Why $k[[x^2,x^3]]$ is dimension 1 obvious?

Question

This is related to example 4.4.4. of Weibel, Homological Algebra.

Consider $k[[x^2,x^3]]\subset k[[x]]$. The book says it is 1-dimensional. It seems obvious but I could not be convinced so. I would have to invoke some non-trivial results though not too hard to check.

$\textbf{Q:}$ Why $k[[x^2,x^3]]$ is of 1 dimension? I would consider $k[x,y]/(x^3-y^2)\to k[t]$ by $x\to t^2,y\to t^3$ map. Identify the image. Now complete on both sides at $(0,0)$ to identify the image. Since completion does not change ideal containment and in particular primes, I see 1 dimensional. How would I prove directly by elements if $(0)\subset P\subset (x^2,x^3)\subset k[[x^2,y^3]]$, then prime ideal $P$ must be either $(0)$ or $(x^2,x^3)$?

Answer 1

Let $f\in k[[x^2,x^3]]$ be nonzero and let $n$ be the smallest power of $x$ appearing in $f$. Note then that $f$ divides $x^{n+2}$ in $k[[x^2,x^3]]$: the usual procedure to divide by solving for the coefficients one by one will not require a linear term. So if $P\subset k[[x^2,x^3]]$ is a nonzero prime ideal, applying this to some nonzero element of $P$ gives that $x^{n+2}\in P$. Writing $x^{n+2}$ as a product of $x^2$'s and $x^3$'s this implies either $x^2$ or $x^3$ is in $P$, and then the other must be as well since $x^6$ is.

Answer 2

Set $f(Y):=Y^2-x^2$. Hence, $f(Y)$ is a monic polynomial in the polynomial ring $k[[x^2,x^3]][Y]$(the polynomial ring over $k[[x^2,x^3]]$). Now since $f(x)=0$, $k[[x]]$ is integral over $k[[x^2,x^3]]$. Thus, $\dim(k[[x^2,x^3]])=\dim(k[[x]])$. On the other hand, since $k[[x]]$ is a PID (it is a discrete valuation ring), it is 1 dimensional. Hence, $\dim(k[[x^2,x^3]])=\dim(k[[x]])=1$.