Detecting whether some rings are Dedekind domains

Question

Consider the three rings $\mathbb{C}[x,y] / \langle x^4 + xy -1\rangle$, $\mathbb{Z}[x,y] / \langle x^4 + xy -1\rangle$ and $\mathbb{F}_2[x,y] /\langle x^4- y^3 \rangle$. I am supposed to detect whether these are Dedekind domains or not.

However I've no idea how to do this. I know that a Dedekind domain is normal, noetherian, and of dimension $1$.

So I can at least see that each of these is noetherian, because they are quotients of noetherian rings. But I have no idea how to verify the other things, or unverify them. Are there any standard methods or tricks to work this out? I would really appreciate any help on this, thank you.

Answer 1

The trickiest part is probably proving that $f(x,y) = x^4 + xy - 1$ is irreducible over $\mathbb{C}$, and in particular over $\mathbb{Z}$. I'll leave this to you, but ask me if you have trouble.

Thus $A = \mathbb{C}[x,y]/(f)$ is a domain. You already saw that $A$ is noetherian. Now the prime ideal $(f)$ has height one (because it is principal) and $\mathbb{C}[x,y]$ is two-dimensional, so $A$ is one-dimensional. Geometrically, $A$ is the algebra of polynomial functions on the curve $X = \{ f = 0 \} \subset \mathbb{C}^2$. As for normality, take the partial derivatives $f_x(x,y) = 4x^3 + y$ and $f_y(x,y) = x$. The set $X \cap \{ f_x = f_y = 0 \}$ is empty, which means $X$ is nonsingular. This implies that $A$ is normal.

As for $B = \mathbb{Z}[x,y]/(f)$, again $(f)$ has height one, but $\mathbb{Z}[x,y]$ is three-dimensional, so $B$ is two-dimensional and in particular cannot be a Dedekind domain.

Finally, $C = \mathbb{F}_2[x,y]/(x^4 - y^3)$ is not normal because $y/x$, which lies in the fraction field of $C$ but not $C$ itself, satisfies the monic polynomial $t^3 - x$.

Answer 2

As said in the comments, $\mathbb{Z}[x,y]/(f)$ cannot have dimension $1$ so cannot be a Dedekind domain.

For any field $k$ and any nonzero polynomial $f(x,y) \in k[x,y]$, the quotient $k[x,y]/(f)$ is the coordinate ring of the plane algebraic curve $f(x,y) = 0$. This ring is a domain iff $f$ is irreducible. It is always be Noetherian and one-dimensional. It is a Dedekind domain iff it is nonsingular (nonsingular $\implies$ normal $\implies$ nonsingular in codimension one).

Assuming that the ground field $k$ is perfect, nonsingular is equivalent to smooth, and the latter is much easier to detect. The way to do it goes back essentially to multivariable calculus: the curve defined by $f(x,y) = 0$ is smooth at a point $P = (x,y)$ (such that $f(P) = 0$!) iff at least one of $\partial f / \partial x, \partial f / \partial y$ is nonzero at $P$.

Thus for instance in your third example both partial derivatives vanish at $P = (0,0)$. What about the first example?