AliReza Olfati
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Registered User
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Everyone thinks of changing the world, but no one thinks of changing himself. "Tolstoy"
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Apr 30 |
awarded | ● Yearling |
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Nov 28 |
accepted | Relative extremely disconnected space |
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Nov 27 |
comment |
Ideals generated by idempotent elements Dear all, I think if $R$ is an integral domain, in the ring $M_2(R)$ your statement is trivially true. because in this case, we could characterize all of idempotents. all of idempotents have the form $0$, $I_2$ or a matrix whose entries are $a_{11}=a$, $a_{12}=b$ , $a_{21}=c$, $a_{22}=1-a$ and $a,b,c$ satisfy in the relation $bc=a-a^2$. so in this case the only nonzero ideal in $M_2(R)$ which contains an idempotent is the total ring. so in this case, your claim is true. |
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Nov 26 |
comment |
Relative extremely disconnected space Dear Alex. Please check it more precisely. You could not change the intervals in x-axis to them. please look at the basis of $(0,0)$ and $(1,0)$. each of neighborhoods of $(0,0)$ geometrically should contains the rectangular rigion which has the fixed length equal to $\frac{1}{2}$. also each of neighborhoods of $(1,0)$ contains the rectangular rigion which has the fixed length equal to $\frac{1}{2}$. so you could not change them in your favor. |
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Nov 26 |
comment |
Idempotent elements in matrix ring Hi.Let $F$ be an infinite field, then there exist infinitely many distinct pair $(I,J)$ of minimal left ideals of $M_2(F)$ such that $M_2(F)=I \oplus J$. so this shows that in this case $F$ has only two trivial idempotent, But $M_2(F)$ has infinitely many nontrivial idempotent. you could find this point at exercise [11.b] page 443 of Hubgerford. So I think you should determine the property of your ring$R$ which entries of matrix ring come from it. |
