Understanding Units in a Ring

In abstract algebra, a ring is a set equipped with two binary operations, usually called addition and multiplication, satisfying certain axioms. A ring R is said to have unity (or identity) if there exists a multiplicative identity element, often denoted by $1$, such that for any element $a$ in R, $a \cdot 1 = 1 \cdot a = a$.

What is a Unit?

An element $a$ in a ring R with unity is called a unit if there exists an element $b$ in R such that $a \cdot b = b \cdot a = 1$. This element $b$ is called the multiplicative inverse of $a$ and is often denoted as $a^{-1}$. The set of all units in a ring R is often denoted by $R^{\times}$ or $U(R)$.

The Set of Units Forms a Group

Let's examine the set of all units in a ring R with unity under the operation of multiplication inherited from the ring. We need to check if this set satisfies the axioms of a group:

1. Closure: If $a$ and $b$ are units in R, does $a \cdot b$ belong to the set of units? Since $a$ and $b$ are units, their inverses $a^{-1}$ and $b^{-1}$ exist in R. Consider the element $b^{-1} \cdot a^{-1}$. We have $(a \cdot b) \cdot (b^{-1} \cdot a^{-1}) = a \cdot (b \cdot b^{-1}) \cdot a^{-1} = a \cdot 1 \cdot a^{-1} = a \cdot a^{-1} = 1$. Similarly, $(b^{-1} \cdot a^{-1}) \cdot (a \cdot b) = b^{-1} \cdot (a^{-1} \cdot a) \cdot b = b^{-1} \cdot 1 \cdot b = b^{-1} \cdot b = 1$. Since $a \cdot b$ has a multiplicative inverse $b^{-1} \cdot a^{-1}$ in R, $a \cdot b$ is a unit. Thus, the set of units is closed under multiplication.
2. Associativity: The multiplication operation in a ring is associative, i.e., for any $a, b, c$ in R, $(a \cdot b) \cdot c = a \cdot (b \cdot c)$. Since the set of units is a subset of R, the multiplication is associative for elements in the set of units as well.
3. Identity Element: The unity element $1$ of the ring R is always a unit because $1 \cdot 1 = 1$. So, $1$ is in the set of units. Also, for any unit $a$, $a \cdot 1 = 1 \cdot a = a$. Thus, $1$ serves as the identity element for multiplication in the set of units.
4. Inverse Element: By definition, every unit $a$ has a multiplicative inverse $a^{-1}$ in R. We need to check if this $a^{-1}$ is also a unit. Since $a \cdot a^{-1} = a^{-1} \cdot a = 1$, $a^{-1}$ has $a$ as its multiplicative inverse. Therefore, $a^{-1}$ is also a unit and belongs to the set of units.

Since the set of units in a ring with unity satisfies all four group axioms under the operation of multiplication, it forms a group with respect to multiplication.

Analyzing the Options

• a ring with unity: The set of units is generally not closed under addition (e.g., in the ring of integers $\mathbb{Z}$, $1$ and $-1$ are units, but $1 + (-1) = 0$, which is not a unit). Thus, it does not form a ring.
• a group with respect to multiplication: As shown above, the set of units satisfies the group axioms under multiplication.
• a field: A field is a commutative ring where every non-zero element is a unit. The set of units is only the set of invertible elements under multiplication. It does not involve the addition operation required for a field.
• an integral domain: An integral domain is a commutative ring with unity and no zero divisors. The set of units is a multiplicative group and does not encompass the entire structure of an integral domain.

Therefore, the set of all units in a ring R with unity forms a group with respect to multiplication.