Let $K$ be a field and $A$ be an affine $K$-algebra ($K$ is then identified in $A$), i.e. there exists a surjective ring homomorphism $\phi: K[X_1, \ldots, X_n] \rightarrow A$ which "fixes" $K$ in the sense that it identifies $K$ in the same way that it's identified in $A$.
I am dealing with the particular case where $A=K[[x]]$, the formal power series ring. In the context of a proof by contradiction, it was assumed that $A$ is an affine $K$-algebra, and it is claimed then that the field of fractions of $A$, which is the field of formal Laurent series $L$, is also an affine $K$-algebra. I can't see why this is the case. It was suggested by notation that $L=A[x^{-1}]$, but I don't see how this could mean that $L$ is simply the polynomial ring of $A$ having indeterminate $x^{-1}$, because $x^{-1}$ interacts with the elements of $A$, it's not just a "free" indeterminate.
Hence the general question. I'm more interested in why it's true in the particular case I mentioned, in case it isn't true in the general case.
Thank you for your time.
The commutative ring with unity $R$ is a Jacobson Ring if for every proper ideal $I\subset R$, we have $$ \sqrt{I}=\bigcap_{m \,\text{maximal}\\ \,\,\,\,I\subset m}m. $$
The only maximal ideal of $K[[x]]$ is $(x)$, the ideal generated by $x$. So this ring is not Jacobson, because $$ (0)=\sqrt{0}\subsetneq(x)=\bigcap_{m \,\text{maximal}\\ \,\,(0)\subset m}m. $$
As every affine $K$-algebra is Jacobson, $K[[x]]$ cannot be an affine $K$-algebra.
See Lemma 1.1 and exercice 1.2 of Kemper's book A Course in Commutative Algebra.