attachments/Pasted image 20210428000241.png

Chapter 1

Nice mnemonic: Maximal \(\implies\) prime \(\implies\) radical Field \(\implies\) domain \(\implies\) reduced

Proposition 1.1: Fix an ideal \({\mathfrak{a}}{~\trianglelefteq~}R\). There is a correspondence

\begin{align*}
\left\{{{\mathfrak{b}}{~\mathrel{\Big\vert}~}{\mathfrak{a}}\subseteq {\mathfrak{b}}{~\trianglelefteq~}R}\right\} \iff
\left\{{\tilde {\mathfrak{b}}{~\trianglelefteq~}R/{\mathfrak{a}}}\right\}
.\end{align*}

Proof: Adapted from proof for groups here: https://math.stackexchange.com/a/955413/147053.

Let \(f: R \to T\) be any ring homomorphism and let \(S(R), S(T)\) denote the lattices of subrings of \(R, T\) respectively. Then \(f\) induces two maps:

\begin{align*}
F: S(R) &\to S(T) \\
H  &\mapsto f(H) \\
\\
\\
F^{-1}: S(T) &\to S(R) \\
K &\mapsto f^{-1}(K)
.\end{align*}

It follows that

\(H \leq R \implies F(H) \leq \operatorname{im}f\), by the subring test
- Subring test: contains 1, closed under multiplication/subtraction.
- Properties of ring homomorphisms: \(f(sa + b) = sf(a) + f(b)\) and \(f(1) = 1\).
- Thus if \(f\) is not surjective, \(F\) is not surjective either.
\(K \leq T \implies \ker f\subseteq F^{-1}(K)\).
- Follows because subrings contain 0, and \(H\in \ker F\implies f(H) = 0_T \in K\).
- Thus if there is any subring \(H\) that doesn’t contain \(\ker f\), \(F^{-1}\) is not surjective.

The claim is that if you restrict to

\(S'(R) \coloneqq\left\{{H \leq R {~\mathrel{\Big\vert}~}\ker f \subseteq H}\right\}\) and
\(S'(T) \coloneqq\left\{{K\leq T {~\mathrel{\Big\vert}~}K \subseteq \operatorname{im}f}\right\}\),

this is a bijection.

This follows from the fact that

\((F\circ F^{-1})(K) = K \cap\operatorname{im}f \leq T\)
- No clear motivation for why it’s this specific thing, but the inclusions are easy to check.
\((F^{-1}\circ F)(H) = \left\langle{H, \ker f}\right\rangle \leq S\).
- Inclusions easy to check, need to take subring generated since \(F(H)\) is a pushforward/direct image, which don’t preserve sub-structures in general.

So we take the projection \(f = \pi: R \to R/{\mathfrak{a}}\), then

\(K \subseteq \operatorname{im}\pi \implies K \cap\operatorname{im}\pi = K \implies (F\circ F^{-1})(K) = K\),
\(\ker \pi \subseteq H \implies \left\langle{H, \ker \pi}\right\rangle = H \implies (F^{-1}\circ F)(H) = H\),

so both directions are surjections. Restricting to just those subrings that are ideals preserves this bijection. Moreover, \(\ker \pi = {\mathfrak{a}}\) so \(S'(R)\) is the set of ideals containing \({\mathfrak{a}}\), and \(\operatorname{im}\pi = R/{\mathfrak{a}}\), so \(S'(T)\) is the set of ideals of the quotient.

Proposition 1.2: TFAE

\(R\) is a field
\(R\) is simple, i.e. the only ideals of \(R\) are \(0, R\).
Every nonzero homomorphism \(\phi: R\to S\) for \(S\) an arbitrary ring is injective.

Proof:

Lemma: \(I {~\trianglelefteq~}R\) and \(1\in I \implies I = R\). This is because \(RI \subseteq I\), and \(r\in R \implies r\cdot 1 \in I \implies r\in I \implies R \subseteq I\).

\(1 \implies 2\):

Let \(0\neq I {~\trianglelefteq~}R\) for \(R\) a field, then pick any \(x\in I\), since \(x^{-1}\in R\), we have \(x^{-1}x = 1 \in I \implies I = R\).

\(\not 1 \implies \not 2\):

If \(R\) is not a field, pick a non-unit element \(r\); then \((r) {~\trianglelefteq~}R\) is a proper ideal.

\(2\implies 3\):

\(\ker \phi {~\trianglelefteq~}R\) is an ideal, so \(\ker \phi = 0\).

