_2 = \int_{0}^{\infty} e^{-y^2} dy$$  How does this dummy variable system work?   Since $y$ is the dependent variable for $I_1$ How can $y$ itself be and independent variable for $I_2$  ??  Thanks! Is it true that $A^A\simeq P(A)$? Let $A$ be an infinite set and $$A^A=\{f | f:A\to A\,\,\, \text{is a function}\}.$$ Is there any natural bijection between $A^A$ and $P(A)$? What are the origins of the word "nice"? "Specially" vs "especially" When should I use specially and when especially? . It is a mathematical expression/sequence/set of numbers with the arrow exactly under any one of the numbers (with or without n= 0). How to do it in latex? Any of the up arrow types (single line or double line) shown in the figure is okay. Hello can someone give me short tutorial how to:   Remove Windows 7 from my dual boot system (Ubuntu/Windows7) Let me use free space as new partition for data (so I don't need to extend Ubuntu itself here) Remove windows 7 option from boot screen   Update:  Thanks to Lekensteyn answer I successfully deleted windows. Now I created new partition (formatted to ext4), but I can not use it. I cannot copy files there, only as root.  Also when booting, Ubuntu complains it didn't find /windows partition (of course it didn't, I removed it). I feel like I should update some tables or what, any ideas? What does active mean? Buying shares when the price goes down 2% and selling shares when it goes up 2% I've been having an argument with a colleague about this sentence, could you please let me know which one of us is correct:     There are no shortage of applications for our product in this   space.   She is convinced that are should be replaced by is, and I think it should stand as it is. Thanks for your help! Polynomial ring $R[X] $ has zero Jacobson radical, for $R$ a domain This question pertains to L1 &amp; L2 regularization parameters in Light GBM. As per official documentation:  reg_alpha (float, optional (default=0.)) – L1 regularization term on weights.  reg_lambda (float, optional (default=0.)) – L2 regularization term on weights  I have seen data scientists using both of these parameters at the same time, ideally either you use L1 or L2 not both together.   While reading about tuning LGBM parameters I cam across one such case: Kaggle official GBDT Specification and Optimization Workshop in Paris where Instructors are ML experts. And these experts have used positive values of both L1 &amp; L2 params in LGBM model. Link below (Ctrl+F 'search_spaces' to directly reach parameter grid in this long kernel)      I have seen same in XGBoost implementations.  My question is why use both at the same time in LGBM/XGBoost.  Thanks. Can't upgrade due to low disk space on /boot I try to do do-release-upgrade but then I get:       Not enough free disk space      The upgrade has aborted. The upgrade needs a total of 25.7 M free   space on disk '/boot'. Please free at least an additional 25.7 M of   disk space on '/boot'. Empty your trash and remove temporary packages   of former installations using 'sudo apt-get clean'.   Output of df:                      237251272 214797108  10402504  96% / udev                    488120         4    488116   1% /dev tmpfs                   198676       668    198008   1% /run none                      5120         0      5120   0% /run/lock none                    496684         0    496684   0% /run/shm /dev/sda1               233191    225867         0 100% /boot   How come there is no space left on boot? Here's the output of ls -as /boot:  total 221839     4 .     