8.1: Definitions and Examples
Recall that a group is a set together with a single binary operation, which together satisfy a few modest properties. Loosely speaking, a ring is a set together with two binary operations (called addition and multiplication) that are related via a distributive property.
A ring \(R\) is a set together with two binary operations \(+\) and \(\cdot\) (called addition and multiplication , respectively) satisfying the following:
- \((R,+)\) is an abelian group.
- \(\cdot\) is associative: \((a\cdot b)\cdot c=a\cdot (b\cdot c)\) for all \(a,b,c\in R\) .
- The distributive property holds: \(a\cdot (b+c)=(a\cdot b)+(a\cdot c)\) and \((a+b)\cdot c = (a\cdot c)+(b\cdot c)\) for all \(a,b,c\in R\) .
We make a couple comments about notation.
- We often write \(ab\) in place of \(a\cdot b\) .
- The additive inverse of the ring element \(a\in R\) is denoted \(-a\) .
If \(R\) is a ring, then for all \(a,b\in R\) :
- \(0a=a0=0\)
- \((-a)b=a(-b)=-(ab)\)
- \((-a)(-b)=ab\)
A ring \(R\) is called commutative if multiplication is commutative.
A ring \(R\) is said to have an identity (or called a ring with 1 ) if there is an element \(1\in R\) such that \(1a=a 1=a\) for all \(a\in R\) .
Justify that \(\mathbb{Z}\) is a commutative ring with 1 under the usual operations of addition and multiplication. Which elements have multiplicative inverses in \(\mathbb{Z}\) ?
Justify that \(\mathbb{Z}_n\) is a commutative ring with 1 under addition and multiplication mod \(n\) .
[prob:Z10Ring] Consider the set \(\mathbb{Z}_{10}=\{0,1,2,3,4,5,6,7,8,9\}\) . Which elements have multiplicative inverses in \(\mathbb{Z}_{10}\) ?
For each of the following, find a positive integer \(n\) such that the ring \(\mathbb{Z}_n\) does not have the stated property.
- \(a^2=a\) implies \(a=0\) or \(a=1\) .
- \(ab=0\) implies \(a=0\) or \(b=0\) .
- \(ab=ac\) and \(a\neq 0\) imply \(b=c\) .
If \(R\) is a ring with 1, then the multiplicative identity is unique and \(-a=(-1)a\) .
Requiring \((R,+)\) to be a group is fairly natural, but why require \((R,+)\) to be abelian? Suppose \(R\) has a 1. Compute \((1+1)(a+b)\) in two different ways.
A ring \(R\) with 1 (with \(1\neq 0\) ) is called a division ring if every nonzero element in \(R\) has a multiplicative inverse: if \(a\in R\setminus\{0\}\) , then there exists \(b\in R\) such that \(ab=ba=1\) .
A commutative division ring is called a field .
A nonzero element \(a\) in a ring \(R\) is called a zero divisor if there is a nonzero element \(b\in R\) such that either \(ab=0\) or \(ba=0\) .
Are there any zero divisors in \(\mathbb{Z}_{10}\) ? If so, find all of them.
Are there any zero divisors in \(\mathbb{Z}_5\) ? If so, find all of them.
Provide an example of a ring \(R\) and elements \(a,b\in R\) such that \(ax=b\) has more than one solution. How does this compare with groups?
Assume \(a,b,c\in R\) such that \(a\) is not a zero divisor. If \(ab=ac\) , then either \(a=0\) or \(b=c\) .
Assume \(R\) is a ring with \(1\) with \(1\neq 0\) . An element \(u\in R\) is called a unit in \(R\) if \(u\) has a multiplicative inverse (i.e., there exists \(v\in R\) such that \(uv=vu=1\) ). The set of units in \(R\) is denoted \(U(R)\) .
Consider the ring \(\mathbb{Z}_{20}\) .
- Find \(U(\mathbb{Z}_{20})\) .
- Find the zero divisors of \(\mathbb{Z}_{20}\) .
- Any observations?
If \(U(R)\neq\emptyset\) , then \(U(R)\) forms a group under multiplication.
We make a few observations.
- A field is a commutative ring \(F\) with identity \(1\neq 0\) in which every nonzero element is a unit, i.e., \(U(F)=F\setminus\{0\}\) .
- Zero divisors can never be units.
- Fields never have zero divisors.
A commutative ring with identity \(1\neq 0\) is called an integral domain if it has no zero divisors.
The Cancellation Law (Theorem \(\PageIndex{3}\)) holds in integral domains for any three elements.
Any finite integral domain is a field.
Here are a few examples. Details left as an exercise.
- Zero Ring : If \(R=\{0\}\) , we can turn \(R\) into a ring in the obvious way. The zero ring is a finite commutative ring with 1. It is the only ring where the additive and multiplicative identities are equal. The zero ring is not a division ring, not a field, and not an integral domain.
- Trivial Ring: Given any abelian group \(R\) , we can turn \(R\) into a ring by defining multiplication via \(ab=0\) for all \(a,b\in R\) . Trivial rings are commutative rings in which every nonzero element is a zero divisor. Hence a trivial ring is not a division ring, not a field, and not a integral domain.
