Subrings and Ring Extensions

# Subrings and Ring Extensions

 Definition: Let $(R, +, \cdot)$ be a ring and let $S \subseteq R$. Then $(S, +, \cdot)$ is said to be a Subring of $R$ is $S$ is a ring with respect to the $+$ and $\cdot$ defined on $R$ and $(R, +, \cdot)$ is said to be a Ring Extension of $(S, +, \cdot)$.

Recall that the set of real numbers with respect to standard addition $+$ and standard multiplication $\cdot$ formed a ring $(\mathbb{R}, +, \cdot)$. Furthermore, we noted that the set of complex number with respect to the same standard addition and standard multiplication formed a ring $(\mathbb{C}, +, \cdot)$. We know that $\mathbb{R} \subset \mathbb{C}$ since every real number is a complex number of the form $a + 0i$ where $a \in \mathbb{R}$. Therefore, $(\mathbb{R}, +, \cdot)$ is a subring of $(\mathbb{C}, +, \cdot)$.

The following proposition gives us criterion for when a subset of a ring is a subring.

 Proposition 1: Let $(R, +, \cdot)$ be a ring and let $S \subseteq R$. Then $(S, +, \cdot)$ is a subring of $(R, +, \cdot)$ if: 1. $S$ is closed under $+$. 2. For every element $a \in S$ we have that $-a \in S$. 3. $S$ is closed under $\cdot$. 4. $1 \in S$.
• Proof: Let $(R, +, \cdot)$ be a ring and let $S \subseteq R$. To show that $(S, +, \cdot)$ is a subring of $(R, +, \cdot)$ we only need to verify the remaining ring axioms.
• Let $a, b, c \in S$. We must have that associativity holds for $+$ on elements in $S$. If we had that $a + (b + c) \neq (a + b) + c$ then this inequality would hold in $R$ as well, implying that $+$ is not associative on $R$, so $R$ would not be a ring which is a contradiction.
• Furthermore, the identity $0$ for $+$ must be contained in $S$ because for all $a \in S$ we have that $-a \in S$ and since $S$ is closed under addition that means that $(a + (-a)) = 0 \in S$.
• We must also have the commutativity holds for $+$ on elements in $S$. If we had that $a + b \neq b + c$ then this inequality would hold in $R$ as well, implying that $+$ is not commutative on $R$, so $R$ would not be a ring which is a contradiction.
• Furthermore, associativity holds for $\cdot$ on elements in $S$ because if we have that $a \cdot b \neq b \cdot a$ then this inequality would hold in $R$ as well, implying that $\cdot$ is not associative on $R$, so $R$ would not be a ring which is a contradiction.
• We are given that $1 \in S$, so $S$ contains the multiplicative identity.
• Lastly, we must have that distributivity holds for the same reasons as associativity holds for $+$ and $\cdot$ and commutativity holds for $+$ since $a \cdot (b + c) \neq (a \cdot b) + (a \cdot c)$ would imply this inequality exists for elements in $R$ contradicting the fact that $R$ is a ring.
• Therefore all of the ring axioms hold and so $(S, +, \cdot)$ is a subring of $(R, +, \cdot)$. $\blacksquare$