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Sunday, March 31, 2019

A few exercises about Associated Primes.

I'm trying to learn about Stanley-Reisner rings, and these are some exercises previous to the theory. These are taken from Villarreal's book: Graphs, Rings and Polyhedra.
There will be more.

Exercise 1.
Let $I$ be an ideal of a ring $R$. Prove that $\rad I=I$ if and only if for any $x\in R$ such that $x^2\in I$ we have $x\in I$.

Proof.
First suppose $I$ is a radical ideal. Let $x\in R$ be such that $x^2\in I$. Then $x\in \rad I=$. Conversely, suppose that for any $x\in R$ such that $x^2\in I$ we have $x\in I$. We want to prove that if $y^n\in I$ for some $n$ then $y\in I$, and we'll try using induction. For $n=1,2$ the result already holds. Take $n>2$, suppose that $y^k\in I\Rightarrow y\in I$ for $k<n$ and suppose $y^n\in I$. If $n$ is even, then by the $k=2$ case it means that $y^{\frac{n}{2}}\in I$ from what by induction hypothesis $y\in I$. If $n=2j+1$ is odd then $y^{2j+1}\in I$. So $y^{2j+2}\in I$ and by the $k=2$ case we get $y^{j+1}\in I$. Then, by induction hypothesis $y\in I$.

Exercise 2.
Let $I_1,\dots,I_n,P$ be ideals of a ring $R$ with $P$ prime. Suppose that
$$J=\bigcap_{i=1}^n I_i\subseteq P.$$
Prove that there is some $i\in \{1,\dots,n\}$ such that $I_i\subseteq P$. In particular, if $\bigcap_{i=1}^n I_i=P$ then prove that there is some $i$ such that $I_i=P$.

Proof.
Suppose not. Then, for each $i$ there are elements $f_i\in I_i$ such that $f_i\notin P$. Then $f_1\cdots f_n\in J\subseteq P$. Since $P$ is prime and $f_1\notin P$ then $f_2\cdots f_n\in P$. Since $f_2\notin P$ then $f_3\cdots f_n\in P$. Thus, $f_n\in P$, a contradiction. So there is some $i$ such that $I_i\subseteq P$.

For the second case, consider, by a permutation, that $I_n\subseteq P$. Let $x\in P$, then $x\in J$, so $x\in I_n$, therefore $I_n=P$.

Exercise 3.
Let $R$ be a ring and $P$ a proper ideal of $R$. If every element of $R/P$ is nilpotent or invertible, prove that $P$ is a primary ideal. 

Proof.
Suppose $fg\in P$. Then $fg+P=(f+P)(g+P)=0$. Suppose $f+P$ is invertible. Then $g+P=0$, and $g\in P$. Now suppose $f+P$ is not invertible. Then $f+P$ is nilpotent and there is some $n$ such that $f^n+P=0$, i.e. $f^n\in P$. Therefore, $P$ is primary.


Exercise 4.
Let $R$ be a ring and $I$ be an ideal. If $x\in R-I$ then there is an exact sequence of $R-$modules
$$0\to R/(I:x)\stackrel{\psi}{\to} R/I\stackrel{\phi}{\to} R/(I,x)\to 0$$
where $\psi(r+(I:x))=xr+I$ and $\phi(x+I)=x+(I,x)$.

Proof.
First we'll prove that $\psi$ is injective. Suppose $\psi(r+(I:x))=xr+I=0$. Then $xr\in I$ so that $r\in (I:x)$ and $r+(I:x)=0$.

Now we'll prove that $\phi$ is a well defined surjective map. It's just that since $I\subseteq (I,x)$, then the canonical map $\pi:R\to R/(I,x)$ factors through $I$. And for $r+(I,x)\in R/(I,x)$ just take $r+I\in R/I$ and we're done, for $\phi(r+I)=r+(I,x)$.

Now we'll prove that $\im \psi=\ker \phi$.
Take $r+I\in R/I$. Then
$r+(I,x)=0$ iff $r\in (I,x)$, iff there exist $a\in I,b\in R$ such that $r=a+bx$, but this occurs iff $r+I=a+bx+I=bx+I\in \im \psi$. So $\im\psi=\ker\phi$ and the sequence is exact.

From now, a map will be a morphism in a corresponding category (Only if it is not ambiguous, in which case we'll talk about the corresponding kind of morphism).

Exercise 5 (Localization defines a covariant functor).
If $\phi:M\to N$ is a map of $R-$modules then there is an induced map of $S^{-1}R-$módules:
$$S^{-1}\phi:S^{-1}M\to S^{-1}N$$
given by $\frac{m}{s}\mapsto \frac{\phi(m)}{s}$.

Proof.
$S^{-1}\phi$ is a well defined function. Indeed, $\frac{m}{s}=\frac{m'}{s'}$ implies $t(s'm-sm')=0$ for some $t\in S$, which implies $t(s'\phi(m)-s\phi(m'))=0$, which means $\frac{\phi(m)}{s}=\frac{\phi(m')}{s'}$.
Now, take Take $f=\frac{r_1}{s_1}\frac{m_1}{s_1'}+\frac{r_2}{s_2}\frac{m_2}{s_2'}\in S^{-1}M$. Then $$f=\frac{r_1s_2s_2'm_1+r_2s_1 s_1'm_2}{s_1s_1's_2s_2'}$$
and
$$S^{-1}\phi(f)=\frac{\phi(r_1s_2s_2'm_1+r_2s_1 s_1'm_2)}{s_1s_1's_2s_2'}=\frac{r_1s_2s_2'\phi(m_1)+r_2s_1 s_1'\phi(m_2)}{s_1s_1's_2s_2'}$$
which is clearly 
$$\frac{r_1}{s_1}S^{-1}\phi\left(\frac{m_1}{s_1'}\right)+\frac{r_2}{s_2}S^{-1}\phi\left(\frac{m_2}{s_2'}\right)$$
so $S^{-1}\phi$ is a map of $S^{-1}R-$modules.

