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I've managed to prove the following exercise:

(21.) Let $\mathfrak{m}$ be a maximal ideal of a commutative ring $R$. Prove the following: if $A$ is a finitely generated $R$-module, and $x_1, x_2, \ldots, x_n$ is a minimal generating subset of $A$, then $x_1 + \mathfrak{m} A, \ldots, x_n + \mathfrak{m} A$ is a basis of $A/\mathfrak{m}A$ over $R / \mathfrak{m}$.

in Grillet's Abstract Algebra, page 377, only under the hypothesis that $\mathfrak{m}$ is the only maximal ideal, i.e., that $R$ is a local ring.

Weak proof: Firstly, the scalar multiplication is defined as $(r\!+\!\mathfrak{m})(a\!+\!\mathfrak{m}A) \!:=\! ra\!+\!\mathfrak{m}A$. To see this is well defined, notice that if $r\!+\!\mathfrak{m} \!=\! 0$, i.e. $r\!\in\!\mathfrak{m}$, then $ra\!\in\!\mathfrak{m}A$, i.e. $ra\!+\!\mathfrak{m}A\!=\!0$; and also that if $a\!+\!\mathfrak{m}A\!=\!0$, i.e. $a\!=\!ma'\!\in\!\mathfrak{m}A$, then $ra\!=\!rma'\!\in\!\mathfrak{m}A$, i.e. $ra\!+\!\mathfrak{m}A\!=\!0$. Actually, the $R/\mathfrak{m}$-module $A/\mathfrak{m}A$ is constructed by first creating the $R$-module $A/\mathfrak{m}A$ (this is possible because $\mathfrak{m}A$ is a submodule of $A$, since $\mathfrak{m}$ is an ideal of $R$) and then turning it into an $R/\mathfrak{m}$-module (this is possible because $\mathrm{Ann}_R(A/\mathfrak{m}A)\supseteq\mathfrak{m}$).

To prove $x_1\!+\!\mathfrak{m}A,\ldots,x_n\!+\!\mathfrak{m}A$ generate $A/\mathfrak{m}A$, we must show that $(R/\mathfrak{m})(x_1\!+\!\mathfrak{m}A)\!+\!\ldots\!+\!(R/\mathfrak{m})(x_n\!+\!\mathfrak{m}A)\!=\!A/\mathfrak{m}A$, which means $Rx_1\!+\!\ldots\!+\!Rx_n\!+\!\mathfrak{m}A\!=\!A/\mathfrak{m}A$, or equivalently, $Rx_1\!+\!\ldots\!+\!Rx_n\!+\!\mathfrak{m}A\!=\!A$, but this is true since $Rx_1\!+\!\ldots\!+\!Rx_n\!=\!A$ by hypothesis. Alternatively, we could argue that since $x_1,\ldots,x_n$ generate the $R$-module $A$, they generate the $R$-module $A/\mathfrak{m}A$, and therefore generate the $R/\mathfrak{m}$-module $A/\mathfrak{m}A$.

If $x_1\!+\!\mathfrak{m}A,\ldots,x_n\!+\!\mathfrak{m}A$ were $R/\mathfrak{m}$-linearly dependent, then WLOG $x_n\!+\!\mathfrak{m}A$ could be expressed as a $R/\mathfrak{m}$-linear combination of the others, so already $x_1\!+\!\mathfrak{m}A,\ldots,x_{n-1}\!+\!\mathfrak{m}A$ would generate $A/\mathfrak{m}A$, which would mean $Rx_1\!+\!\ldots\!+\!Rx_{n-1}\!+\!\mathfrak{m}A\!=\!A$. But since $J(R)\!=\!\mathfrak{m}$, by Nakayama's Lemma this would mean $Rx_1\!+\!\ldots\!+\!Rx_{n-1}\!=\!A$, a contradiction with the hypothesis on minimality of $x_1,\ldots,x_n$. $\blacksquare$

Question: how can I prove the general version of the exercise? I am somewhat skeptical of the claim...

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2  
What happens if you localize? –  Arturo Magidin Dec 1 '11 at 20:45
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I'm just worried, because in general I don't think that minimal generating sets all have the same size (I think I learned this from Matsumura's book). Like if I take a non-trivial ring $R$ and form the ring $S = R \times R$, then both $\{(1,1)\}$ and $\{(1,0), (0,1)\}$ are minimal in the sense that you can't remove anything and still get a generating set for $S$ as a module over $S$. Have I done something stupid? –  Dylan Moreland Dec 1 '11 at 20:57
    
Localize at $\mathfrak{m}$? Like $R_\mathfrak{m}$? Umm, I get a ring with only one maximal ideal, namely $\mathfrak{m}^E\!=\!\{\frac{r}{s}\!\in\!R_\mathfrak{m};r\!\in\!\mathfrak{m}\}$. Hmm... –  Leon Lampret Dec 1 '11 at 20:59

2 Answers 2

up vote 3 down vote accepted

You are quite right to be skeptical of the claim, Leon, because I think it is false. Here is a counterexample.

