I need to show that if R is an integral domain and it has unity then the only elements of R which are inverse to itself are 1 and -1 (with respect to multiplication). But I don't know where to start. Can anyone help? Thanks..
I like to be thorough, but if you feel confident you can skip the first paragraph.  Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds.  If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain.  We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1).    Here's what I did:   For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$.  Assume $a^2 = 1$.  We want to show that letting $b^2 = 1$ as well will lead to a contradiction.  So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution?  If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction.  If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction.  Similarly, $ab = b$ won't work either.  So it must be that $ab = c$.  So by "admitting" $b$ as a solution, we're forced to admit $c$ as well.  So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$.  We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution.  From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$.  So $abc$ is not a new solution; it's just one.  At this point I'm stuck.  I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$.  What am I missing?
As far as I know, the diamond symbol is given to the folks who  are the community elected moderators, for example,  are employees of the Stack Exchange, Inc., for example,   In those regards,  is neither a , nor is he a part of the  since March 1st, 2012. So my question is, other than the fact that he is the , is there any other reason why he still has the diamond symbol next to his name and whether he can still exercise his moderation abilities? :) Note: I do not have any problems with him having the Diamond Symbol. I just wanted to know the reason and whether he still has those privileges or not. In fact, I'd be very happy to know that he still has those privileges because, you know, he knows the site, the Best. :) Note: as of October 2017, . Must I enter Europe with Schengen visa from the country where I applied? 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I'm wondering if these special forms, the Quasit, Imp, and Pseudodragon in particular, allow the player character to share the familiar's Magic Resistance feature. In the Monster Manual each of these creatures has a sidebar that states that the familiar shares its Magic Resistance feature with the companion they are bonded to, but the PHB doesn't mention this in any of the creatures' stat blocks.   For example, in the Variant: Pseudodragon Familiar sidebar on page 254 of the MM it says:     "While the pseudodragon is within 10 feet of its companion,        the companion shares the pseudodragon's Magic Resistance trait."   This feature seems clear in the MM but I'm led to believe that it wasn't intended for Player Characters since mention of it is absent from the stat blocks in Appendix D of the PHB (pages 307-309).  Would a 3rd level warlock who chose one of these familiar forms through the Pact of the Chain feature benefit from advantage from saving throws from spells and other magical effects due to these special forms, or was this feature only meant for powerful NPCs and enemy spellcasters who had made this link with these familiars themselves? Why is it dangerous use a coiled extension cord Is a small size block cipher usable? Consider a block cipher with a key of size 128 bits but a small block size, say 32 bits.  Is this kind of block cipher secure?  I would like to know to what extent I can use a small block cipher, like a 32 bit block cipher. What are the constraints so that the use is acceptable in a real life application.  For instance, if we use a one-time key, that is, to encrypt only one message. Besides, the message could be small.  Suppose that when we want so send one file, we perform a key agreement so that we use the key for one message. 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Would it be correct to just say that if $\frac{r_1}{r_2}$ is the ratio of radii between the two circles, then   1) The radii are invariant under translation, $z \mapsto z+a$, so $\frac{r_1}{r_2}$ stays the same  2) Under scaling by a factor $z \mapsto az$, $\frac{ar_1}{ar_2} = \frac{r_1}{r_2}$   3) Under inversion, $z \mapsto \frac{1}{z}$, $\frac{1/r_1}{1/r_2} = \frac{r_2}{r_1}$  Or is there a different/better way to think about this problem? How do I make a Lightning Bow in vanilla Minecraft? I'm trying to make a setup with command blocks that summons lightning at the location of a shot arrow. The setup should recognize the arrow that is shot or the bow that is used by its name or another property, so that lightning only strikes at that specific arrow.   Also, the lightning should only strike once when the arrow hits the ground. I know this is somehow possible with scoreboard, but I'm no good with that. 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Let $F$ be a Galois extension over $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^n$, then all elements in $F$ are constructible Let $F\subseteq\mathbb{C}$ be a Galois extension of $\mathbb{Q}$ such that $[F:\mathbb{Q}]=2^n$; then all elements in $F$ are constructible.    Added.  Here is what I have so far.  Since there exist the finite normal extension $F\subseteq\mathbb{C}$ over $\mathbb{Q}$ which contains an element say, s, then s has a degree of power of 2. From group theory we have 2 propositons we could use. 'A group $G$ is solvable iff it has a normal series whose factors are abelian', and also 'If $G$ is solvable and $N\triangleleft G$, then $G/N$ is solvable'. Then we can say that the galois group F has a normal series $\{e\}=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_n=\mathrm{Gal}(F/\mathbb{Q})$, whose factors have order 2. Thus there must exist intermediate fields $\mathbb{Q}=F_0 \subseteq F_1 \subseteq F_2 \subseteq\cdots\subseteq F_n=F$, such that $[F_{j+1}:F_j]=2$ for all $j$. Now this is where I got stuck. Next I need to somehow show that $F_{j+1}=F_j(s_j)$, where $s^2_j$ is an element in $F_j$. Then I think I can conclude that if $s_j$ is constructible than every element in $F_{j+1}$ is constructible... then with $F_n$ is constructible... Prove: If $a$ and $b$ are odd integers, then $8\mid (a^2-b^2)$  So far I have: $$8\mid \left((2n+1)^2-(2m+1)^2\right)\Longleftrightarrow 8\mid \left(4n^2+4n+1-4m^2-4m-1\right) $$ Is this right so far? Where do I go from here? Step by step explanation please! Internal keyboard and trackpad stop working after plugging in USB device I got a new MacBook Pro with OS X Mavericks. When connecting my phone via USB, the internal keyboard and the trackpad immediately stop working -- not always but in most cases. The bluetooth mouse and an external USB keyboard are not affected. Everything else works also fine. The power button of the internal keyboard still works. Reboot helps. Any ideas? LaTeX templates for writing a thesis Aside from CTAN, what are good resources/repositories for Latex templates? In particular, I'm looking for some pretty Thesis templates (I'm familiar with classicthesis) Graph of a continuous function is closed I was given a MySQL database file that I need to restore as a database on my Windows Server 2008 machine.  I tried using MySQL Administrator, but I got the following error:     The selected file was generated by   mysqldump and cannot be restored by   this application.   How do I get this working?