4 ..   645 abi-2.6.32-34-generic-pae   698 abi-2.6.35-30-generic-pae   727 abi-2.6.38-12-generic-pae   727 abi-3.0.0-12-generic-pae   727 abi-3.0.0-13-generic-pae   727 abi-3.0.0-14-generic-pae   727 abi-3.0.0-15-generic-pae   727 abi-3.0.0-16-generic-pae   727 abi-3.0.0-17-generic-pae   727 abi-3.0.0-19-generic-pae   761 abi-3.0.0-20-generic-pae   115 config-2.6.32-34-generic-pae   128 config-2.6.35-30-generic-pae   136 config-2.6.38-12-generic-pae   140 config-3.0.0-12-generic-pae   140 config-3.0.0-13-generic-pae   140 config-3.0.0-14-generic-pae   140 config-3.0.0-15-generic-pae   140 config-3.0.0-16-generic-pae   140 config-3.0.0-17-generic-pae   140 config-3.0.0-19-generic-pae   140 config-3.0.0-20-generic-pae     5 grub 10773 initrd.img-2.6.32-34-generic-pae 13619 initrd.img-2.6.35-30-generic-pae 15365 initrd.img-2.6.38-12-generic-pae 16481 initrd.img-3.0.0-12-generic-pae 16487 initrd.img-3.0.0-13-generic-pae 16501 initrd.img-3.0.0-14-generic-pae 16476 initrd.img-3.0.0-15-generic-pae 16481 initrd.img-3.0.0-16-generic-pae 16478 initrd.img-3.0.0-17-generic-pae    12 lost+found   174 memtest86+.bin   176 memtest86+_multiboot.bin  1700 System.map-2.6.32-34-generic-pae  1841 System.map-2.6.35-30-generic-pae  2115 System.map-2.6.38-12-generic-pae  2141 System.map-3.0.0-12-generic-pae  2141 System.map-3.0.0-13-generic-pae  2143 System.map-3.0.0-14-generic-pae  2146 System.map-3.0.0-15-generic-pae  2147 System.map-3.0.0-16-generic-pae  2147 System.map-3.0.0-17-generic-pae  2148 System.map-3.0.0-19-generic-pae  2149 System.map-3.0.0-20-generic-pae     2 vmcoreinfo-2.6.32-34-generic-pae     2 vmcoreinfo-2.6.35-30-generic-pae     2 vmcoreinfo-2.6.38-12-generic-pae     2 vmcoreinfo-3.0.0-12-generic-pae     2 vmcoreinfo-3.0.0-13-generic-pae     2 vmcoreinfo-3.0.0-14-generic-pae     2 vmcoreinfo-3.0.0-15-generic-pae     2 vmcoreinfo-3.0.0-16-generic-pae     2 vmcoreinfo-3.0.0-17-generic-pae     2 vmcoreinfo-3.0.0-19-generic-pae     2 vmcoreinfo-3.0.0-20-generic-pae  4092 vmlinuz-2.6.32-34-generic-pae  4347 vmlinuz-2.6.35-30-generic-pae  4567 vmlinuz-2.6.38-12-generic-pae  4675 vmlinuz-3.0.0-12-generic-pae  4676 vmlinuz-3.0.0-13-generic-pae  4681 vmlinuz-3.0.0-14-generic-pae  4698 vmlinuz-3.0.0-15-generic-pae  4700 vmlinuz-3.0.0-16-generic-pae  4700 vmlinuz-3.0.0-17-generic-pae  4703 vmlinuz-3.0.0-19-generic-pae  4705 vmlinuz-3.0.0-20-generic-pae   Output of uname -a:   Linux kitsch 3.0.0-17-generic-pae #30-Ubuntu SMP Thu Mar 8 17:53:35 UTC 2012 i686 i686 i386 GNU/Linux I often use  to keep my sentences gender neutral. Many of my friends (none of them are native speakers), however, consider this to be grammatically incorrect, and suggest me not to use it in formal English exams.  My questions are:   Is singular they widely accepted in English speaking countries such as America and Britain? Is it formal enough to be used in formal situations, such as a public speech? Is it casual enough to be used in informal situations, such as gossiping among friends? How to find out what's accessing a particular hard drive on Windows? What will be the value of the following Infinite Product :   $$\displaystyle \prod_{k=0}^\infty \left(1+\dfrac{1}{2^{2^k}}\right)$$  It would be nice if anyone could spare the time and boil down to the absolute basics and tell how they reached the solution. The isomorphism $\mathbb{Z}[i]/(a+bi) \cong \Bbb Z/(a^2+b^2)\Bbb Z$ is , when the integers $a$ and $b$ are coprime. But what happens when they are not coprime, say $(a,b)=d&gt;1$?      — For instance if $p$ is prime (which is not coprime with $0$) then $$\mathbb{Z}[i]/(p) \cong \mathbb{F}_p[X]/(X^2+1) \cong \begin{cases} \mathbb{F}_{p^2} &amp;\text{if } p \equiv 3 \pmod 4\\ \mathbb{F}_{p} \times \mathbb{F}_{p} &amp;\text{if } p \equiv 1 \pmod 4 \end{cases}$$ (because $-1$ is a square mod $p$ iff $(-1)^{(p-1)/2}=1$).  — More generally, if $n=p_1^{r_1} \cdots p_m^{r_m} \in \Bbb N$, then  each pair of integers $p_j^{r_j}$ are coprime, so that by CRT we get $$\mathbb{Z}[i]/(n) \cong \mathbb{Z}[i]/(p_1^{r_1}) \times \cdots \times \mathbb{Z}[i]/(p_m^{r_m})$$  I was not sure how to find the structure of $\mathbb{Z}[i]/(p^{r}) \cong (\Bbb Z/p^r \Bbb Z)[X] \,/\, (X^2+1)$ when $p$ is prime and $r&gt;1$.  — Even more generally, in order to determine the structure of $\mathbb{Z}[i]/(a+bi)$ with $a+bi=d(x+iy)$ and $(x,y)=1$, we could try to use the CRT, provided that $d$ is coprime with $x+iy$ in $\Bbb Z[i]$. But this is not always true: for $d=13$ and $x+iy=2+3i$, we can't find Gauss integers $u$ and $v$ such that $du + (x+iy)v=1$, because this would mean that $(2+3i)[(2-3i)u+v]=1$, i.e. $2+3i$ is a unit in $\Bbb Z[i]$ which is not because its norm is $13 \neq ±1$.  — I was not able to go further. I recall that my general question is to known what $\mathbb{Z}[i]/(a+bi)$ is isomorphic to, when $a$ and $b$ are integers which are not coprime (for instance $a=p^r,b=0$ or $d=(a,b) = a^2+b^2&gt;1$).  Thank you for your help! I'm starting on mono blue mill deck and I saw . I'm not sure about its ability.     {tap}: put a charge counter on Grindclock      {tap}: target player puts the top X cards of his or her library into his or her graveyard, where X is the number of charge counters on Grindclock.   If I just cast  and use its first ability, will the second ability take effect, or do I just put a charge counter on it? If there's already one charge counter on it and I use the first ability again, it will have two charge counters on it? Then the second ability will take effect with two cards being sent to graveyard?  If I use spells or ability to untap it after being used can I use its ability again on the same turn? Navigate folder structure in Open/Save File Dialog with keyboard? After hitting ⌘-O, how can I navigate my folders and select a file to open using only my keyboard?   In particular, how can I get the focus into the folder list? (From there I can use arrows or start typing the first letters of the desired file/folder.)  In some programs I can rely on hitting TAB 5 times, but this isn't reliable and I figure there must be a better way. How can one expand the function $f_1(x) = x$ on $(−π, π)$in terms of the functions $\cos nx, n = 0, 1, 2, ...$ and $\sin nx, n = 1, 2, ...,$ in a way so that $$f_2(x) = a_0 + \sum_{n=1}^{\infty}(a_n \cos nx + b_n \sin nx)$$ is the expansion of the $f_1$, how is that? I have asked  related to the same issue but didn't get full answer, I have been told that this is a standard application of Fourier series, $f_1$ and $f_2$ are equivalent in the sense that $f_1(x)=f_2(x)$ for all $x∈(−π,π)$. For any $x$ in this interval, the Fourier series will converge to $f(x)$, but I didn't get any rigorous proof  with elaborate explanation. So, plz provide rigorous proof of $f_1=f_2$ with elaborate explanation as I am new to the subject. Thanks You. I have attached the excerpt below -