\(3 \implies 2\):

Take \({\mathfrak{a}}\mathrel{\reflectbox{\)\mathrel{\ooalign{\raisebox{-0.5ex}{\reflectbox{\rotatebox{90}{\(\nshortmid\)}}}\cr\(\triangleright\)\cr}\mkern-3mu}\(}}R\) a proper ideal and let \(S = R/{\mathfrak{a}}\) with \(\phi: R\to S\) the projection. \(\phi\) is a bijection, since it’s always a surjection and assumed injective. So \(R \cong S = R/{\mathfrak{a}}\), forcing \({\mathfrak{a}}= (0)\).

Proposition: If \({\mathfrak{m}}{~\trianglelefteq~}R\) is maximal iff \(R/{\mathfrak{m}}\) is a field.

Proof:

\(R/{\mathfrak{m}}\) is a field \(\iff\) \(R/{\mathfrak{m}}\) is simple \(\iff\) there are no nontrivial ideals \({\mathfrak{a}}\) such that \({\mathfrak{m}}\subset {\mathfrak{a}}\) (correspondence) \(\iff\) \({\mathfrak{m}}\) is maximal.

Proposition: \({\mathfrak{p}}{~\trianglelefteq~}R\) is prime iff \(R/{\mathfrak{p}}\) is a domain.

Proof:

\(\implies\):

WLOG, \((x + {\operatorname{pr}})(y+{\operatorname{pr}}) = xy + {\operatorname{pr}}= 0 \iff xy\in {\operatorname{pr}}\iff x\in {\operatorname{pr}}\iff (x+{\operatorname{pr}}) = 0\).

\(\impliedby\):

WLOG, \(xy\in {\operatorname{pr}}\implies (x+{\operatorname{pr}})(y+{\operatorname{pr}}) = 0 \implies x+{\operatorname{pr}}= 0 \implies x\in {\operatorname{pr}}\).

Proposition: Maximal ideals are prime.

Proof: Let \({\mathfrak{m}}{~\trianglelefteq~}A\) be maximal, then \(R/{\mathfrak{m}}\) is simple and thus a field, so \({\mathfrak{m}}\) is prime.

Proposition: Prime does not imply maximal in general.

Proof: Take \((0) \in {\mathbb{Z}}\), then \(ab = 0 \implies a=0\) or \(b=0\), so this is prime. It is not maximal, because \((0) \in (n)\) for any \(n\).

Theorem 1.3: Every ring \(R\) has a nontrivial nonzero maximal ideal, and every ideal is contained in a maximal ideal.

Proof: Take the sublattice of the ideal lattice given by proper ideals; every chain has an upper bound given by union, so apply Zorn’s lemma. Similarly, for a fixed \({\mathfrak{a}}\), take the sublattice of ideals containing \({\mathfrak{a}}\).

Corollary 1.5: Every non-unit of \(R\) is contained in a maximal ideal.

Proof: ?

Proposition 1.6: If \(A\setminus {\mathfrak{m}}\subset R^{\times}\), then \(A\) is a local ring with \({\mathfrak{m}}\) its maximal ideal. If \({\mathfrak{m}}\) is maximal and \(1+m \in R^{\times}\) for all \(m\in {\mathfrak{m}}\), then \(A\) is a local ring.

Proof: ?

Proposition: If \(f\in k[x_1, \cdots x_n]\) is irreducible over \(k\), then \((f)\) is prime.

Proposition: \({\mathbb{Z}}\) is a PID, and \((p)\) is prime iff \(p\) is zero or a prime number, and every such ideal is maximal.

Proposition: \(k[\left\{{x_i}\right\}]\) has maximal ideals that are not principal iff \(n>1\).

Exercise: Characterize the maximal and prime ideals of \(k[x_1, \cdots, x_n]\)? Is this a field, domain, PID, UFD, a local ring, …?

Proposition: Every nonzero prime ideal in a PID is maximal.

Proof: ?

Definition: The set \({\operatorname{nil}}(A)\) of all nilpotent elements in a ring \(A\) is the nilradical of \(A\). The set \(J(A) = \cap_{{\mathfrak{m}}\in \operatorname{Spec}_{\text{max}}(A)} {\mathfrak{m}}\) is the Jacobson radical.,

Proposition 1.7: \({\operatorname{nil}}(A) {~\trianglelefteq~}R\) is an ideal and \(A/\mathfrak{R}\) has no nonzero nilpotent elements.

Proof: ?

Proposition 1.8: \({\operatorname{nil}}(A) = \cap{{\operatorname{pr}}\in \operatorname{Spec}(A)} {\operatorname{pr}}\) is the intersection of all prime ideals of \(A\).

Proof: ?

Proposition 1.9: \(x\in J(A)\) iff \(1-xa \in A^{\times}\) for all \(a\in A\).

