Ideal of a ring
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Ideal of a ring
Let \(\displaystyle{\left(R,+,\cdot\right)}\) be an associative ring with unity \(\displaystyle{1=1_{R}}\) . Prove that the subset \[\displaystyle{J\,(R)=\left\{x\in R: 1-a\,x\,b\in U\,(R)\,,\forall\,a\,,b\in R\right\}}\] of \(\displaystyle{R}\) is an ideal of the ring \(\displaystyle{\left(R,+,\cdot\right)}\), where \(\displaystyle{U\,(R)}\) is the set of all the invertible elements of \(\displaystyle{\left(R,+,\cdot\right)}\) .
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Re: Ideal of a ring
Here is a solution.
If \(\displaystyle{a\,,b\in R}\), then \(\displaystyle{1-a\cdot 0\cdot b=1\in U(R)}\), so \(\displaystyle{J(R)\neq \varnothing}\) since \(\displaystyle{0\in J(R)}\) .
Let \(\displaystyle{x\,,y\in J(R)}\) and \(\displaystyle{a\,,b\in U(R)}\) . Since \(\displaystyle{x\,,y\in J(R)}\), we have that
\(\displaystyle{1-a\,x\,b=w}\) and \(\displaystyle{1-\left(w^{-1}\,a\right)\,y\,b=t}\) for some \(\displaystyle{w\,,t\in U(R)}\). Then,
\(\displaystyle{w-a\,y\,b=w\,t}\) and \(\displaystyle{1-a\,(x+y)\,b=1-a\,x\,b-a\,y\,b=w-a\,y\,b=w\,t\in U(R)}\) because \(\displaystyle{w\,,t\in U(R)}\) and the
pair \(\displaystyle{\left(U(R),\cdot\right)}\) is a group, which means that \(\displaystyle{x+y\in J(R)}\) .
Therefore, we get \(\displaystyle{J(R)\leq \left(R,+\right)}\). Let now \(\displaystyle{x\in J(R)}\) and \(\displaystyle{r\in R}\) .
\(\displaystyle{\forall\,a\,,b\in R: 1-a\,(r\cdot x)\,b=1-(a\,r)\cdot x\,b\in U(R)\,\,,1-a\,(x\cdot r)\,b=1-a\,x\,(r\cdot b)\in U(R)}\)
since \(\displaystyle{x\in J(R)}\), so : \(\displaystyle{r\cdot x\,,x\cdot r\in J(R)}\) and finally, we have that \(\displaystyle{J(R)}\) is an ideal of the ring \(\displaystyle{\left(R,+,\cdot\right)}\) .
Application
Consider the ring \(\displaystyle{\left(R,+,\cdot\right)=\left(\mathbb{Z},+,\cdot\right)}\) . It's known that \(\displaystyle{U(\mathbb{Z})=\left\{-1,1\right\}}\).
Let \(\displaystyle{x\in J(\mathbb{Z})}\). For \(\displaystyle{a=b=1}\) we have that :
\(\displaystyle{1-x\in U(\mathbb{Z})\implies 1-x\in\left\{-1,1\right\}\implies x\in\left\{0,2\right\}}\).
Obviously, \(\displaystyle{0\in J(\mathbb{Z})}\) , but for \(\displaystyle{a=1\,,b=2\in\mathbb{Z}}\) holds :
\(\displaystyle{1-1\cdot 2\cdot 2=1-4=-3\notin U(\mathbb{Z})}\) .
So, \(\displaystyle{J(\mathbb{Z})=\bigcap_{p\in \mathbb{P}}p\,\mathbb{Z}=\left\{0\right\}}\), where \(\displaystyle{\mathbb{P}}\) is the set of all
the prime numbers and \(\displaystyle{p\,\mathbb{Z}}\) is a maximal ideal of \(\displaystyle{\left(\mathbb{Z},+,\cdot\right)}\) .
Check here : Jacobson and maximal ideals.
If \(\displaystyle{a\,,b\in R}\), then \(\displaystyle{1-a\cdot 0\cdot b=1\in U(R)}\), so \(\displaystyle{J(R)\neq \varnothing}\) since \(\displaystyle{0\in J(R)}\) .
Let \(\displaystyle{x\,,y\in J(R)}\) and \(\displaystyle{a\,,b\in U(R)}\) . Since \(\displaystyle{x\,,y\in J(R)}\), we have that
\(\displaystyle{1-a\,x\,b=w}\) and \(\displaystyle{1-\left(w^{-1}\,a\right)\,y\,b=t}\) for some \(\displaystyle{w\,,t\in U(R)}\). Then,
\(\displaystyle{w-a\,y\,b=w\,t}\) and \(\displaystyle{1-a\,(x+y)\,b=1-a\,x\,b-a\,y\,b=w-a\,y\,b=w\,t\in U(R)}\) because \(\displaystyle{w\,,t\in U(R)}\) and the
pair \(\displaystyle{\left(U(R),\cdot\right)}\) is a group, which means that \(\displaystyle{x+y\in J(R)}\) .
Therefore, we get \(\displaystyle{J(R)\leq \left(R,+\right)}\). Let now \(\displaystyle{x\in J(R)}\) and \(\displaystyle{r\in R}\) .
\(\displaystyle{\forall\,a\,,b\in R: 1-a\,(r\cdot x)\,b=1-(a\,r)\cdot x\,b\in U(R)\,\,,1-a\,(x\cdot r)\,b=1-a\,x\,(r\cdot b)\in U(R)}\)
since \(\displaystyle{x\in J(R)}\), so : \(\displaystyle{r\cdot x\,,x\cdot r\in J(R)}\) and finally, we have that \(\displaystyle{J(R)}\) is an ideal of the ring \(\displaystyle{\left(R,+,\cdot\right)}\) .
Application
Consider the ring \(\displaystyle{\left(R,+,\cdot\right)=\left(\mathbb{Z},+,\cdot\right)}\) . It's known that \(\displaystyle{U(\mathbb{Z})=\left\{-1,1\right\}}\).
Let \(\displaystyle{x\in J(\mathbb{Z})}\). For \(\displaystyle{a=b=1}\) we have that :
\(\displaystyle{1-x\in U(\mathbb{Z})\implies 1-x\in\left\{-1,1\right\}\implies x\in\left\{0,2\right\}}\).
Obviously, \(\displaystyle{0\in J(\mathbb{Z})}\) , but for \(\displaystyle{a=1\,,b=2\in\mathbb{Z}}\) holds :
\(\displaystyle{1-1\cdot 2\cdot 2=1-4=-3\notin U(\mathbb{Z})}\) .
So, \(\displaystyle{J(\mathbb{Z})=\bigcap_{p\in \mathbb{P}}p\,\mathbb{Z}=\left\{0\right\}}\), where \(\displaystyle{\mathbb{P}}\) is the set of all
the prime numbers and \(\displaystyle{p\,\mathbb{Z}}\) is a maximal ideal of \(\displaystyle{\left(\mathbb{Z},+,\cdot\right)}\) .
Check here : Jacobson and maximal ideals.
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