In Greuel & Pfister's A Singular Introduction to Commutative Algebra, p. 38, there is written:

(1) Consider the two rings $$ A=\mathbb C[x,y]/\langle x^2-y^3\rangle\text{ and } B=\mathbb C[x,y]/\langle xy\rangle$$ and the multiplicative sets:

  • $S$ the set of non-zerodivisors of $A$, respecrively of $B$, and
  • $T:= A\backslash \langle x,y\rangle_A$, respectively $T:=B\backslash\langle x,y\rangle_B$.

Determine the localizations of $A$ and $ B$ with respect to $T$ and $S$.

(2) Are any two of the six rings $A,B,S^{-1}A,S^{-1}B,T^{-1}A,T^{-1}B$ isomorphic?

So we have rings $$\begin{array}{l l} R_1:= K[x,y|x^2\!-\!y^3], & R_4:= K[x,y|xy],\\ R_2:= K[x,y|x^2\!-\!y^3]_{\langle x,y\rangle}, & R_5:= K[x,y|xy]_{\langle x,y\rangle},\\ R_3:= Q(K[x,y|x^2\!-\!y^3]), & R_6:= Q(K[x,y|xy]).\\ \end{array}$$

Question: How can one determine these localizations, whatever that means? How can I simplify?

The polynomial $x^2\!-\!y^3$ is irreducible, so $\langle x^2\!-\!y^3\rangle \unlhd K[x,y]$ is a prime ideal. Hence $R_1$ is a domain, $R_3$ is a field, and $R_1 \subseteq R_2 \subseteq R_3$, so all three rings are domains. But $R_4,R_5,R_6$ contain $x,y$ with $xy=0$, so they are not domains, hence $R_i \ncong R_{3+j}$ for any $i,j\!\in\!\{1,2,3\}$. Furthermore, $R_1$ contains maximal ideals $\langle x,y\rangle$ and $\langle x\!-\!1,y\!-\!1\rangle$ (we have $x^2\!-\!y^3= (x\!+\!1)(x\!-\!1)-(y^2\!+\!y\!+\!1)(y\!-\!1)$), but $R_2$ is local, i.e. contains only one maximal ideal, hence $R_1\ncong R_2$. Similarly, $R_1\ncong R_3$, since $R_3$ contains only one prime ideals (namely $0$), and $R_2\ncong R_3$, by the argument of the number of ideals.

Question: how can I distinguish between $R_4,R_5,R_6$?

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    $\begingroup$ $R_4$ is not isomorphic to $R_5$ because one is not local and the other one is local :) Btw, why invent new notation for factor rings? $\endgroup$ – user26857 May 5 '13 at 15:33
  • $\begingroup$ Ah, you're right. One down, two to go :). Regarding notation: it is painful to write $(K[x,y]/(x^2-y^3))_{(x,y)}$. Besides, analogous notation $\langle x,y|xyx^{-3}\rangle$ for a quotient of the free group $\langle x,y\rangle$ and notation $K\langle x,y|x^2-y^3\rangle$ for a quotient of the free algebra $K\langle x,y\rangle$ is already present, so I feel that I didn't really impose anything, the notation for a quotient of the free commutative algebra $K[x,y]$ is already present. $\endgroup$ – Leo May 5 '13 at 15:44
  • $\begingroup$ Your notation for $R_6$ is a little bit unfortunate, since it suggests that you want to form the quotient field of $K[x,y]/(xy)$, which is not an integral domain, and hence has no quotient field at all. Indeed $R_6$ is the localization of $R_4$ at the set of all non-zero divisors. Can you figure exactly what that set is? This will help you comparing $R_5$ with $R_6$. $\endgroup$ – Nils Matthes May 5 '13 at 15:59
  • $\begingroup$ $Q(R)$ denotes the total quotient ring, which is the usual quotient field whenever $R$ is a domain. An element $f\in K[x,y|xy]$ is a zero divisor iff $\exists g,h: fg=xyh$, iff $f\in \langle x\rangle\cup\langle y\rangle$. $\endgroup$ – Leo May 5 '13 at 16:24

It is not hard to show that the set of zerodivisors of the ring $\mathbb C[x,y]$ with $xy=0$ is the union of the prime ideals $(x)$ and $(y)$. (One knows this without any proof since the set of zerodivisors is the union of associated prime ideals and it's easily seen that $(XY)=(X)\cap(Y)$.) Then $S$, the set of non-zerodivisors of $\mathbb C[x,y]$, is nothing but $\mathbb C[x,y]-(x)\cup(y)$. In particular, this shows that the total ring of fractions $S^{-1}\mathbb C[x,y]$ is a semilocal ring having exactly two maximal ideals. Now all it's done.

