ring of endomorphisms of a commutative ring

30 Views Asked by At

I have a field $E$. In Basic Algebra II by Jacobson, it is stated that

Since $E$ is commutative, $\text{End}_EE = E_E$ where $E_E$ is the set of multiplications $a_E:E\rightarrow E:x\mapsto ax$

But much more than that is true surely?

Indeed, I would like to say that $\text{End}_EE = \{\text{Id}\}$ since if $\phi\in \text{End}_EE$, $\phi(x) = \phi(x\cdot1) = x\phi(1) = x$.

Or is there a mistake here.

1

There are 1 best solutions below

0
On BEST ANSWER

Jacobson probably means endomorphisms of $E$ as an $E$-vector space, not as a ring. In this case, you don't have $\phi(1)=1$.