Is $\mathbb{C}[x,y] / (xy)$ a local ring?

Question

Is $\mathbb{C}[x,y] / (xy)$ a local ring? Also, is its localization at $(x,y)$ isomorphic to itself? I was thinking that the localization at $(x,y)$ is isomorphic to itself and the maximal ideal is $(x,y)$. Any feedback would be appreciated.

What about the ideal $(x, y-1)$? Or more generally, ideals of the form $(x - a, y)$ and $(x, y - b)$? — Viktor Vaughn, Mar 24 '20 at 19:57
$(x,y)$ is one of the maximal ideals. All ideals $(x-r,y)$ and $(x,y-r)$, for $r\in\mathbb{C}$ are also maximal. — , Mar 24 '20 at 19:58
If we localize at $(x,y)$, then $\mathbb{C}[x,y] / (xy)$ becomes a local ring, no? — user100101212, Mar 24 '20 at 19:59
@user100101212 It's well known that localizing at any prime ideal ($(x,y)$ would qualify) results in a local ring. — rschwieb, Mar 24 '20 at 20:00
@user100101212 Yes. Open any book on commutative algebra. Have a great day. — rschwieb, Mar 24 '20 at 20:01
@user100101212 https://math.stackexchange.com/questions/300446/why-is-the-localization-at-a-prime-ideal-a-local-ring — Viktor Vaughn, Mar 24 '20 at 20:02

score 4 · Answer 1 · answered Mar 24 '20 at 19:59

No, $\mathbb{C}[x,y]/(xy)$ is not local. Indeed, there are many maximal ideals in this ring, of the form $(x,y-\alpha)$ and $(x-\beta,y)$ for $\alpha,\beta \in \mathbb{C}$. Indeed, $(xy)\subsetneq (x,y-\alpha)$ for instance because $x\mid xy$ while $x\not \in (xy)$.

You can verify that $(x,y-\alpha)$ is maximal by calculating that the quotient by this ideal is $\mathbb{C}$ and similarly for $(x-\beta, y)$. You should not believe that this ring is local because the vanishing locus of $xy$ in $\mathbb{C}^2$ is the union of the axes, which contains many points $(0,\alpha)$ and $(\beta,0)$, which correspond to the family of maximal ideals proposed above.

Because this ring is not local, it cannot be isomorphic to any of its localizations, so it is not isomorphic to its localization at $(x,y)$.

It's worth pointing out that the maximal spectrum of this ring is the union of two lines. — WhatsUp, Mar 24 '20 at 20:01

score 3 · Answer 2 · answered Mar 24 '20 at 20:08

Clearly, $x$ is not a unit, because it is a zero divisor. Also, $1-x$ is not a unit either, because it is sent to zero under the ring homomorphism $\mathbb{C}[x,y]/(xy) \to \mathbb{C}$ that sends $x$ to $1$ and $y$ to $0$. Hence, $\mathbb{C}[x,y]/(xy)$ is not a local ring, because there are two nonunits whose sum is a unit.

score 1 · Answer 3 · answered Mar 25 '20 at 12:08

If you want to think about this geometrically (perhaps a bit early, but still):

To see what $\mathbb C[x,y]/(xy)$ corresponds to geometrically, consider what are all the zeroes of the equation $xy=0$. This is zero exactly when $x=0$ or $y=0$.

Thus it has many points $p$, corresponding to maximal ideals $\mathfrak m_p$, each with its own local ring $\simeq \mathbb C[x,y]/\mathfrak m_p$.

Is $\mathbb{C}[x,y] / (xy)$ a local ring?

3 Answers3