What does $\mathbb{C}[f]$ mean?

Question

Let $f\colon E \to E$ be an endomorphism of stable bundles over a variety $X$ over $\mathbb{C}$.

What does the notation $\mathbb{C}[f]$ mean? (This is supposed to be a field if $f$ is an isomorphism.)

Reference: Le Potier's book on Vector Bundles.

Answer 1

On page 75, the author says

Consider the commutative algebra $\mathbb{C}[f]$ generated by $f$.

This answers your question, then. We know that $f \in \text{End}(E)$, where $\text{End}(E)$ is an endomorphism ring. So we can look at the smallest subring containing $f$ and closed under scalar multiplication.

Why is $\mathbb{C}[f]$ the notation for this? Well think about what an element of "the ring generated by $f$" should look like! Certainly $f$ should be there. But it's closed under multiplication, so we should also have $f^2$,$f^3$, etc. It's closed under scalar multiplication and addition, so actually we should get all polynomials in $f$. But it's reasonable to refer to "polynomials in $f$" by the notation $\mathbb{C}[f]$.

Now, since we're talking about coherent sheaves, we know our modules are finite dimensional. Then if $f$ is an isomorphism, actually $f^{-1}$ is a polynomial in $f$ (basically since $P(f) = 0$ by the cayley-hamilton theorem, so if $f$ is invertible we can rearrange this to get $f^{-1} = Q(f)$). Thus when $f$ is invertible, $f^{-1} \in \mathbb{C}[f]$ so that actually $\mathbb{C}[f]$ is a field.

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I hope this helps ^_^