- The integers form an integral domain, but \(\mathbb{Z}\) is not a division ring, and hence not a field.
- The rational numbers \(\mathbb{Q}\) , the real numbers \(\mathbb{R}\) , and the complex numbers \(\mathbb{C}\) are fields under the usual operations of addition and multiplication.
- The group of units \(U(\mathbb{Z}_n)\) is the set of elements in \(\mathbb{Z}_n\) that are relatively prime to \(n\) . That is, \(U(\mathbb{Z}_n)=U_n\) . All other nonzero elements are zero divisors. It turns out that \(\mathbb{Z}_n\) forms a finite field if and only if \(n\) is prime.
- The set of even integers \(2\mathbb{Z}\) forms a commutative ring under the usual operations of addition and multiplication. However, \(2\mathbb{Z}\) does not have a 1, and hence cannot be a division ring nor a field. Moreover, if you look closely at the definition of integral domain, you’ll see that \(2\mathbb{Z}\) is also not an integral domain since \(2\mathbb{Z}\) does not contain a multiplicative identity.
- Polynomial Ring: Fix a commutative ring \(R\) . Let \(R[x]\) denote the set of polynomials in the variable \(x\) with coefficients in \(R\) . Then \(R[x]\) is a commutative ring. Moreover, \(R[x]\) is a ring with 1 if and only if \(R\) is a ring with 1. The units of \(R[x]\) are exactly the units of \(R\) (if there are any). So, \(R[x]\) is never a division ring nor a field. However, if \(R\) is an integral domain, then so is \(R[x]\) .
- Matrix Ring: Fix a ring \(R\) and let \(n\) be a positive integer. Let \(M_n(R)\) be the set of \(n\times n\) matrices with entries from \(R\) . Then \(M_n(R)\) forms a ring under ordinary matrix addition and multiplication. If \(R\) is nontrivial and \(n\geq 2\) , then \(M_n(R)\) always has zero divisors and \(M_n(R)\) is not commutative even if \(R\) is. If \(R\) has a 1, then the matrix with 1’s down the diagonal and 0’s elsewhere is the multiplicative identity in \(M_n(R)\) . In this case, the group of units is the set of invertible \(n\times n\) matrices, denoted \(GL_n(R)\) and called the general linear group of degree \(n\) over \(R\) .
- Quadratic Field: Define \(\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}\) . It turns out that \(\mathbb{Q}(\sqrt{2})\) is a field. In fact, we can replace 2 with any rational number that is not a perfect square in \(\mathbb{Q}\) .
- Hamilton Quaternions: Define \(\mathbb{H}=\{a+bi+cj+dk\mid a,b,c,d\in\mathbb{R}, i,j,k\in Q_8\}\) Then \(\mathbb{H}\) forms a ring, where addition is definite componentwise in \(i\) , \(j\) , and \(k\) and multiplication is defined by expanding products and the simplifying using the relations of \(Q_8\) . It turns out that \(\mathbb{H}\) is a non-commutative ring with 1.
Find an example of a ring \(R\) and an element \(a\in R\setminus\{0\}\) such that \(a\) is neither a zero divisor nor a unit.
A subring of a ring \(R\) is a subgroup of \(R\) under addition that is also closed under multiplication.
The property “is a subring" is clearly transitive. To show that a subset \(S\) of a ring \(R\) is a subring, it suffices to show that \(S\neq \emptyset\) , \(S\) is closed under subtraction, and \(S\) is closed under multiplication.
Here are a few quick examples.
- \(\mathbb{Z}\) is a subring of \(\mathbb{Q}\) , which is a subring of \(\mathbb{R}\) , which in turn is a subring of \(\mathbb{C}\) .
- \(2\mathbb{Z}\) is a subring of \(\mathbb{Z}\) .
- The set \(\mathbb{Z}(\sqrt{2})=\{a+b\sqrt{2}\mid a,b\in\mathbb{Z}\}\) is a subring of \(\mathbb{Q}(\sqrt{2})\) .
- The ring \(R\) is a subring of \(R[x]\) if we identify \(R\) with set of constant functions.
- The set of polynomials with zero constant term in \(R[x]\) is a subring of \(R[x]\) .
- \(\mathbb{Z}[x]\) is a subring of \(\mathbb{Q}[x]\) .
- \(\mathbb{Z}_n\) is not a subring of \(\mathbb{Z}\) as the operations are different.
Consider the ring \(\mathbb{Z}_{10}\) from Problem \(\PageIndex{3}\). Let \(S=\{0,2,4,6,8\}\) .
- Argue that \(S\) is a subring of \(\mathbb{Z}_{10}\) .
- Is \(S\) a ring with 1? If so, find the multiplicative identity. If not, explain why.
- Is \(S\) a field? Justify your answer.
Suppose \(R\) is a ring and let \(a\in R\) . Define \(S=\{x\in R\mid ax=0\}\) . Prove that \(S\) is a subring of \(R\) .
Consider the ring \(\mathbb{Z}\) . It turns out that \(2\mathbb{Z}\) and \(3\mathbb{Z}\) are subrings (but you don’t need to prove this). Determine whether \(2\mathbb{Z}\cup 3\mathbb{Z}\) is a subring of \(\mathbb{Z}\) . Justify your answer.