I am confused about how to show whether a ring is normal or not. For example, consider the $k$-algebra $k[x,y] /\langle x^2 - y^3 \rangle$, which is a domain. How do I show it is not normal? Are there any standard techniques? I know that I want to show it is not integrally closed in its fraction field, but I can't work out how to do this, i.e. find some $z$ in the fraction field, integral over the ring but is not in the ring.

Also how do I work out what the normalisation actually is? Sorry I have no working to offer because I am completely stumped.

  • 5
    $\begingroup$ $y^3=x^2$ so $(x/y)^2=y$, i.e., $x/y$ works. $\endgroup$ – wxu May 21 '12 at 17:09
  • $\begingroup$ OK, that's brilliant! Thanks. what about calculating the actual normalisation though? $\endgroup$ – Paul Slevin May 21 '12 at 17:12

As wxu explained in a comment, $x/y$ is integral over your ring, which therefore cannot be normal. To compute the normalization, observe that the map $x \mapsto t^2$, $y \mapsto t^3$ sends your algebra isomorphically onto $k[t^2,t^3] = k \oplus (t^2) \subset k[t]$. This subalgebra has codimension one, whence $k[t]$ is its normalization.

If you know any algebraic geometry, this is the algebra of polynomial functions on the cuspidal cubic curve $X = \{ (x,y) \in \mathbb{A}^2 \ | \ y^2 = x^3 \}$, pictured here. A curve is nonsingular if and only if its algebra of functions is normal, and it is geometrically obvious that this curve is singular at the origin. The normalization map corresponds to the parameterization $\mathbb{A}^1 \to X$ given by $t \mapsto (t^2,t^3)$.

  • $\begingroup$ Sorry about the long delay in an upvote/reponse. I understand that my ring is isomorphic to $k[t^2, t^3]$, but I am not sure why $k[t^2, t^3] = k \oplus (t^2)$. What do you mean by codimension of a subalgebra - I understand what it means for prime ideals - and why does codimension 1 imply that $k[t]$ is the normalisation? $\endgroup$ – Paul Slevin Jun 1 '12 at 22:34
  • 2
    $\begingroup$ For any $n \geq 2$, there exist $k,\ell \in \mathbb{Z}$ such that $2k+3\ell = n$, which explains why $k[t^2,t^3] = k \oplus (t^2)$. The codimension of a subalgebra is the codimension as a $k$-vector subspace, just as with ideals. Since $k[t]$ is a normal algebra containing $k[t^2,t^3]$ and there are no algebras strictly in between, $k[t]$ must be the normalization (look at the universal property). $\endgroup$ – Justin Campbell Jun 2 '12 at 1:08

There is an easy way to tell where the ring $k[x,y] / f(x,y)$ is not regular: it is precisely at the ideal $$ \langle f(x,y), \partial_x f(x,y), \partial_y f(x,y) \rangle$$ This remains true in more variables and more equations -- take all first-derivatives of all equations. The geometric interpretation is that this ideal describes the singular set of your system of equations.

Since your ring is one-dimensional, being non-regular is synonymous with being non-normal.

Applying the test and assuming we aren't in characteristic 2 or 3, we get the ideal $$ \langle x^2 - y^3, 2x, 3y^2 \rangle = \langle x, y^2 \rangle$$ so the ring is not normal. The normalization will involve adjoining fractions $u/v$ of things in this ideal. Alas, I haven't gone through this procedure often enough to be able to do it systematically, so I can't offer advice on that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.