# Is the multiplication beetween even numbers an associative algebra?

We were discussing about the possibility of having an algebra over a field which is associative but has not the unity. Does it exist? It has been proposed as a counterexample the set of even numbers. We believe it is wrong because it can't be a vector space. If an associative but not unital algebra exists, how can we write its associative property? We usually write it as a commutative diagram or in terms of maps composition. In these terms the identity seams unavoidable. i.e. $$m\circ\left(m\otimes id\right) = m\circ\left(id\otimes m\right)\qquad associativity$$ with $m : A\otimes A\rightarrow A$

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Try the following: take your field to be R (the real numbers) and take $A$ to be the $R$-algebra consisting of all $R$-valued sequences $(a_n)_{n=1}^\infty$ which satisfy $\lim_n |a_n| =0$. –  Yemon Choi Apr 12 '10 at 18:48
Does this help? en.wikipedia.org/wiki/Associative_algebra Also, you should check spelling/grammar before posting. Especially for a new user, people are likely to discount your question without even reading it if it looks like not much effort was put into writing it. –  Steven Gubkin Apr 12 '10 at 18:49
Also you can have an identity MAP without having an identity ELEMENT. –  Steven Gubkin Apr 12 '10 at 18:50
To simplify Yemon's example a bit: just consider any vector space $V$ over your field, and define the product to be zero, for all pairs of elements of $V$ . This is an associative algebra over the field which does not have an identity –  Mariano Suárez-Alvarez Apr 12 '10 at 19:08
@Mariano - I did consider a drastic example of the type you describe. The one I gave was motivated by the fact that functional analysis has quite a lot of natural examples which do not have identity elements, but which "share many of the nice properties of algebras with identity". (Specifically, $c_0$ has a bounded approximate identity, from which much niceness follows.) –  Yemon Choi Apr 12 '10 at 19:09

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.
Toy example: 6 is the unit for $(2) \subset Z/_{10}$. –  Sammy Black Apr 13 '10 at 6:50