Also note that $S^{-1}I_M=I_{S^{-1}M}$ and $S^{-1}(\phi)\circ S^{-1}(\psi)(\frac{m}{s})=\frac{\phi\circ\psi(m)}{s}=S^{-1}(\phi\circ\psi)$, so localization is a functor.

Exercise 6.
Following on the previous exercise, $S^{-1}(-)$ is an exact functor.

Proof.
Take an exact sequence
$$0\to L\stackrel{\psi}{\to} M\stackrel{\phi}{\to} N\to 0$$
Now, $S^{-1}\psi$ is injective.
Indeed, suppose $S^{-1}\psi(\frac{l}{s})=\frac{\psi(l)}{s}=0$. Then there is $t\in S$ such that $t\psi(l)=\psi(tl)=0$ so $tl=0$ thus $\frac{l}{s}=0$ and $S^{-1}\psi$ is injective.

Now, to prove $S^{-1}\phi$ is surjective, just notice that every $\frac{n}{s}\in S^{-1}N$ has the form $\frac{\phi(m)}{s}=S^{-1}\phi(\frac{m}{s})$ for some $m\in M$. Therefore $S^{-1\phi}$ is surjective.

Now we want to prove that $\ker S^{-1}\phi=\im S^{-1}\psi$.

Take $\frac{m}{s}\in S^{-1}M$. Then $S^{-1}\phi(\frac{m}{s})=\frac{\phi(m)}{s}=0$ iff there is some $t\in S$ such that $tm=0$, iff there is some $l\in L$ such that $\psi(l)=tm$, iff $S^{-1}\psi(\frac{l}{ts})=\frac{psi(l)}{ts}=\frac{tm}{ts}=\frac{m}{s}$, which is the desired result, that the induced sequence
$$0\to S^{-1}L\stackrel{S^{-1}\psi}{\to} S^{-1}M\stackrel{S^{-1}\phi}{\to} S^{-1}N\to 0$$
is exact, and $S^{-1}(-)$ is an exact functor.

Exercise 7.
If $M,N$ are $R-$modules then $\Ass(M\oplus N)=\Ass M\cup \Ass N$.

Proof.
Let $P=\Ann(m+n)\in\Ass(M\oplus N)$ for $m+n\neq 0$. Then $x\in P$ iff $xm=0$ and $xn=0$ iff $x\in \Ann(m)\cap\Ann(n)$, so $P=\Ann(m)\cap \Ann(n)$. By a previous exercise, $P=\Ann(m)$ or $P=\Ann(n)$. Thus, $P\in \Ass(M)\cup \Ass(N)$.

Conversely if $P\in\Ass(M)\cup\Ass(N)$ then $P=\Ann(m)$ for some $m\in M\cup N$. Then since $m\in M\oplus N$, $P\in \Ass(M\oplus N)$.

Therefore, $\Ass(M\oplus N)=\Ass(M)\cup \Ass(N)$.

Exercise 8.
If $\rad Q$ is maximal then $Q$ is primary.

Proof.
Suppose $ab\in Q$ with $b^n\notin Q$ for every $n$. Then $b\notin \rad Q$. Since $\rad Q$ is maximal then $\rad Q+\langle b\rangle=R$, so there exists $c\in R$ and $q\in\rad Q$ such that $q+cb=1$. Since $q\in\rad Q$ there exists some $n$ such that $q^n\in Q$. So $(q+cb)^n=q^n+rb=1$ where $r\in R$. But $q^n\in Q$, so $a(q^n+rb)=aq^n+rab=a\in Q$.
Therefore $Q$ is primary.

Exercise 9 (Lemma).
If $0\to M'\stackrel{\phi}{\to} M\stackrel{\psi}{\to} M''\to 0$ is an exact sequence of $R-$modules then
$$\Ass(M)\subseteq \Ass M'\cup \Ass M''.$$
Proof.
Let $P\in\Ass M$. Then there is a monomorphism $\rho:R/P\to M$. Set $N=\rho(M)$. Then $\rho|_{N}$ is an isomorphism, thus for every $m\in N-0$ we have $\Ann(m)=P$\footnote{This is because for $m\in N-0$, $xm=0$ iff $x\rho^{-1}(m)=0$ iff $x\in P$, because $P$ is prime and $\rho|_{N}^{-1}(m)\neq 0$.}.
Set $N'=\phi(M')$ and suppose $N'\cap N\neq 0$. Then, take $0\neq s\in N'\cap N$, then, since $\Ann(s)=P$, and $\phi$ is a monomorphism, $\Ann(\phi|_{M}^{-1}(s))=P$, thus $P\in \Ass M'$. Now, suppose $N\cap N'=0$ so $N\cap \ker\psi=0$, thus the composition $\overline{\rho}=\pi\circ\rho:R/P\to M/\ker\psi\cong M''$ is also a monomorphism ($\pi$ is the canonical map), therefore $P\in\Ass M''$. 
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