Let $(R,M)$ be a local ring. Suppose that $R$ is noetherian and let $I,J \unlhd R$ such that $J \subseteq I$. Prove that the following are equivalent.

Question

Let $R$ be a local ring with maximal ideal $M$.

Suppose that $R$ is noetherian and let $I,J$ be ideals of $R$ such that $J \subseteq I$.

Consider the following statements:

1) Every minimal set of generators of $J$ can be extended to a minimal set of generators of $I$

2) $MI \cap J = MJ$

3) There exists $L \unlhd R$ such that $\frac{I/J}{L(I/J)}$ is a free $R/L$-module and $LI \cap J = LJ$

4) (minimal number of generators of $I$) = (minimal number of generators of $J$) + (minimal number of generators of $I/J$)

I want to prove that $1) \Rightarrow 2)$ and $3) \Rightarrow 4)$.

In fact, now I only want to prove that $3) \Rightarrow 4)$ due to @Rafael Holanda answer.

The idea is to prove that all these statements are equivalent, and in fact I proved $2) \Rightarrow 3)$ and $4) \Rightarrow 1)$.

More ideas: the sequences $$0 \rightarrow \frac{J}{IL \cap J} \rightarrow \frac{I}{IL} \rightarrow \frac{I}{IL+J} \rightarrow 0$$ $$0 \rightarrow \frac{IL+J}{IL} \rightarrow I \otimes \frac{R}{L} \rightarrow \frac{I}{J} \otimes \frac{R}{L} \rightarrow 0$$ $$0 \rightarrow \frac{J}{IL \cap J} \rightarrow \frac{I}{IL} \rightarrow \frac{I/J}{L(I/J)} \rightarrow 0$$ are exact (remark that the 3rd one is using well-known isomorphisms).

I have no clue how to do this, can someone help me with this?

Thanks!

Answer 1

Let $R$ be a local ring. We denote by $\mu_R(M)$ the minimal number of generators of a finitely generated $R$-module. This turns out to be $\dim_{R/\mathfrak m}M/\mathfrak mM$, where $\mathfrak m$ denotes the maximal ideal of $R$. If $\mathfrak a$ is an ideal of $R$, $\mathfrak a\subseteq\mathfrak m$, then $\mu_R(M)=\mu_{R/\mathfrak a}(M/\mathfrak aM)$ (why?).

1) $\Rightarrow$ 2) Let $x_1\dots,x_m$ be a minimal system of generators in $J$. This is equivalent to $\overline x_1,\dots,\overline x_m$ is an $R/\mathfrak m$-basis in $J/\mathfrak mJ$. Now define $\varphi:J/\mathfrak mJ\to I/\mathfrak mI$ by $\varphi(\overline a)=\widehat a$. Since there is $y_1,\dots,y_n$ in $I$ such that $x_1\dots,x_m,y_1,\dots,y_n$ is a minimal system of generators for $I$ we get that $\widehat x_1\dots,\widehat x_m,\widehat y_1,\dots,\widehat y_n$ is an $R/\mathfrak m$-basis in $I/\mathfrak mI$. This shows that $\varphi$ is injective (why?), so $\ker\varphi=(0)$, that is, $J\cap\mathfrak mI=\mathfrak mJ$.

2) $\Rightarrow$ 3) Set $L=\mathfrak m$.

3) $\Rightarrow$ 4) Let $x_1,\dots,x_m$ be a minimal system of generators in $J$. Let $y_1,\dots,y_n\in I$ be such that their classes form an $R/\mathfrak a$-basis in $I/(\mathfrak aI+J)=\dfrac{I/J}{\mathfrak a(I/J)}$. They are also a minimal system of generators for $I/J$ and then $x_1\dots,x_m,y_1,\dots,y_n$ is a system of generators for $I$. Let's show that this is minimal.

Since $\mathfrak aI\cap J=\mathfrak aJ$ we have a short exact sequence $$0\to J/\mathfrak aJ\to I/\mathfrak aI\to I/(\mathfrak aI+J)\to 0.$$ But $I/(\mathfrak aI+J)$ is a free $R/\mathfrak a$-module, so the sequence is split. Then $$\mu_{R/\mathfrak a}(I/\mathfrak aI)=\mu_{R/\mathfrak a}(J/\mathfrak aJ)+\mu_{R/\mathfrak a}(I/(\mathfrak aI+J)),$$ so $\mu_R(I)=\mu_R(J)+\mu_R(I/J)$, and we are done.

4) $\Rightarrow$ 1) Suppose $\mu(I)=\mu(J)+\mu(I/J)$ and let $x_1,\dots,x_m$ be a minimal system of generators for $J$ and $\overline y_1,\dots,\overline y_m$ a minimal system of generators for $I/J$. Then $x_1\dots,x_m,y_1,\dots,y_n$ is a system of generators for $I$ and it is minimal since $\mu(I)=m+n$.

Answer 2

$1)\Rightarrow 2)$

It is clear that $MJ\subseteq J$. Since $J\subseteq I$, $MJ\subseteq MI$. Therefore $MJ\subseteq MI\cap J$. Let $\{r_{1},...,r_{j}\}$ be a minimal set of generators of $J$. Then there are $r_{j+1},...,r_{n}\in I$ such that $\{r_{1},...,r_{n}\}$ is a minimal set of generators of $I$. For all $x\in MI\cap J$ there are $m_{x}\in M, a_{x1},...,a_{xn}\in I$ and $b_{x1},...,b_{xj}\in J$ such that $$x=m_{x}(a_{x1}r_{1}+...+a_{xn}r_{n})=m_{x}a_{x1}r_{1}+...+m_{x}a_{xn}r_{n}$$ and $$x=b_{x1}r_{1}+...+b_{xj}r_{j},$$ so $$b_{x1}r_{1}+...+b_{xj}r_{j}=m_{x}a_{x1}r_{1}+...+m_{x}a_{xn}r_{n}.$$ Therefore $a_{x(j+1)}=...=a_{xn}=0$ and $x=m_{x}(a_{x1}r_{1}+...+a_{xj}r_{j})\in MJ$ (remember that $r_{i}\in J\Rightarrow a_{i}r_{i}\in J$). We prove that $MI\cap J=MJ$.