If $B$ is a f.g. module over a ring $A\subset R$, and $B\subset R$, then is $B$ a ring? Here $R$ is a commutative ring with identity.
If $B$ is a f.g. module over a ring $A\subset R$, and $B\subset R$, then is $B$ a ring?
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abstract-algebra
commutative-algebra
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0Do you mean: is $B$ stable under the multiplication of $R$ ? Is the addition of $B$ assumed to be the same as the addition of $R$ ? – 2017-01-19
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1@Watson I don't think $B$ should be a $R$-module, but I think the addition of $B$ is the same as the addition of $R$. – 2017-01-19
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4$\mathbb{Z} \subseteq \mathbb{Z} + \mathbb{Z} \sqrt[3]{2} \subseteq \mathbb{Z}\left[\sqrt[3]2\right]$. – 2017-01-19
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3If the module $B$ is given by $B=\mathbb{Z} + \mathbb{Z} \sqrt[3]{2}$, then B is not a ring -- it's not closed under multiplication. For example, $(\sqrt[3]{2})^2$ is not in $B$. – 2017-01-19