Let $k$ be a field, $R$ be the ring $R=k[X,Y]/\langle Y^2-X^3\rangle=:k[x,y]$ and $A$ the ideal $A=\langle x,y\rangle\subset R$.
The chosen set of generators $x_1=x,x_2=y$ of the ideal $A$ is already minimal since $A$ is not principal.
Now choose the maximal ideal ${\frak m}=\langle x-1,y-1 \rangle$.
We have $R/{\frak m}=k$ and ${\frak m}A=\langle x-1,y-1 \rangle \langle x,y \rangle= \langle x^2-x,xy-y,yx-x,y^2-y\rangle $ so that
$A/{\frak m}A=\langle x,y\rangle/\langle x^2-x,xy-y,yx-x,y^2-y\rangle= \langle \bar x \rangle$, a one-dimensional $k$ vector space.
But then the two classes $x+{\frak m}A ,\; y+{\frak m}A$ can obviously not be a basis of this one-dimensional vector space.

Edit
At Leon's request I'll explain in detail why the ideal $A\subset R$ is not principal (this should be skipped by whoever is familiar with the underlying algebraic geometry of the situation).

Suppose $A=\langle f(x,y)\rangle $ for some $ f(X,Y)\in k[X,Y]$ with $f(X,Y)=0$. Since $f\in A$ and since $y^2=x^3$, we can write $f(x,y)=xg(x)+yh(x)$ with $g(X),h(X)\in k[X,Y]$ .

i) Since $x\in A=\langle f(x,y) \rangle$ we must have $x=f(x,y) a(x,y)$ for some $a(X,Y)\in k[X,Y] \:$ . Lifting to actual polynomials we get $X=(Xg(X)+Yh(X))a(X,Y)+(Y^2-X^3)P(X,Y)$.
Looking at linear terms, we see that this is only possible if $g(0)=:c\neq0$ and $h(0)=0$ so that $$g(X)=cX+X^2.l(X) ,\quad h(X)=Xm(X) \quad (*)$$

ii) But then $y\in A=\langle f(x,y) \rangle$ similarly translates into $Y=(Xg(X)+YhX))b(X,Y)+(Y^2-X^3)Q(X,Y)$ and so, using $(*)$ ,
$Y=[X.(cX+X^2l(X))+YXm(X)].b(X,Y)+(Y^2-X^3)Q(X,Y)$
But this is impossible because the right-hand side has zero as its linear term.
We have proved that the assumption that $A$ is principal is contradictory.

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If $I\!=\!\langle Y^2\!-\!X^3\rangle$, we denote $x\!=\!X+I$ and $y\!=\!Y\!+\!I$ in $R$. How can I make sure $\langle x,y\rangle$ is not a principal ideal? What contradiction do I get with $\langle x,y\rangle\!=\!\langle z\rangle$? –  Leon Lampret Dec 1 '11 at 22:43
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@Leon: the geometric reason is that it is the ideal of the origin, a singular point on the curve $y^2=x^3$. You can also check by hand that it is impossible to have a polynomial f(X,Y) without constant term such that x and y are both multiples of f(x,y). You can assume $f(x,y)=xg(x)+yh(x) $ to ease calculations. –  Georges Elencwajg Dec 1 '11 at 23:02
    
Sorry, I am still a little puzzled. Could you please write down an explicit proof why $(x,y)$ isn't principal. $R$ is a domain, since $X^2\!-\!Y^3$ is irreducible. If $f(x,y)\!=\!xg(x,y)\!+\!yh(x,y)$ and $x\!=\!a(x,y)f(x,y)$ and $y\!=\!b(x,y)f(x,y)$, where do I get a contradiction. I'm bothered because $x,y$ are equivalence classes, not variables. Where does $X^2\!-\!Y^3$ come into play? Why wouldn't it be o.k. to take $X\!-\!Y^3$ or any other polynomial? –  Leon Lampret Dec 2 '11 at 4:00
    
Dear @Leon, I have added a very detailed explanation of non-principality of $(x,y)$ in an edit . –  Georges Elencwajg Dec 2 '11 at 9:13
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Dear @Leon: you write "I wonder what you see geometrically". In a nutshell: the curve $y^2=x^3$ is singular at the origin, which corresponds to the ideal $A$. There is an algebraic characterization of singular points on a curve which implies that $A$ then cannot be principal (even locally). The beauty of this characterization (due to Zariski) is precisely that it works on arbitrary fields. But this point of view is completely optional: it is the algebraic geometer's way of seeing things. –  Georges Elencwajg Dec 2 '11 at 19:50

Using Dylan's comment, we get the following counterexample. Let $K$ and $L$ be fields, put $$ R=K\times L,\quad A=\mathfrak m=K,\quad n=1,\quad x_1=(1,0), $$ and observe $$ A/\mathfrak mA=K/KK=K/K=0. $$

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very lucid, thank you –  Leon Lampret Dec 2 '11 at 2:25
    
Just one more question: if $K_1,\ldots,K_n$ are fields, then $K_1\!\times\!\ldots\!\times\!K_n$ has $2^n$ ideals, yes? –  Leon Lampret Dec 2 '11 at 2:28
    
Dear @Leon: Thank you! Yes. Any ideal of $R\times S$ can be written in one and only one way as $\mathfrak a\times\mathfrak b$, where $\mathfrak a$ is an ideal of $R$ and $\mathfrak b$ is an ideal of $S$. In fact, any $R\times S$-module can be written in one and only one way as $M\times N$, where $M$ is an $R$-module and $N$ is an $S$-module. - About lucidity, I admire your's, Dylan's, and Georges's! –  Pierre-Yves Gaillard Dec 2 '11 at 2:40
    
Very nice, as usual, @Pierre-Yves. –  Georges Elencwajg Dec 2 '11 at 9:10

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