Proposition: If \((m), (n) {~\trianglelefteq~}{\mathbb{Z}}\) then \((m)\cap(n) = (\gcd(m, n))\) and \((m)(n) = (mn)\).

Exercise: If \({\mathfrak{a}}{~\trianglelefteq~}k[x_1, \cdots, x_m]\), characterize \({\mathfrak{a}}^n\).

Exercise: Show that \({\mathfrak{a}}, {\mathfrak{b}}{~\trianglelefteq~}A\) are coprime iff there exist \(a\in {\mathfrak{a}}, b\in {\mathfrak{b}}\) such that \(a+b = 1\).

Proposition 1.10: Let \(\left\{{mfa_i}\right\} {~\trianglelefteq~}A\) be a family of ideals and define \(\phi: A \to \prod A/{\mathfrak{a}}_i\).

If \(\left\{{{\mathfrak{a}}_i}\right\}\) are pairwise coprime, then \(\prod {\mathfrak{a}}_i = \cap{\mathfrak{a}}_i\)
\(\phi\) is surjective iff \(\left\{{{\mathfrak{a}}_i}\right\}\) are pairwise coprime.
\(\phi\) is injective iff \(\cap{\mathfrak{a}}_i = (0)\).

Exercise: Show that the union of ideals is not necessarily an ideal.

Proposition 1.11:

Let \(\left\{{{\operatorname{pr}}_i}\right\}\) be a set of prime ideals and let \({\mathfrak{a}}\in \cup{\operatorname{pr}}\). Then \({\mathfrak{a}}\subseteq {\operatorname{pr}}_i\) for some \(i\).
Let \(\left\{{{\mathfrak{a}}_i}\right\}\) be ideals and \({\operatorname{pr}}\supseteq \cap{\mathfrak{a}}_i\) be prime. \({\operatorname{pr}}\supseteq {\mathfrak{a}}_i\) for some \(i\), and if \({\operatorname{pr}}= \cap{\mathfrak{a}}_i\), then \({\operatorname{pr}}= {\mathfrak{a}}_i\) for some \(i\).

Exercise: Let \(A = {\mathbb{Z}}\), and characterize the ideal quotient \((m : n)\).

Exercise 1.12:

\({\mathfrak{a}}\subseteq ({\mathfrak{a}}: {\mathfrak{b}})\)
\(({\mathfrak{a}}: {\mathfrak{b}}){\mathfrak{b}}\subseteq {\mathfrak{a}}\)
\((({\mathfrak{a}}:{\mathfrak{b}}): {\mathfrak{c}}) = ({\mathfrak{a}}: {\mathfrak{b}}{\mathfrak{c}}) = (({\mathfrak{a}}:{\mathfrak{c}}): {\mathfrak{b}})\)
\((\cap{\mathfrak{a}}_i: {\mathfrak{b}}) = \cap({\mathfrak{a}}_i: {\mathfrak{b}})\)
\(({\mathfrak{a}}: \sum {\mathfrak{b}}_i) = \cap({\mathfrak{a}}: {\mathfrak{b}}_i)\)

Proposition: For \({\mathfrak{a}}{~\trianglelefteq~}A\), \(\sqrt{{\mathfrak{a}}}\) is an ideal.

Exercise 1.13:

\(\sqrt{\mathfrak{a}}\supseteq {\mathfrak{a}}\)
\(\sqrt{\sqrt {\mathfrak{a}}} = \sqrt {\mathfrak{a}}\)
\(\sqrt{{\mathfrak{a}}{\mathfrak{b}}} = \sqrt{{\mathfrak{a}}\cap{\mathfrak{b}}} = \sqrt{\mathfrak{a}}\cap\sqrt{\mathfrak{b}}\)
\(\sqrt{\mathfrak{a}}= (1) \iff {\mathfrak{a}}= (1)\)
\(\sqrt{{\mathfrak{a}}+ {\mathfrak{b}}} = \sqrt{\sqrt{\mathfrak{a}}+ \sqrt {\mathfrak{b}}}\).
For \({\mathfrak{p}}\) prime, \(\sqrt{{\mathfrak{p}}^n} = {\mathfrak{p}}\) for all \(n\geq 1\).

Proposition 1.14: \(\sqrt{{\mathfrak{a}}} = \cap_{{\mathfrak{a}}\subseteq {\operatorname{pr}}\in \operatorname{Spec}(A)} {\operatorname{pr}}\)

Proposition 1.15: Let \(D\) be the set of zero-divisors in \(A\). Then \(D = \cup_{x\neq 0}\sqrt{\operatorname{Ann}(x)}\).

