Q: Let K be a field and let $A=K[X^3,X^{11}]\subset K[X]$. Show that A is an Noetherian Domain.
My proof:
$[X^3,X^{11}] \simeq K[X,Y]/(x^{11}-y^3)$
K is a field so K is noetherian. From Hilbert's base theorem we get K[X] is also noetherian and, by using it again, (K[X])[Y]=K[X,Y] is noetherian as well.
$x^{11}-y^3$ is a polynomial of degree 3 over K[X] with no roots in K[X], thus it is irreducible over K[X]. Therefore $x^{11}-y^3$ is irreducible over K[X,Y]. This implies that $(x^{11}-y^{3}$ is a prime and finitely generated (1) ideal in K[X,Y] and such $K[X,Y]/(x^{11}-y^3)$ is both a domain and is noetherian.
PS: now trying to figure out how I can find A's Krull Dimension from here; could use a hint
Your ring is a subring of a domain, so it is clearly a domain. Your ring is manifestly finitely generated over a field, so it is noetherian. There is no need to have an explicit description of the ring to conclude these things.
On the other hand, since your ring $A$ is contained in $K[X]$, its fraction field $F$ is contained in the fraction field of $K[X]$, which is the field of rational functions $K(X)$. Since $X^3$ and $X^{11}$ are in $A$ and $X=(X^3)^4/X^{11}$ is an element of $F$, we see that in fact $F=K(X)$.
Now the Krull dimension of a domain which is an algebra over a field $K$ is equal to the transcendence degree of its fraction field over $K$, so we see at once that $\dim A=1$.