I am studying localization right now from the lecture notes given by our instructor. In this notes he describes an example of localization what he told us yesterday in the class. Here's this $:$

Let $\mathcal k$ be a field. Let $A= \mathcal k [X_1,X_2, \cdots, X_n].$ Let $\mathfrak p$ be a prime ideal of $A.$ Consider the variety $V$ of $\mathfrak p$ defined by $$V := \{ (a_1,a_2,\cdots,a_n) \in \mathcal {k^n} : f(a_1,a_2, \cdots , a_n) = 0\ \forall f \in \mathfrak p \}.$$ Let $S = A \setminus \mathfrak p.$ Then clearly $S$ is a multiplicatively closed set. Let $A_{\mathfrak p}$ denote the localization of $A$ w.r.t. the multiplicatively closed set $S$ (which we usually denote by $S^{-1} A.$ Our instructor gave $A_{\mathfrak p}$ a name which is "$A$ localized at $\mathfrak p$"). Then $$A_{\mathfrak p} = \left \{\frac f g : g \notin \mathfrak p \right \}.$$

Now what is meant by $g \notin \mathfrak p$? What can I say about $g$ in terms of $V$? Our instructor told us that $g \notin \mathfrak p \implies$ $g$ is nowhere vanishing on $V.$ But how do I say that? If $g \in \mathfrak p$ then it is clear that $g$ will vanish at all points of $V.$ But what will happen if $g \notin \mathfrak p$? Can $g$ then vanish at all or some points of $V$? I am very confused at this stage and I got stuck here. Can somebody please help me to clear my confusion? Then it will be really very helpful for me.

Thank you very much for your valuable time.

  • $\begingroup$ There has to be a mistake somewhere. Take $K[X_1, X_2]$ and $\mathfrak{p}$ the ideal generated by $X_1$ (which is clearly prime). Then $X_2 \notin \mathfrak{p}$ but $X_2$ vanishes at $(0,0) \in V$. $\endgroup$ May 2, 2019 at 8:28
  • $\begingroup$ So according to your statement @Sebastian Schoennenbeck it comes out that if $g \notin \mathfrak p$ then $g$ may not be necessarily non vanishing at every point of $V.$ But if $g \notin \mathfrak p$ then can we conclude that $g$ doesn't vanish at some points of $V$? $\endgroup$
    – little o
    May 2, 2019 at 8:37

1 Answer 1


Your question contains many questions.

Now what is meant by $g\notin\mathfrak{p}$?

In general, the localization $S^{-1}R$ of a commutative ring $R$ at a multiplicatively closed subset $S\subset R$ consists of fractions of the form $\tfrac rs$ with $r\in R$ and $s\in S$. In this case $R=A$ and $S=A\setminus\mathfrak{p}$, so $s\in S$ is equivalent to $s\notin\mathfrak{p}$. So indeed $$A_{\mathfrak{p}} =S^{-1}A =\left\{\frac fg:\ f\in R,\ g\in S\right\} =\left\{\frac fg:\ f,g\in A,\ g\notin\mathfrak{p}\ \right\}.$$

What can I say about $g$ in terms of $V$?

Unfortunately, not much in general. A simple classical example is given by $k=\Bbb{R}$ and $n=1$. The ring $A=\Bbb{R}[X]$ has the two prime ideals $\mathfrak{p}_1=(X^2+1)$ and $\mathfrak{p}_2=(X^2+2)$, and the corresponding varieties are both empty, i.e. $V_1=V_2=\varnothing$. We have $$X^2+2\notin\mathfrak{p}_1,\ X^2+2\in\mathfrak{p}_2, \qquad\text{ and }\qquad X^2+1\in\mathfrak{p}_1,\ X^2+1\notin\mathfrak{p}_2,$$ but there is clearly no way to tell from the corresponding varieties because $V_1=V_2$.

But what will happen if $g\notin\mathfrak{p}$? Can $g$ then vanish at all or some points of $V$?

The example given in the comments illustrates that for some $\mathfrak{p}$ there exist $g\notin\mathfrak{p}$ that vanish on all of $V$. Indeed the example above also illustrates this, albeit vacuously; every $g\in\Bbb{R}[X]$ vanishes on $V_1$ and $V_2$ because $V_1=V_2=\varnothing$.

However, this does not provide a counterexample to the claim that

Our instructor told us that $g\notin\mathfrak{p}\ \implies\ g$ is nowhere vanishing on $V$.

The example in the comments does give a counterexample; take $n=2$ and $\mathfrak{p}=(X_1)\subset k[X_1,X_2]$. Then $V=\{(0,x):\ x\in k\}$ and $g=X_2\notin\mathfrak{p}$, but $g(0,0)=0$, so $g$ is not nowhere vanishing on $V$.

Even worse; if we take $k=\Bbb{R}$ again and $\mathfrak{p}=(X_1^2+X_2^2)\subset\Bbb{R}[X_1,X_2]$ then $V=\{(0,0)\}$, and $g=X_1\notin\mathfrak{p}$ but $g(0,0)=0$ so $g$ vanishes everywhere on $V$.

The situation is better if $k$ is algebraically closed. Then it follows from the Nullstellensatz that if $\mathfrak{p}\subset A$ is prime and $g\in A$ vanishes on all of $V$, then $g\in\mathfrak{p}$. So conversely, if $g\notin\mathfrak{p}$ then $g(a)\neq0$ for some $a\in V$.

  • $\begingroup$ In the comments @Servaes $X_2$ will not vanish on all of $V.$ $V$ clearly contains $(0,1)$ but $X_2=1 \neq 0$ at $(0,1).$ But your last example illustrates the fact that it may so happen that $g \notin \mathfrak p$ but $g$ is identically zero on $V.$ Your vacuous argument was also pretty nice and convincing one. Thank you very much for your help. $\endgroup$
    – little o
    May 2, 2019 at 9:49
  • $\begingroup$ But how did our professor do such a crucial mistake? It's completely unexpected. $\endgroup$
    – little o
    May 2, 2019 at 9:54
  • 1
    $\begingroup$ @Dbchatto67 Indeed in that example $X_2$ does not vanish on all of $V$, but only somewhere on $V$. In some sense this is typical. What your professor should have said is that if $k$ is algebraically closed, then $$g\notin\mathfrak{p}\qquad\implies\qquad g \text{ is not identically } 0 \text{ on } V.$$ Perhaps the assumption of $k$ being algebraically closed was made somewhere earlier? Perhaps even at the outset of the course... (continued) $\endgroup$
    – Servaes
    May 2, 2019 at 10:03
  • 1
    $\begingroup$ (continued)... And one might have phrased the above as $$g\notin\mathfrak{p}\qquad\implies\qquad g \text{ does not vanish on all of } V,$$ which is rather ambiguous; it can be interpreted as the correct implication above, or as the incorrect implication in your question. Depending on the language the ambiguity may be greater. Either way, it does seem very sloppy not to make this crucial statement sufficiently precise. $\endgroup$
    – Servaes
    May 2, 2019 at 10:05
  • $\begingroup$ As far as I remember @Servaes our professor neither mentioned that $\mathcal k$ is algebraically closed anywhere in his lecture yesterday nor he assumed it at the outset of the course. Then on the very first day when the course began he would have mentioned earlier that whatever field is considered in this course is implicitly assumed to be algebraically closed. But I don't think that he did it. $\endgroup$
    – little o
    May 2, 2019 at 15:05

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