Subalgebra

Remarks

Clause 3 with $k = -1$ gives $-b \in B$, so $B$ is a subring of $A$ and a submodule of the $K$-module $A$. With the operations of $A$, the set $B$ is then a $K$-algebra. Let $\eta \colon K \to A$, $k \mapsto k1$, be the structure map. A subset $B$ is a subalgebra exactly when it is a subring of $A$ that contains $\eta(K)$, because $kb = \eta(k)b$.

Example: in the $\R$-algebra $M_2(\R)$ of real $2 \times 2$ matrices, the set of upper triangular matrices is a subalgebra. The identity matrix is upper triangular, and sums and real multiples of upper triangular matrices are upper triangular. For upper triangular $P$ and $Q$, the $(2,1)$ entry of $PQ$ is $p_{21} q_{11} + p_{22} q_{21} = 0$, since $p_{21} = q_{21} = 0$.

Non-example: in the $\R$-algebra $\R[x]$, the set $\Z[x]$ of polynomials with integer coefficients satisfies clauses 1 and 2 but not clause 3. For $k = 1/2$ and $b = 1$ the product $kb = 1/2$ is not in $\Z[x]$. So $\Z[x]$ is a subring of $\R[x]$ but not an $\R$-subalgebra.

Some authors work with algebras without a unit element and drop clause 1. For a field $K$, a subalgebra is a linear subspace of $A$ that contains $1$ and is closed under multiplication.