The coface map on Amitsur's complex is an isomorphism.

54 Views Asked by At

Let $S$ be a commutative algebra over a commutative ring $K$. Put $S^0=K$ and $S^n=S\otimes_K\cdots\otimes_K S$ ($n$ factors). In the Basic Algebra volume II, Jacbson says that the map $\delta^i: \otimes_{i=1}^nS\to\otimes_{i=1}^{n+1}S$ given by $$x_1\otimes \cdots x_n\mapsto x_1\otimes \cdots \otimes x_{i-1}\otimes 1\otimes x_i\otimes\cdots x_n$$ is an algebra isomorphism. (exercise 3 section 6.2.)

Is it really true? It seems to define a homomorphism, but are $S$ and $S\otimes_K S$ really isomorphic?

For example. Let's see $\mathbb{R}$ as an algebra over $\mathbb{Q}$. I don't see how $(\pi\otimes 1)\in\operatorname{Im}\delta^1:\mathbb{R}\to \mathbb{R}\otimes_\mathbb{Q} \mathbb{R}$.

1

There are 1 best solutions below

0
On BEST ANSWER

The confusion arises from Jacobson's outdated terminology. He says "isomorphism into", and by this he means an injection, i.e. he identifies $S^n$ as a subalgebra of $S^{n+1}$ via that map.