This is a problem in K. Hulek's Elementary Algebraic Geometry. I figured out that $k[X]$ is the collection of polynomials of the form $f(x) + g(y)$ and also the local ring of an affine line at the point $0$ seems like the form $h(t)/j(t)$ where $j(t)$ does not have $t$ as its factor.
But I can't solve (a). To me the two local rings seem totally different, because $k[X]$ consists of polynomials of two independent variables whereas $\mathbb C[t]$ is one-variable polynomial ring. Could anyone tell me how to solve the problems...
Also, according to (a) the local ring of an affine line at a point is not an integral domain. But the local ring is obtained from $\mathbb C[t]$ and it seems to be identified in $\mathbb C(t)$, which is the field of fractions of $\mathbb C[t]$. Could anyone explain this also?
$k[X]$ and $\mathbf{C}[t]$ are totally different: but the question isn't asking about them, it's asking about their localizations.
The ring $k[X]$ is not a domain, since it has nonzero zero divisors. Among the things inverted when you construct the local ring at $(0,u)$ is the polynomial $y$, which is a zero divisor. Our initial intuition about localization is built up around inverting things that are not zero divisors, and without examples and experience, it is quite faulty in situations where you invert things that are zero divisors.
A good exercise, then, is to go back to the definition of localizing a ring by inverting a multiplicative subset, and work through what that means for in this particular example, paying careful attention to the fact that there is a zero divisor in the multiplicative subset. (and yes, I do mean to review the definition again: there's a subtlety to it that is only relevant when you invert zero divisors, and it's quite likely you've forgotten it)
A different approach is to do a (higher) algebraic calculation: localization and modding out commute:
$$ (S^{-1} R) / (S^{-1}I) = S^{-1} (R/I) $$
Or, in particular,
$$ (\mathbf{C}[x,y] / (xy))[\frac{1}{y}] \cong C[x,y,1/y] / (xy) $$
and you can simplify the ring presented on the right hand side.