Equivalence of Definitions of Unital Associative Commutative Algebra/Correspondence

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 $A$.


 * 1) $C$ is the underlying ring of $B$ and $f : A \to C$ is the canonical mapping to the unital algebra $B$.
 * 2) $B$ 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 $b \in B$:

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 $a \in A$: