# Monday February 24th

## Divisible Modules

We know that injective implies divisible, and uniquely divisible implies injective.
Fact: quotients of divisible modules are divisible

Exercise
:   If $R$ is a domain that is not a field and $M$ is a finitely-generated divisible $R\dash$module, then $M = 0$.

Proof (of exercise)
:   Claim: for any ring $R$, any nonzero f.g. $R\dash$module $M$ has a nonzero cyclic (monogenic) quotient given by modding out by all but one of the generators.
    Thus if $M$ admits a finitely generated divisible $R\dash$module, it admits a cyclic module.

    Then $M \cong R/\ann M$, and there are two cases:

    - $\ann M = 0$, in which case $M\cong R$.
      Then choosing $r\in R^\bullet\setminus R\units$, then $[r]: R\to R$ is *not* a surjection.

    - Otherwise, choose $x\in \ann(M) \setminus \theset{0}$.
      Then $\times x: R \to R$ is the same map as $\times 0: R\to R$, so it is not surjective.

Fact: there is a classification of divisible (iff injective) $\ZZ\dash$modules:

- $(\QQ, +)$, since the fraction field of any domain is divisible.
- $(\QQ/\ZZ, +) = \bigoplus_{\text{primes}} \QQ_p / \ZZ_p$, where $\QQ_p / \ZZ_p = \lim \ZZ/p^n\ZZ$.
    This is isomorphic to the group of $p$ power roots of unity.
    On the other hand, $\QQ/\ZZ$ is the group of *all* roots of unity

Fact (Classification of Divisible $\ZZ\dash$Modules: 
:   Any divisible $\ZZ\dash$module is isomorphic to a direct sum of copies of

  - $(\QQ, +)$
  - $(\QQ_p/\ZZ_p, +)$

  Note that any direct sum of divisible groups is still divisible.
  Moreover, this decomposition is unique.

## Toward Localization

Proposition (Multiplicative Avoidance)
:   Let $S\subset R$ with $SS \subset S$, $1\in S$, $0\not\in S$.
    Define $\mci(S)  = \theset{I\normal R \suchthat I\intersect S = \emptyset}$.
    Then

    1. $\mci(S) \neq \emptyset$
    2. Every element of $\mci(S)$ is contained in a maximal element of $\mci(S)$.
    3. Every maximal element of $\mci(S)$ is prime. 

Proof
:   In parts:

    a. $(0) \in \mci(S)$ by construction.

    b. Standard Zorn's lemma argument. 

    c. Let $I \in \mci(S)$ be a maximal element, and let $x, y \in R$ such that $xy\in I$ with $x\not\in I$.
      Then $\generators{x, I} \supsetneq I$, so $S\intersect \generators{x, I} \neq \emptyset$ by maximality.
      I.e., there exists $s_1 \in S$ such that $s_1 = i_1 + ax$ for some $a\in R$.
      Continuing this way, if $y\not \in I$, produce an $s_2  = i_2 + by_1$ for some $b\in R$.
      Since $S$ is multiplicatively closed, $s_1 s_2 \in S$.
      But we also have $s_1 s_2 = (i_1 + ax)(i_2 + by) \in I$, a contradiction.

> See Kaplansky's Commutative Algebra book.

Proposition (Prime Ideals Behave Like Primes)
:   Let $\mfp \in \spec(R)$ and $I_1, \cdots, I_n \normal R$, then if $\mfp \supset \prod I_i$, then $\mfp \supset I_i$ for some $i$.

Proof
:   Suppose $\mfp \supsetneq I_i$ for any $i$, and let $x_i \in I_i \setminus \mfp$.
    Consider $x \definedas \prod x_i \in \prod I_i \subset \mfp$; then since $\mfp$ is prime, some $x_i \in \mfp$.
    
Corollary:
:   If $\mfp \supset I^n$, then $\mfp \supset I$.

> I.e. prime ideals are radical.

## Radicals

Definition (Nilpotent Elements)
:   An *element* $x\in R$ is *nilpotent* iff $x^n = 0$ for some $n\in \ZZ$.
    An *ideal* is *nilpotent* iff $I^n = (0)$ for some $n$, and is *nil* iff every element $x\in I$ is nilpotent.

Proposition (Nilpotent Implies Nil)
:   Nilpotent $\implies$ nil.

Proof
:   If $I^n = (0)$, then for any $x\in I$, $x^n \in I^n = (0)$ so $x^n = 0$.

Proposition (Nil and FG implies nilpotent)
:   If $I$ is finitely generated and nil, then $I$ is nilpotent.

Proof
:   Let $I = \generators{x_1, \cdots, x_n}$.
    Then for each $i$, choose $e_i \in \ZZ$ such that $x_i^{e_i} = 0$.
    The (check) $I^{\sum e_i} = (0)$.

Definition (Nil)
: An ideal is nil iff all generators are nilpotent.

Corollary:
:   If $R$ is Noetherian, $I$ is nilpotent iff $I$ is nil.

Exercise
:   Exhibit a ring with an ideal that is nil but not nilpotent.
    
    > Need to choose a non-Noetherian ring, e.g. a polynomial ring in infinitely many indeterminates $\theset{t_i}$, and consider $\generators{t_n^n \suchthat n\in \NN}$.

Definition (Nilradical)
:   The *nilradical* of $R$, $\nil(R)$, is the set of all nilpotent elements.

Proposition (Universal Property of Nil)
:   \hfill
    
    a. $\nil(R) \normal R$, since $a^n = b^n = 0\implies (xa + yb)^{2n} = 0$.
    b. $R/\nil(R)$ is reduced, and this quotient map is universal wrt morphism into a reduced ring.
        I.e., if $R \to S$ with $S$ reduced, there is commutative diagram
        \begin{center}
        \begin{tikzcd}
        R \arrow[rr, "f"] \arrow[rdd, "\pi"] &                                                   & S \\
        &                                                   &   \\
        & R/\nil(R) \arrow[ruu, "\exists \tilde f", dashed] &  
        \end{tikzcd}
        \end{center}

    c. $\nil(R) = \intersect_{\text{prime ideal}} \mfp$.

Proof (of c)
:   $\subseteq$: If $x\in \nil(R)$, then $x^n =0$ for some $n$, so $x^n \in \mfp$ and since $\mfp$ is prime, $x\in \mfp$.
    
    $\subseteq$: We'll show that if $x$ is not nilpotent, then it avoids some prime ideal.
    Define $S\definedas \theset{x^n \suchthat n\in\NN}$; since $x$ is not nilpotent, $S$ is multiplicatively closed and does not contain zero, so by a previous result, there is some $\mfp \in \spec(R)$ such that $S\intersect \mfp = \emptyset$.

Definition (Radical Ideals)
:   An ideal $I\normal R$ is *radical* iff for all $x\in R$ there exists an $n$ such that $x^n \in I \implies x\in I$.

Proposition (Prime Implies Radical)
:   Prime ideals are radical.

> Idea: the set of radical ideals is much easier to work with than the set of prime ideals.