# Friday February 28th

## Radicals: The Jacobson Radical

Definition (The Jacobson Radical)
: $\mcj(R) = \intersect{\mfm \in \maxspec(R)} \mfm$.
  
For a noncommutative ring, instead of intersecting just two-sided ideals, need to intersect either left ideals *or* right ideals (the intersections turn out to be equivalent).

Fact
: If $R$ is finite dimensional over a field, then $\mcj(R) = 0 \iff R$ is semisimple.
  By Wedderburn, this happens iff $R = \prod M_{n_i}(D_i)$.

Definition (Semiprimitive)
: A ring is *semiprimitive* *(or $\mcj\dash$semisimple or Jacobson-semisimple)* iff $\mcj(R) = 0$.

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

Proof
: Let $x\in \mcj(R)$ and suppose $1-xy \not\in R\units$, so $1-xy\in\mfm$ for some maximal ideal.
  But then $x\in \mfm$, so $xy \in \mfm$, so $1 = \mfm + xy \in \mfm$, a contradiction.
  
  Suppose instead that $x\not\in\mcj(R)$, so there exists some maximal such that $\generators{m, x} = R$.
  Thus for $y\in R, m\in\mfm$, we have $1 = m+xy$ so $1-xy = m \in \mfm$ and thus $1-xy \not\in R\units$.

In other words, $R\units + \mcj(R) \subset R\units$, and is the largest ideal with this property.
Thus the elements are "close to zero" in the sense that it doesn't take you outside of the unit group.

## Proposition (Commutative Algebra Analog of Euclid IX.20: Infinitely Many Primes)

Let $R$ be a domain, then recall that $p\in R^\bullet$ is irreducible iff $p\not\in R\units$ and $p=xy \implies x\in R\units$ or $y\in R\units$.
If $p$ is irreducible and $u\in R\units$, then $up$ is irreducible and associate to $p$, and $(up) = (p)$.

Define an *atom* to be the principal ideal generated by an irreducible element.

Define a *Fursentenberg domain* to be a domain such that $x\in R^\bullet\setminus R\units$ has an irreducible divisor.
Note that we have a chain of implication, UFD $\implies$ Noetherian = ACC $\implies$ ACC on principal ideals $\implies$ nonzero nonunits factor into irreducibles (atomic domain) $\implies$ Fursentenberg.
So this is a weak factorization condition.

Exercise
: Let $R = \mathrm{Hol}(\CC)$ be the ring of holomorphic functions, which is a domain by the identity theorem.
  Show that $R$ is semiprimitive, Furstenberg but not atomic.

Theorem (Euclidean Criterion)
:   Let $R$ be a domain, not a field, and semiprimitive.
  
    a. There exists a sequence of pairwise comaximal elements $\theset{a_n}_n^\infty$ such that $\generators{a_m, a_N} = R$ for $m\neq n$.

    b. If $R$ is Forstenburg, then there is a sequence of primitive pairwise comaximal *irreducible* elements, and thus infinitely many atoms.

Note that applying this to $R = \ZZ$, the only unit ideals are generated by $\pm 1$, and the result follows immediately.

Proof
: Exercise.

> For what class of rings does this criterion apply?

Application
: For $R$ a Noetherian domain, then by Hilbert's basis theorem $R[t]$ is Noetherian and semiprimitive.
  So by the above result, $R[t]$ is has infinitely many elements.
  Most interesting for $R = \FF_q$, since for e.g. $R=\RR$ we can consider ideals $(x-r)$.

Fact
:   a. If $I, J \normal R$ and $r(I) + r(J) = R$, then $I+J = R$, and $r(r(I)) + r(J) = r(I + J)$.
    b. If $I, J_1, \cdots, J_n \normal R$ and $I + J_i = R$ for each $i$, then $I + \prod I_i = R$.
    c. Suppose $I_1, \cdots, I_n$ are pairwise comaximal, then $\prod I_i = \intersect I_i$ (note: could be smaller and general).

Theorem (Chinese Remainder)
:   Suppose $R$ is arbitrary with $I_1, \cdots, I_n \normal R$ pairwise comaximal.
    Then there is a natural map 

    \begin{align*}
    \Phi: R &\to \prod_{i=1}^n R/I_i \\
    r  &\mapsto (r+ I_1, \cdots, r+ I_i)
    .\end{align*}

    1. $\Phi$ is surjective, and $\ker \Phi = \intersect I_i$.
    2. By pairwise primality, $R/\prod I_i \cong \prod R/I_i$.

Note that as modules, both sides are cyclic.

Proof
:   By induction on $n$, with trivial base case.

    Let $R' \definedas \prod_{i=1}^{n-1} R/I_i$ and assume by induction that $\Phi': R\to R'$ is surjective by induction.
    Let $(r', \bar s) \in R' \cross R/I_n$.
    By hypothesis, $\ker \Phi' = \prod_{i=1}^{n-1}$.
    So there exists an $r\in R$ such that $\Phi'(r) = r'$.
    Lifting to $s\in R$ such that $s + I_n = \bar s + I_n$.
    
    By assumption, $$I' + I \definedas \qty{ \prod_{i=1}^{n-1} I_i} + I_n = R.$$
    So there exist $x\in I', y\in I_n$ such that $s-r = x+y$.
    Note that $\Phi'(r+x) = r'$ since $x\in \ker \Phi$, so
    $$
    r_x = r+x + y = x \mod I_n
    .$$

    But then $\Phi(r+x) = (r', s)$.

Exercise (Converse to CRT (Good for Problem Sessions))
:   Let $I_1, \cdots, I_n \normal R$.
    If $\prod R/ I_i$ is a cyclic $R\dash$module, then the $I_i$ are pairwise comaximal.

    > Immediately reduce to $n=2$ case. Also a nice proof using tensor products, use characterization of $R/I \tensor R/J$.

## Monoid Rings

Here let $R$ be a ring$^*$ (potentially noncommutative) and $(M, \cdot)$ a monoid (i.e. a group without requiring inverses).

Goal: we want to define a *monoid ring* $R[M]$.

If $M$ is finite, the definition is unambiguous, but for infinite $M$ we require an extra condition.
In this case we define the *big monoid ring* $R[[M]]$.

Example
: For $R$ a nonzero ring and $M = (\NN, +)$, $R[M] = R[t]$, and $R[[M]] = R[[t]]$.

Step 1: suppose $M$ is finite, then $$R[M] \stackrel[\rmod]{=} R^M = \theset{f: M\to R},$$ the set of *all* functions.
Note that $(f+g)(m) = f(m) + g(m)$, and define a new multiplication $$(f\ast g)(m) \definedas \sum_{(x, y) \in M^2,~ xy = m} f(x) g(y),$$ the *convolution product*.
One must check that this actually satisfies the axiom of a ring, since we are building this by hand.
This is a ring iff $R$ is a ring and $(M, \ast)$ is commutative.

There is an identity, namely $1 \mapsto 1$ and $x\mapsto 0$ for $x\neq 1$.
Distributivity isn't difficult, but we need to check that $\ast$ is associative.
This follows from 
$$
((f\ast g) \ast h)(m) = \sum_{x,y,z\in M^3,~ xyz = m} f(x) g(y) h(z) = (f\ast (g\ast h))
.$$

Define $[m] \cdot [n] = [mn]$, then check that $\qty{ \sum_{m\in M} r_m [m]  } \qty{ \sum_{m\in M}s_m [m] } = ?$.