“I made a lot of mistakes out of the ring, but I never made any in it.”
―Jack Johnson
Definition8.1.Ring.
A ring is a set \(R\) equipped with two binary operations, \(+\) and \(\cdot\text{,}\) satisfying:
\((R,+)\) is an abelian group with identity element denoted \(0\text{,}\)
Associative Multiplication.
\(\cdot\) is associative (making \((R,\cdot)\) a monoid)
Distributive Property.
\(a \cdot (b + c) = a \cdot b + a \cdot c\) and \((a + b) \cdot c = a \cdot c + b \cdot c\) hold for all \(a,b,c \in R\text{.}\)
\(R\) is a unital ring (or a ring with identity) if, in addition to (1), (2), (3)
there is a multiplicative identity element written as \(1\) such that \(1 \cdot a = a = a \cdot 1\) for all \(a \in R\text{.}\)
\(R\) is commutative if in addition to (1)–(3)
\(a \cdot b = b \cdot a\) holds for all \(a,b \in R\text{.}\)
\(R\) is a division ring if \(1\neq 0\text{,}\) (1)–(4) and (6) hold
\((R \sm \{0\})\) is a group under \(\cdot\) (i.e. every \(r\in (R - \{0\})\) has a multiplicative inverse)
\(R\) is a field if \(1\neq 0\) and (1)–(6) hold (i.e. a field is a commutative division ring).
Discussion8.1.
One could argue that commutative rings should really be called abelian rings to remain consistent with abelian groups. Discuss.
Here are some basic consequences of the axioms.
Proposition8.2.Ring Arithmetic.
For any ring \(R\) and all \(a,b \in R\) we have:
\(a \cdot 0 = 0 = 0 \cdot a\text{,}\)
\((-a)b = -(ab) = a(-b)\text{,}\)
\((-a)(-b) = ab\text{.}\)
For a unital ring \(R\) and all \(a\in R\) we have:
\(1\) is unique, and
\((-1)a = -a\text{.}\)
Proof.
Let \(R\) be a ring with identity.
Let \(a\in R\text{.}\) Observe that \(0a=(0+0)a=0a+0a\) by the distributive law. Subtracting \(0a\) from both sides we have \(0=0a\text{.}\) A similar argument shows that \(0=a0\text{.}\)
Let \(a\in R\text{.}\) Note that \(-a+a=0\text{.}\) Consider \((-1)a+a\text{.}\) Using the distributive law we see that \((-1)a+a=a(-1+1)=a(0)=0\text{.}\) Thus \(-a\) and \((-1)a\) are both additive inverses of \(a,\) making them unique.
Phew. Let’s take a breather and see some examples.
Example8.3.Relevant Rings.
\(R=\{0\}\) is called the trivial ring, or zero ring. 2
Personally, I am in favor of calling this the bor-ring.
\(\mathbb{Z}\) is a commutative ring.
\(\mathbb{Z}/n\) is a commutative ring under addition and multiplication modulo \(n\text{.}\)
The familiar sets \(\mathbb{Q}, {\mathbb{R}}, \mathbb{C}\) are fields.
The set \(\Z[i] = \{a + bi \mid a,b \in \Z\}\) is a commutative ring known as the Gaussian integers.
If \(R\) is any ring (not necessarily commutative), so is \(\operatorname{M}_{n}(R)\) for any natural number \(n\text{,}\) using the usual rules for addition and multiplication of square matrices.
The Cartesian product \(R\times R'\) of two rings \(R\) and \(R'\) has a natural ring structure with addition and multiplication defined componentwise:
Let \(R\) be a ring. Prove that \(R=\{0\}\) if and only if \(1=0\text{.}\)
Solution.
Notice that in the trivial ring \(0=1\text{.}\) Conversely, if \(1 = 0\) in a ring, then \(R = \{0\}\text{,}\) since in this case for all \(a\text{,}\) we have \(a \cdot 0 = 0\) and hence \(a = a \cdot 1 = a \cdot 0 = 0\text{.}\)
Exercise8.5.Direct Product Preservations.
Let \(R\) and \(S\) be rings. The Cartesian product \(R\times S\) is commutative if and only if \(R\) and \(S\) are commutative. Similarly, \(R\times S\) has identity if and only if both \(R\) and \(S\) do as well.
Exercise8.6.Less Relevant Rings.
Opposite Day.
Given a ring \(R\text{,}\) let \(R^{op}\) denote the “opposite ring’’. This is the same underlying set as \(R\) equipped with the same rule for \(+\) as \(R\text{,}\) but with multiplication rule (which I will write here as \(\cdot_{op}\)) redefined to be \(a \cdot_{op} b := b a\) (where \(ba\) refers to the original multiplication rule for \(R\)). Then \(R^{op}\) is also a ring
Function Ring.
If \(X\) is a set and \(R\) is a ring, let \(\operatorname{Fun}(X, R)\) be the collection of set theoretic functions from \(X\) to \(R\text{,}\) and define \((f + g)(x) = f(x) + g(x)\) and \((f \cdot g)(x) := f(x) \cdot g(x)\text{.}\) Then \(\operatorname{Fun}(X,R)\) is a ring. If \(X\) is a finite set and \(|X|=n\text{,}\) then \(\operatorname{Fun}(X,R)\) may be identified with \(R^n=\underbrace{R \times \cdots \times R}_n\text{,}\) the direct product of \(n\) copies of \(R\text{.}\)
Endomorphism Ring.
