# Equivalence of Definitions of Unital Associative Commutative Algebra/Correspondence

## Contents

## Theorem

Let $A$ be a commutative ring with unity.

Let $B$ be a algebra over $A$ that is unital, associative and commutative.

Let $(C, f)$ be a ring under $A$.

The following are equivalent:

- $C$ is the underlying ring of $B$ and $f : A \to C$ is the canonical mapping to the unital algebra $B$.
- $B$ is the algebra defined by $f$.

## Proof

Let $\cdot : A \times B \to B$ the ring action of $B$.

### 1 implies 2

Let $C$ equal the underlying ring of $B$ and $f : A \to C$ equal the canonical mapping to the unital algebra $B$.

We show that $B$ is the algebra defined by $f$.

#### Addition

By definition of the underlying ring of $B$, the addition of $C$ is the addition of $B$, say $+$.

By definition of the module defined by $f$, the addition of the algebra defined by $f$ is also $+$.

#### Multiplication

By definition of the underlying ring of $B$, the multiplication of $C$ is the ring product of $B$, say $\times$.

By definition of the algebra defined by $f$, its multiplication is also $\times$.

#### Ring action

It remains to show that the ring action $\cdot$ of $B$ is the ring action $*$ of the module defined by $f$.

We have, for $a \in A$ and $b \in B$:

\(\displaystyle a * b\) | \(=\) | \(\displaystyle f(a) \times b\) | Definition of Module Defined by Ring Homomorphism | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle (a \cdot 1_B) \times b\) | Definition of Canonical Homomorphism from Ring to Unital Algebra | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \cdot (1_B \times b)\) | Definition of Bilinear Mapping | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \cdot b\) | Definition of Unit of Algebra |

$\Box$

### 2 implies 1

Let $B$ equal the algebra defined by $f$.

#### Addition

By definition of the module defined by $f$, the addition of $B$ is the addition of $C$, say $+$.

By definition of the underlying ring of $B$, its addition is also $+$.

#### Multiplication

By definition of the algebra defined by $f$, the multiplication of $B$ is the ring product of $C$, say $\times$.

By definition of the underlying ring of $B$, its multiplication is also $\times$.

Thus $C$ is the underlying ring of $B$.

#### Homomorphism

By Identity is Unique, the unit $1_B$ of $B$ equals the unity $1_C$ of $C$.

Let $g : A \to B$ be the canonical mapping.

We show that $g=f$.

We have, for $a \in A$:

\(\displaystyle g(a)\) | \(=\) | \(\displaystyle a \cdot 1_B\) | Definition of Canonical Homomorphism from Ring to Unital Algebra | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \cdot 1_C\) | $1_B = 1_C$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle f(a) \times 1_C\) | Definition of Module Defined by Ring Homomorphism | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle f(a)\) | Definition of Unity of Ring |

$\blacksquare$