Another way to prove that is to notice that \( \displaystyle 1-r \) is also an idempotent and then apply the method used in the first case.Papapetros Vaggelis wrote: If \(\displaystyle{1-r}\) is invertible, then
\(\displaystyle{r^2=r\implies r(1-r)=0\implies r(1-r)\,(1-r)^{-1}=0\implies r=0}\) .
Search found 284 matches
- Fri Nov 27, 2015 1:14 am
- Forum: Algebraic Structures
- Topic: On Ring Theory
- Replies: 2
- Views: 2557
Re: On Ring Theory
- Thu Nov 26, 2015 1:57 am
- Forum: Algebraic Structures
- Topic: On Ring Theory (A Long One)
- Replies: 0
- Views: 1601
On Ring Theory (A Long One)
Let \( \displaystyle R \) be an associative ring with unity. Prove that the following are equivalent: \( \displaystyle R \) is a local ring. The set of non-invertible elements of \( \displaystyle R \) forms an ideal of \( \displaystyle R \). There is a proper left ideal of \( \displaystyle R \) cont...
- Wed Nov 25, 2015 11:22 pm
- Forum: Algebraic Structures
- Topic: On Ring Theory
- Replies: 2
- Views: 2557
On Ring Theory
Let \( \displaystyle R \) be a ring. If \( \displaystyle r \in R \) is nilpotent, then show that \( \displaystyle 1-r \) is invertible. Let \( \displaystyle R \) be a local ring. Show that the only idempotents in \( \displaystyle R \) are the trivial ones. Let \( \displaystyle R \) be a ring whose ...
- Mon Nov 23, 2015 10:05 pm
- Forum: Algebraic Structures
- Topic: On Ring Theory (An Easy One)
- Replies: 8
- Views: 6528
Re: On Ring Theory (An Easy One)
Suppose that \( \displaystyle x^{n} = 0 \) and \( \displaystyle y^{m} = 0 \). If you expand the expression \( \displaystyle (x+y)^{n+m-1} \), you will obtain \( \displaystyle (x+y)^{n+m-1} =0 \). Can you see why?Papapetros Vaggelis wrote:What about \(\displaystyle{x+y}\) if \(\displaystyle{x\,,y\in I}\) ?
- Mon Nov 23, 2015 9:57 pm
- Forum: Algebraic Structures
- Topic: On Ring Theory (An Easy One)
- Replies: 8
- Views: 6528
Re: On Ring Theory (An Easy One)
Ιn fact, I had in mind that the rings are commutative (and associative, of course) with unity - influenced by my recent studying. Well done!Papapetros Vaggelis wrote:I suppose that you mean associative rings with unity.
- Mon Nov 23, 2015 2:09 am
- Forum: Complex Analysis
- Topic: Are These Sets Biholomorphic?
- Replies: 2
- Views: 3157
Re: Are These Sets Biholomorphic?
Suppose that \( \displaystyle f : \mathbb{C} \smallsetminus \{ 0 \} \longrightarrow \mathbb{D}^{*} \) is a biholomorphic map. Observe that \( \displaystyle f \) is a holomorphic function (into \( \mathbb{C} \)), bounded on a neighborhood of \( 0 \) - which is a singular point of \( \displaystyle f \...
- Sun Nov 22, 2015 12:34 am
- Forum: Algebraic Geometry
- Topic: On The Spectrum Of A Ring
- Replies: 0
- Views: 1895
On The Spectrum Of A Ring
Let \( A \) be a ring. Show that \( Spec(A) \) is not connected if and only if \( A \) is isomorphic to the product of two non-zero rings \( R \) and \( S \), if and only if \( A \) contains non-trivial idempotents. If \( A \) is an integral domain, then show that \( Spec(A) \) is irreducible. Show...
- Sat Nov 21, 2015 4:48 pm
- Forum: Algebraic Structures
- Topic: On Ring Theory (An Easy One)
- Replies: 8
- Views: 6528
On Ring Theory (An Easy One)
- Let \( \displaystyle \phi : B \longrightarrow A \) be a ring homomorphism. If \( \displaystyle \mathscr{p} \) is a prime ideal of \( A \), then show that \( \phi^{-1}(\mathscr{p}) \) is a prime ideal of \( B \).
- Show that the nilpotents of a ring \( R \) form an ideal.
- Mon Nov 16, 2015 9:20 pm
- Forum: Algebraic Topology
- Topic: On Fundamental Group
- Replies: 0
- Views: 3862
On Fundamental Group
Let \( \displaystyle f : X \longrightarrow Y \) be a continuous map between two topological spaces. Prove that \( \displaystyle f \) induces a mapping \( \displaystyle f_{*} : \pi_{1}(X,x_{0}) \longrightarrow \pi_{1}(Y,f(x_{0})) \) of the fundamental groups, which is a group homomorphism. Additiona...
Re: Integral
Thank you, Apostolos, for your hint! Here is the requested solution: \begin{align*} \int_{0}^{+\infty} \frac{x-1}{ \sqrt{2^{x}-1} \log(2^{x}-1) } \mathrm{d}x &\overset{2^{x}-1=u \, , \, dx = \frac{du}{(\log2)(u+1)}}{\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=} \int_{0}^{+\infty} \frac{ \frac{1}{\...