The set of even numbers is a non-unital ring, in particular it has an associative multiplication, but you are right that it isn't an algebra over a field. For an example of a non-unital algebra, consider the continuous functions $\mathbb{R} \to \mathbb{R}$ which vanish at a particular point, under pointwise product. More generally, any proper two-sided ideal of an algebra is a non-unital algebra, just as any proper two-sided ideal of a ring is a non-unital ring.

You may think that these examples are "unnatural," so here is a "natural" one: the algebra of compactly supported continuous functions $\mathbb{R} \to \mathbb{R}$ under convolution.

For an example of a non-associative algebra, take, for example, the octonions. Your confusion arises because of the following issue: given a vector space $V$ and a bilinear operation $V \times V \to V$ we can associate to any $a \in V$ the linear operator $L_a$ which is left multiplication by $a$, and these linear operators form an associative algebra. However, composition of the operators $L_a$ need not be the same as $\times$: associativity is equivalent to the statement that $L_a L_b = L_{a \times b}$, which need not be the case in general. I think this is the source of your confusion.

The set of even numbers is a non-unital ring, in particular it has an associative multiplication, but you are right that it isn't an algebra over a field. For an example of a non-unital algebra, consider the continuous functions $\mathbb{R} \to \mathbb{R}$ which vanish at a particular point, under pointwise product. More generally, any proper two-sided ideal of an algebra is a non-unital algebra, just as any proper two-sided ideal of a ring is a non-unital ring.

You may think that these examples are "unnatural," so here is a "natural" one: the algebra of compactly supported continuous functions $\mathbb{R} \to \mathbb{R}$ under convolution.

For an example of a non-associative algebra, take, for example, the octonions. Your confusion arises because of the following issue: given a vector space $V$ and a bilinear operation $V \times V \to V$ we can associate to any $a \in V$ the linear operator $L_a$ which is left multiplication by $a$, and these linear operators form an associative algebra. However, composition of the operators $L_a$ need not be the same as $\times$: associativity is equivalent to the statement that $L_a L_b = L_{a \times b}$, which need not be the case in general. I think this is the source of your confusion.

The set of even numbers is a non-unital ring, in particular it has an associative multiplication, but you are right that it isn't an algebra. For an example of a non-unital algebra, consider the set of continuous functions $\mathbb{R} \to \mathbb{R}$ which vanish at a particular point. More generally, any proper ideal of an algebra is a non-unital algebra.

For an example of a non-associative algebra, take, for example, the octonions. Your confusion arises because of the following issue: given a vector space $V$ and a bilinear operation $V \times V \to V$ we can associate to any $a \in V$ the linear operator $L_a$ which is multiplication by $a$, and these linear operators form an associative algebra. However, composition of the operators $L_a$ need not be the same as $\times$: associativity is equivalent to the statement that $L_a L_b = L_{a \times b}$, which need not be the case in general. I think this is the source of your confusion.

The set of even numbers is a non-unital ring, in particular it has an associative multiplication, but you are right that it isn't an algebra over a field. For an example of a non-unital algebra, consider the continuous functions $\mathbb{R} \to \mathbb{R}$ which vanish at a particular point, under pointwise product. More generally, any proper two-sided ideal of an algebra is a non-unital algebra, just as any proper two-sided ideal of a ring is a non-unital ring.

You may think that these examples are "unnatural," so here is a "natural" one: the algebra of compactly supported continuous functions $\mathbb{R} \to \mathbb{R}$ under convolution.

For an example of a non-associative algebra, take, for example, the octonions. Your confusion arises because of the following issue: given a vector space $V$ and a bilinear operation $V \times V \to V$ we can associate to any $a \in V$ the linear operator $L_a$ which is left multiplication by $a$, and these linear operators form an associative algebra. However, composition of the operators $L_a$ need not be the same as $\times$: associativity is equivalent to the statement that $L_a L_b = L_{a \times b}$, which need not be the case in general. I think this is the source of your confusion.