If \(A = (A, +)\) is any abelian group, set \(\operatorname{End}_{Ab}(A)\) to be the collection of endomorphisms of \(A\) — that is, the set of group homomorphisms \(f: A \to A\) from \(A\) to itself.
Then \(\operatorname{End}_{Ab}(A)\) is a ring with addition \((f + g)(a) := f(a) + g(a)\) and multiplication \(f \cdot g := f \circ g\text{.}\) This is almost always a non-commutative ring.
Here is a nice generalization that will prove usful quite a few times down the road.
Theorem8.7.Binomial Theorem for Commutative Rings.
For any commutative ring \(R\) and any elements \(a\) and \(b\) in \(R\text{,}\) we have:
where \({n\choose i}\) is the binomial coefficient, defined as: \({n\choose i} = \frac{n!}{i!(n-i)!}\) with integers \(n\) and \(i\) such that \(0\le i\le n\text{.}\)
SubsectionSubrings
“We all live in a yellow subma-ring”
―(Ring)o Starr
We will spend the first half of our exploration of rings following the same general structure as we did in Group Theory. First, we looked at rings in general, and now we will examine subrings, just as we did with subgroups. Then it’s off to homomorphisms, isomorphisms, and quotients. It’ll be just like old times.
Definition8.8.Subring.
A subring of a ring \(R\) is a subset \(S \subseteq R\) such that \(S\) is a ring under the operations of \(R\text{.}\)
When \(R\) is a field we call \(S\) a subfield of \(R\) 3
We generally do not define subdomains, subcommutativerings, or subdivisionringswithidentity, but if you’re feeling inspired I say go for it.
.
Lemma8.9.Subring Test.
A nonempty subset \(S\) of a ring \(R\) is a subring if and only if either one of the following hold:
\(S\) is a subgroup of \(R\) closed under multiplication.
\(S\) is closed under subtraction and multiplication.
Exercise8.10.Subring Preservations.
Any subring of a commutative ring is a commutative ring. Any unital subring of an integral domain is an integral domain.
Example8.11.Examples of Subrings.
\(\Z\) is a subring of \(\Q\text{,}\) which is a subring of \(\R\text{,}\) which is a subring of \(\C\text{.}\)
\(n\Z\) is a subring without \(1\) of the ring \(\Z\) with \(1\text{.}\)
The set of continuous functions mapping \([0,1]\to \R\) is a subring of \(\Fun([0,1],\R)\text{,}\) denoted \(\cC([0,1])\text{.}\)
\(\Z[i]\text{,}\) the ring of Gaussian Integers, is a subring of \(\C\text{.}\)
Exploration8.2.Nilradical Radishes.
Prove that the set of all nilpotent elements of a ring \(R\) is a subring of \(R\text{,}\) which we call the nilradical.
Exercise8.12.Fancy Subrings.
If \(R\) is a ring and \(S\) is a subring of \(R\text{,}\) it can happen that
\(R\) is unital but \(S\) is not (e.g. \(S=2\Z\subset R= \Z\))
\(S\) is unital but \(R\) is not
both \(R\) and \(S\) are unital but \(1_R\neq 1_S\)
Find examples for each of these situations!
Definition8.13.Center of a Ring.
The center of a ring \(R\) is the set
\begin{equation*}
Z(R)=\{z \in R \mid zr = rz \text{ for all }r \in R\}.
\end{equation*}
Lemma8.14.Center is a Subring.
The center \(Z(R)\) is a subring of \(R\text{.}\) If \(R\) is a ring with identity \(1\) then \(Z(R)\) is a subring that contains the same \(1\text{.}\)
Centers of rings don’t show up as much in ring theory as they did with groups, but that might also be because there hasn’t been a non-commutative ring spotted within twenty miles of UNL since 1977.
Lemma8.15.\(\Q(\sqrt {d})\).
Let \(d\) be a squarefree integer (that is, the prime factorization of \(d\) has no repeated primes). Then the subset
of \(\C\) is a subring that is a field (called a quadratic field), and \(\Z[\sqrt{d}] = \{a + b\sqrt{d} | a,b \in \Z\}\) is a subring of \(\Q(\sqrt{d})\text{.}\)
Proof.
Both \(\Q(\sqrt {d})\) and \(\Z[\sqrt{d}]\) are closed under subtraction and multiplication, so they are subrings of \(\C\text{.}\)
The fact that \(\Q(\sqrt {d})\) is a {} follows since \(\Q(\sqrt {d})\) is also closed under taking inverses. Indeed the inverse of \(r + q \sqrt{d}\) (from \(\C\)) turns out to be
whenever \(r + q \sqrt{d}\neq 0\text{.}\) A slightly subtle point here is that the fraction above makes sense since \(r^2 -d q^2 \ne 0\) provided \(r\) and \(q\) are not simultaneously \(0\text{.}\) This is because, if \(r^2 -d q^2 = 0\) then either \(d=(r/q)^2\text{,}\) which contradicts the assumption that \(d\) is squarefree, or \(r=q=0\text{,}\) which contradicts the assumption \(r + q \sqrt{d}\neq 0\text{.}\)
Remark8.16.
Note the difference in notation between \(\Q(\sqrt d)\) and \(\Z[\sqrt d]\text{:}\) one uses parenthesis and the other brackets. This is to denote that one (the one with parentheses) is a field, whereas the other [the one with brackets] is not. This is also some spicy foreshadowing that won’t pay off until all the way down in Field Theory. Stay tuned.
