# Wednesday February 26th

## Radicals

For $R$ a ring, we defined $\nil(R) \definedas \theset{x\in R \suchthat \exists n\in \NN,~ x^n = 0} \normal R$.
We had a theorem: $$\nil(R) = \intersect_{\mfp\in\spec(R)} \mfp$$

Definition (Radical of an Ideal) (Flat Implies Torsionfree in Domains)
: For $I\normal R$, we define $\rad(I) = \theset{x\in R \suchthat \exists n,~ x^n \in I} \supseteq I$.

Fact
: $I \normal R$.
  To see this, note that for any $I \normal R$, then $\nil(R/I) \normal R = \rad(I)$.

Definition (Radical Ideals)
: $I$ is a *radical ideal* iff $I = \rad(I)$.


Example
: Prime ideals are radical.
  
Definition (Closure Operators)
:   Define a *closure operator* $\ell: I \mapsto \rad(I)$.
    In general, if $(X, \leq)$ is a poset, then a Moore closure operator is a map $c: X\to X$ satisfying

    1. $c(c(x)) = c(x)$
    2. $x \leq c(x)$
    3. $x\leq y \implies c(x) \leq c(y)$.

This is most often applied to $X$ the family of subsets of a set $A$ and $\leq$ subset inclusion.
Note that this doesn't completely correspond to a topological closure, since this would also require preservation of intersections.

> Related to Galois connections, not covering in this class but good for a final topic.

We can produce a nice characterization: 
$$
\rad(I) = \nil(R/I) = \intersect_{\mfp \in R/I} \mfp = \intersect_{\mfp \supseteq I} \mfp
$$

Exercise (easy)
: If $\theset{I_i}$ is any family of radical ideals, then $\intersect_i I_i$ is radical.

Exercise
: Let $R=\ZZ$.
  What are the radical ideals? $(0), (p)$, but $(p^2)$ is not radical -- i.e. $(0), (n)$ for $n$ squarefree.

Fact
: $I$ is radical iff $R/I$ is reduced.
  Noting that by the CRT, $\ZZ/n\ZZ \cong \prod \ZZ/ p_i^{a_i} \ZZ$, which is reduced iff $a_i = 1$ for all $i$.
  If $R$ is a PID, $\pi_1 \cdots \pi_r$ radical ideals, then $(\pi_1 \cdots \pi_r)$ nonassociate prime elements ??

Exercise
: Let $R$ be a ring, $\mfp_1 \neq \mfp_2$ prime ideals.
  
  1. Must $\mfp_1 \mfp_2$ be radical?
  2. If $\mfp_1 + \mfp_2 + R$, then $\mfp_1 \mfp_2 = \mfp_1 \intersect \mfp_2$, and is thus radical.

> Product may be smaller than intersection in general.

Proposition (Algebraic Properties of Radicals)
:   Let $I, J \normal R$.

    a. $I \subset \rad(I)$, $\rad(\rad(I)) \subset \rad(I)$, and $I\subset J \implies \rad(I) \subset \rad(J)$.

    b. $\rad(IJ) = \rad(I\intersect J) = \rad(I) \intersect \rad(J)$

    c. $\rad(I+J) = \rad( \rad(I) + \rad(J) )$

    d. $\rad(I) = R \iff I = R$

    e. $\rad(I^n) = \rad(I)$ for $n\geq 1$

    f. If $R$ is Noetherian and $J\subset \rad(I)$, then $J^n \subset I$ for some $n\geq 1$.

So for Noetherian rings, two radicals are equal iff powers of each ideal land in the other.

Proof (of (b))
: $IJ \subseteq I\intersect J$ and thus $\rad(IJ) \subset \rad(I\intersect J)$.
  If $x\in \rad(I \intersect J)$, there exists an $n$ such that $x^n \in I \intersect J$.
  Then $x^{2n} = x^n x^N \in IJ \implies x\in \rad(IJ)$.

Proof (of b)
: Since $I\intersect J \subset I, J$, we have $\rad(I\intersect J) \subset \rad(I) \intersect \rad(J)$.
  If $x\in \rad(I) \intersect \rad(J)$, then $x^n\in I, x^m \in J$ for some $n, m$, so $x^{m+n} \in I\intersect J \subset \rad(I \intersect J)$.

Proof (of (f))
: By replacing $R$ with $R/I$, assume $I = (0)$, then $J\in \nil(R)$ and since $R$ is Noetherian, $J$ is nilpotent and $J^n=(0)$ for some $n$.

So we simplify things by passing from $I$ to $\rad(I)$. 
There is a class of rings for which it's feasible to understand all *radical* ideals, and hopeless to understand *all* ideals.

Example
: Take $R = \CC[x]$, a PID.
  Suppose $I\normal R$ and $\rad(I) = x^n$, then $I = (x^n)$.
  So this is no big deal, it's just an extra integer parameter.

  Now instead take $R = \CC[x, y]$, and let $I = \generators{x, y}$.
  Note that applying (f) above to $J = \rad(I)$, we find that $I \supset \generators{x, y}^n$ for some $n$.
  But note that $$\generators{x, y}^n = \generators{x^n, x^{n-1}y, \cdots, xy^{n-1}, y^n}.$$

Exercise
: Suppose $I \supset \generators{x ,y}^2$.
  For such $I$, $\dim_\CC R/I < \infty$.
  So for each $d$, try to find all ideals $I$ such that $\rad(I) = \generators{x, y}$ and $\dim_\CC R/I = d$.

> Note that these correspond to "fat points" in algebraic geometry.
> The idea $\generators{x, y}$ corresponds to a fat point at zero.
> When doing AG, we hope to restrict attention entirely to radical ideals.

Definition (Jacobson Radical)
: The *Jacobson radical* is defined by $$\mcj(R) = \intersect_{\mfm \in \maxspec(R)} \mfm.$$

Fact
: $\mcj(R) \supset \nil(R)$, since not every prime ideal is maximal.

Example
: If $(R, \mfm)$ is a local domain, then $\nil(R) = 0$ and $\mcj(R) = \mfm$.

Exercise
: Let $R$ be a domain, show that $\mcj(R[t]) = (0)$.

Proposition (Characterization of Jacobson Radical in Terms of Units)
: $x\in \mcj(R) \iff 1\pm xR \subset R\units$.

Exercise
: Show directly that $x^n = 0 \implies \forall y, 1-xy \in R\units$.