Equivalent Conditions On A Ring
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Equivalent Conditions On A Ring
Show that an associative ring with unity is a division ring if and only if the only left (right) ideals it contains are the trivial ones.
What happens if we additionally suppose that the given ring is commutative?
What happens if we additionally suppose that the given ring is commutative?
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Re: Equivalent Conditions On A Ring
Let \(\displaystyle{\left(R,+,\cdot\right)}\) be an associative ring with unity \(\displaystyle{1}\) .
Suppose that this ring is a division ring. The subsets \(\displaystyle{\left\{0\right\}\,,R}\) are left ideals of \(\displaystyle{\left(R,+,\cdot\right)}\).
Let \(\displaystyle{I}\) be a left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) such that \(\displaystyle{I\neq \left\{0\right\}}\) .
We'' ll prove that \(\displaystyle{I=R}\) . Since \(\displaystyle{I\neq \left\{0\right\}}\), there is \(\displaystyle{r\in R}\) such that
\(\displaystyle{r\in I}\) and \(\displaystyle{r\neq 0}\). Due to the fact that the ring \(\displaystyle{\left(R,+,\cdot\right)}\) is a division ring,
there is \(\displaystyle{r^{-1}\in R}\) and satisfies the relation : \(\displaystyle{r\cdot r^{-1}=1=r^{-1}\cdot r}\) . We have that :
\(\displaystyle{r^{-1}\cdot r=1\in I}\) cause \(\displaystyle{r^{-1}\in R\,,r\in I}\) and \(\displaystyle{I}\) is a left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) .
So, if \(\displaystyle{x\in R}\), then \(\displaystyle{x=x\cdot 1\in I}\) for the same reason and then: \(\displaystyle{R\subseteq I\implies I=R}\) .
On the other hand, suppose that the only left ideals it contains are the trivial ones, \(\displaystyle{\left\{0\right\}\,,R}\) .
It is sufficient to prove that every non-zero element of \(\displaystyle{R}\) is invertible.
For this reason, let \(\displaystyle{x\in R-\left\{0\right\}}\) and consider the main left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) produced
by \(\displaystyle{x\,\,,I=R\,x=\left\{r\cdot x: r\in R\right\}}\). According to the hypothesis, \(\displaystyle{I=\left\{0\right\}}\) or \(\displaystyle{I=R}\) .
If \(\displaystyle{I=\left\{0\right\}}\), then : \(\displaystyle{x\in I\implies x\in\left\{0\right\}\implies x=0}\), which is not true, so \(\displaystyle{I=R}\) an then :
\(\displaystyle{1\in R\implies 1\in I\implies \exists\,y\in R: y\cdot x=1}\). Also, for the main left ideal produced by
\(\displaystyle{y\,,J=R\,y=\left\{r\cdot y\in R: r\in R\right\}}\), holds : \(\displaystyle{J=R}\) cause \(\displaystyle{y\neq 0}\). So,
\(\displaystyle{1\in R\implies 1\in J\implies \exists\,r\in R: 1=r\cdot y}\) and then :
\(\displaystyle{1=r\cdot y\implies 1\cdot x=\left(r\cdot y\right)\cdot x\implies x=r\cdot\left(y\cdot x\right)\implies x=r\cdot 1\implies x=r}\), so
\(\displaystyle{x\cdot y=1=y\cdot x}\) which means that the element \(\displaystyle{x}\) is invertible.
If we additionally suppose that the given ring is commutative, then we have that :
The ring \(\displaystyle{\left(R,+,\cdot\right)}\) is a field if, and only if, the only ideals it contains are the trivial ones.
Suppose that this ring is a division ring. The subsets \(\displaystyle{\left\{0\right\}\,,R}\) are left ideals of \(\displaystyle{\left(R,+,\cdot\right)}\).
Let \(\displaystyle{I}\) be a left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) such that \(\displaystyle{I\neq \left\{0\right\}}\) .
We'' ll prove that \(\displaystyle{I=R}\) . Since \(\displaystyle{I\neq \left\{0\right\}}\), there is \(\displaystyle{r\in R}\) such that
\(\displaystyle{r\in I}\) and \(\displaystyle{r\neq 0}\). Due to the fact that the ring \(\displaystyle{\left(R,+,\cdot\right)}\) is a division ring,
there is \(\displaystyle{r^{-1}\in R}\) and satisfies the relation : \(\displaystyle{r\cdot r^{-1}=1=r^{-1}\cdot r}\) . We have that :
\(\displaystyle{r^{-1}\cdot r=1\in I}\) cause \(\displaystyle{r^{-1}\in R\,,r\in I}\) and \(\displaystyle{I}\) is a left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) .
So, if \(\displaystyle{x\in R}\), then \(\displaystyle{x=x\cdot 1\in I}\) for the same reason and then: \(\displaystyle{R\subseteq I\implies I=R}\) .
On the other hand, suppose that the only left ideals it contains are the trivial ones, \(\displaystyle{\left\{0\right\}\,,R}\) .
It is sufficient to prove that every non-zero element of \(\displaystyle{R}\) is invertible.
For this reason, let \(\displaystyle{x\in R-\left\{0\right\}}\) and consider the main left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) produced
by \(\displaystyle{x\,\,,I=R\,x=\left\{r\cdot x: r\in R\right\}}\). According to the hypothesis, \(\displaystyle{I=\left\{0\right\}}\) or \(\displaystyle{I=R}\) .
If \(\displaystyle{I=\left\{0\right\}}\), then : \(\displaystyle{x\in I\implies x\in\left\{0\right\}\implies x=0}\), which is not true, so \(\displaystyle{I=R}\) an then :
\(\displaystyle{1\in R\implies 1\in I\implies \exists\,y\in R: y\cdot x=1}\). Also, for the main left ideal produced by
\(\displaystyle{y\,,J=R\,y=\left\{r\cdot y\in R: r\in R\right\}}\), holds : \(\displaystyle{J=R}\) cause \(\displaystyle{y\neq 0}\). So,
\(\displaystyle{1\in R\implies 1\in J\implies \exists\,r\in R: 1=r\cdot y}\) and then :
\(\displaystyle{1=r\cdot y\implies 1\cdot x=\left(r\cdot y\right)\cdot x\implies x=r\cdot\left(y\cdot x\right)\implies x=r\cdot 1\implies x=r}\), so
\(\displaystyle{x\cdot y=1=y\cdot x}\) which means that the element \(\displaystyle{x}\) is invertible.
If we additionally suppose that the given ring is commutative, then we have that :
The ring \(\displaystyle{\left(R,+,\cdot\right)}\) is a field if, and only if, the only ideals it contains are the trivial ones.
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