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  • $\begingroup$ Q1: If $\frak{p}_1,\ldots,\frak{p}_k$ are prime ideals of $R$ and $S=R∖(\frak{p}_1\cup\ldots\cup\frak{p}_k)$, does $S^{−1}R$ have precisely $k$ maximal ideals? Q2: For $R=R_2,R_3,R_5,R_6$, can we find $n,f_1,\ldots,f_k$ with $R\cong K[x_1,\ldots,x_n|f_1,\ldots,f_k]$? Q3: Is the following true: if $a_1,\ldots,a_n\in K$ are pairwise distinct, then $K[x\,|\,(x-a_1)^{k_1}\cdots(x-a_n)^{k_n}]_{\langle x-a_1\rangle} \cong K[x\,|\,(x-a_1)^{k_1}]$, since $x-a_2,\ldots,x-a_n$ do not lie in $\langle x-a_1\rangle$, hence are invertible in the localization, thus do not contribute to the ideal? $\endgroup$ – Leo May 5 '13 at 17:33
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    $\begingroup$ Q1: Yes. Q3 No. The last ring must be localized at $x-a_1$. Q2 You can post this as a separate question. $\endgroup$ – user26857 May 5 '13 at 17:46
  • $\begingroup$ I've checked Greuel & Pfister SICA, Eisenbud CAIAG, Gortz & Wedhorn AG, Stacks Project, but ended up empty-handed. Do you have any reference for your claim in Q1? In Kemper CCA, p. 98, Exercise 7.6, it is stated that if $\mathfrak{p}_1,\ldots,\mathfrak{p}_k\!\in\!\mathrm{Spec}R$ and $\forall i,j\!: \mathfrak{p}_i\!\nsubseteq\!\mathfrak{p}_j$ and $S\!=\!R\!\setminus\!(\mathfrak{p}_1\!\cup\!\ldots\!\cup\!\mathfrak{p}_k)$, then $\mathrm{Spec}\,S^{-1}\!R=\{S^{-1}\mathfrak{p}_1,\ldots,S^{-1}\mathfrak{p}_k\}$. Is the assumption $\forall i,j\!: \mathfrak{p}_i\!\nsubseteq\!\mathfrak{p}_j$ necessary? $\endgroup$ – Leo May 7 '13 at 12:51
  • $\begingroup$ Atiyah and Macdonald and any other basic textbooks on Commutative Algebra will learn you that. (Actually you can check this by yourself by using the correspondence between the prime ideals of $S^{-1}R$ and those of $R$ which do not intersect $S$.) In Kemper must be stated that MaxSpec is what you said before, and the condition about primes is obviously necessary, otherwise the union of primes would have less than $k$ terms. $\endgroup$ – user26857 May 7 '13 at 13:13
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    $\begingroup$ I'm seeing these things a little bit different, but the answer to your questions is affirmative: $(k[x]/(x^2(x-1)))_{(x)}\simeq k[x]_{(x)}/(x^2(x-1))_{(x)}$, and $(x^2(x-1))_{(x)}=(x^2)_{(x)}$. So we get $(k[x]/(x^2(x-1)))_{(x)}\simeq k[x]_{(x)}/(x^2)_{(x)}$. Now taking into account that $k[x]/(x^2)$ is local with maximal ideal $(x)/(x^2)$ we obtain $k[x]/(x^2)\simeq (k[x]/(x^2))_{(x)/(x^2)}$ and the last ring is nothing but $k[x]_{(x)}/(x^2)_{(x)}$. $\endgroup$ – user26857 May 9 '13 at 19:19

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