Problem taken from Jacobson's Basic Algebra, 2.1.4
PROBLEM
Let $I$ be the set of complex numbers of the form $m+n\sqrt{-3}$ where either $m, n$ is an integer or both $m$ and $n$ are halves of odd integers. Show that $I$ is a subring of $\mathbb{C}$.
PROOF
An arbitrary element of $I$ has the form $m/2+n/2\sqrt{-3}$ for some integers $m$ and $n$. First note that $1=2/2+0/2\sqrt{-3}\in I$. Next note that $$m_1/2+n_1/2\sqrt{-3}-m_2/2-n_2/2\sqrt{-3}=(m_1-m_2)/2+(n_1-n_2)\sqrt{-3}\in I$$
so $I$ is closed under addition and subtraction. Finally $$(m_1/2+(n_1/2)\sqrt{-3})(m_2/2-(n_2/2)\sqrt{-3})=((m_1m_2-3n_1n_2)/4+((m_1n_2-n_1m_2)/4)\sqrt{-3}\in I$$
Notice that $m_1m_2-3n_1n_2$ and $m_1n_2-n_1m_2$ are always even integers, $I$ Is closed under multiplication and hence $I$ is a subring of the complex numbers.
PROBLEM
Let $I$ be the set of complex numbers of the form $m+n\sqrt{-3}$ where either $m, n$ is an integer or both $m$ and $n$ are halves of odd integers. Show that $I$ is a subring of $\mathbb{C}$.
PROOF
An arbitrary element of $I$ has the form $m/2+n/2\sqrt{-3}$ for some integers $m$ and $n$. First note that $1=2/2+0/2\sqrt{-3}\in I$. Next note that $$m_1/2+n_1/2\sqrt{-3}-m_2/2-n_2/2\sqrt{-3}=(m_1-m_2)/2+(n_1-n_2)\sqrt{-3}\in I$$
so $I$ is closed under addition and subtraction. Finally $$(m_1/2+(n_1/2)\sqrt{-3})(m_2/2-(n_2/2)\sqrt{-3})=((m_1m_2-3n_1n_2)/4+((m_1n_2-n_1m_2)/4)\sqrt{-3}\in I$$
Notice that $m_1m_2-3n_1n_2$ and $m_1n_2-n_1m_2$ are always even integers, $I$ Is closed under multiplication and hence $I$ is a subring of the complex numbers.
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