The strictly upper triangular "subalgebra" of $M_n(K)$ is not unital?

Question

I am reading Alex Bartel's PDF "Introduction to Representation Theory of Finite Groups" (which I, who am by no means an algebraist, have by the way found very helpful), and in Chapter 2 in which he first discusses algebras over a field, he says that he always consider algebras which contain an identity (with respect to the ring multiplication). But then he gives an example: it is clear that given a field $K$, the ring of $n\times n$ matrices $M_n(K)$ is an algebra over $K$. The author then says that the ring of strictly upper triangular matrices (that is, matrices with zeroes on and below the diagonal) is a subalgebra of $M_n(K)$. But this subalgebra is not unital, since all of its elements are nilpotent, thus contradicting his original definition of algebras as unital.

Am I missing anything here? Or is this just a small mistake of the author's?

Answer 1

You are right on the fact that $UT_n(\mathbb{K})$ don’t have a unit.

A lot of algebras you want to study won’t have a unit element.

So I guess that the author was saying that it’s better to work with unital algebras in the sens that if you work with a non-unital algebra $A$ then try to find a unital algebra $B$ such that $A$ is a subalgebra of $B$.

Answer 2

Strictly speaking he writes:

An algebra over a field K, or just a K-algebra, is a ring that is also a K-vector space, such that the ring multiplication commutes with scalar multiplication. We will always assume that our algebras contain 1.

A subalgebra is a subring that is also a sub-K-vector space.

He only claims the strictly upper triangular matrices are a subalgebra and never an algebra.

So one could argue that he simply has defined subalgebras in such a way that they don't satisfy his definition of "algebra," and that everything is perfectly fine if you don't mind your subring not having identity.

This does not seem like a very intuitive choice, but then again I have not read the whole book :)