Subrings and Ring Extensions

Subrings and Ring Extensions

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

Recall that the set of real numbers with respect to standard addition $+$ and standard multiplication $*$ formed a ring $(\mathbb{R}, +, *)$. Furthermore, we noted that the set of complex number with respect to the same standard addition and standard multiplication formed a ring $(\mathbb{C}, +, *)$. 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, $(\mathbb{R}, +, *)$ is a subring of $(\mathbb{C}, +, *)$.

We also noticed that the set of rational numbers $\mathbb{Q}$ under the same standard addition and multiplication formed a ring. Furthermore, $\mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$ so $(\mathbb{Q}, +, *)$ is a subring of $(\mathbb{R}, +, *)$ and $(\mathbb{Q}, +, *)$ is also a subring of $(\mathbb{C}, +, *)$.

Of course, if $(R, +, *)$ is a ring, then not all subsets $S$ of $R$ need to form a ring under the same operations $+$ and $*$. For example, consider the set of all $n \times n$ matrices without the $n \times n$ zero matrix, that is $M_{nn} \setminus \{ 0_{n\times n} \}$, and let $+$ be defined as standard matrix addition and $*$ be defined as standard square matrix multiplication. We've already seen that $(M_{nn}, +, *)$ forms a ring, however, the identity for $+$ was namely the $n \times n$ zero matrix $0_{n \times n}$. Therefore the set $M_{nn} \setminus \{ 0 \times 0 \}$ cannot possibly be a ring as there cannot be a matrix $A \in M_{nn}$ that also satisfies the property of being an identity for $+$.

Of course, if $S \subseteq R$ and $(R, +, *)$ is a ring, then we can always check to see whether $S$ is a ring with respect to $+$ and $*$ by verifying all of the ring axioms for $S$. As we will see in the following lemma - we actually do not need to do all of that work.

Lemma 1: Let $(R, +, *)$ be a ring and let $S \subseteq R$. Then $(S, +, *)$ is a subring of $(R, +, *)$ if:
1. $S$ is closed under $+$.
2. For every element $a \in S$ we have that $-a \in S$.
3. $S$ is closed under $*$.
4. $1 \in S$.

Lemma 1 tells us that to check if $S \subseteq R$ is a subring of $R$, all we need to do is check four out of the original ring axioms.

  • Proof: Let $(R, +, *)$ be a ring and let $S \subseteq R$. To show that $(S, +, *)$ is a subring of $(R, +, *)$ 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 $*$ on elements in $S$ because if we have that $a * b \neq b * a$ 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.
  • 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 $*$ and commutativity holds for $+$ since $a * (b + c) \neq (a * b) + (a * c)$ 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 $(S, +, *)$ is a subring of $(R, +, *)$. $\blacksquare$
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