Exercise: Let \((m) {~\trianglelefteq~}{\mathbb{Z}}\) where \(m = \prod p_i^{k_i}\), and show that \(\sqrt{(m)} = (p_1 p_2 \cdots) = \cap(p_i)\).

Proposition 1.16: If \(\sqrt{\mathfrak{a}}, \sqrt {\mathfrak{b}}\) are coprime then \({\mathfrak{a}}, {\mathfrak{b}}\) are coprime.

Exercise: Show that if \(f: A\to B\) and \({\mathfrak{a}}{~\trianglelefteq~}A\), it is not necessarily the case that \(f({\mathfrak{a}}) {~\trianglelefteq~}B\).

Exercise: Show that if \({\mathfrak{b}}\) is prime then \(A\cdot f^{-1}({\mathfrak{b}})\) is prime, but if \({\mathfrak{a}}\) is prime then \(B\cdot f({\mathfrak{a}})\) need not be prime.

Exercise: Write \({\mathfrak{a}}^e \coloneqq\left\langle{f({\mathfrak{a}})}\right\rangle\) and \({\mathfrak{b}}^c = \left\langle{f^{-1}({\mathfrak{b}})}\right\rangle\). Let \(f: {\mathbb{Z}}\to {\mathbb{Z}}[i]\) be the inclusion, and show that

\((2)^e = \left\langle{(1+i)^2}\right\rangle\), which is not prime in \({\mathbb{Z}}[i]\)
(Nontrivial) If \(p = 1\operatorname{mod}4\), then \({\operatorname{pr}}^e\) is the product of two distinct prime ideals
If \(p=3\operatorname{mod}4\) then \({\operatorname{pr}}^e\) is prime.

Proposition: Let \(C = \left\{{{\mathfrak{b}}^c {~\mathrel{\Big\vert}~}{\mathfrak{b}}{~\trianglelefteq~}B}\right\}\) and \(E = \left\{{{\mathfrak{a}}^e {~\mathrel{\Big\vert}~}{\mathfrak{a}}{~\trianglelefteq~}A}\right\}\). Then

\({\mathfrak{a}}\subseteq {\mathfrak{a}}^{ec}\) and \({\mathfrak{b}}\supset {\mathfrak{b}}^{ce}\),
\({\mathfrak{b}}^c = {\mathfrak{b}}^{cec}\) and \({\mathfrak{a}}^e = {\mathfrak{a}}^{ece}\)
\(C = \left\{{{\mathfrak{a}}{~\trianglelefteq~}A {~\mathrel{\Big\vert}~}{\mathfrak{a}}^{ec} = {\mathfrak{a}}}\right\}\) and \(E = \left\{{{\mathfrak{b}}{~\trianglelefteq~}B {~\mathrel{\Big\vert}~}{\mathfrak{b}}^{ce} = {\mathfrak{b}}}\right\}\).
The map \(\phi: C\to E\) given by \(\phi({\mathfrak{a}}) = {\mathfrak{a}}^{ec}\) is a bijection with inverse \({\mathfrak{b}}\mapsto {\mathfrak{b}}^c\).
If \({\mathfrak{a}}\in C\) then \({\mathfrak{a}}= {\mathfrak{b}}^c = {\mathfrak{b}}^{cec} = {\mathfrak{a}}^{ec}\), and if \({\mathfrak{a}}= {\mathfrak{a}}^{ec}\) then \({\mathfrak{a}}\) is the contraction of \({\mathfrak{a}}^e\).

Exercise 1.18:

[

\begin{array}{ll}{\left(\mathfrak{a}_{1}+\mathfrak{a}_{2}\right)^{\mathfrak{e}}=\mathfrak{a}_{1}^{\mathfrak{e}}+\mathfrak{a}_{2}^{\mathfrak{e}},} & {\left(\mathfrak{b}_{1}+\mathfrak{b}_{2}\right)^{c} \geq \mathfrak{b}_{1}^{\mathfrak{c}}+\mathfrak{b}_{2}^{\mathfrak{c}}} \\ {\left(\mathfrak{a}_{1} \cap \mathfrak{a}_{2}\right)^{e} \subseteq \mathfrak{a}_{1}^{\mathfrak{e}} \cap \mathfrak{a}_{2}^{e},} & {\left(\mathfrak{b}_{1} \cap \mathfrak{b}_{2}\right)^{\mathfrak{c}}=\mathfrak{b}_{1}^{\mathfrak{c}} \cap \mathfrak{b}_{\mathfrak{z}}^{\mathfrak{c}}} \\ {\left(\mathfrak{a}_{1} \mathfrak{a}_{2}\right)^{\mathfrak{e}}=\mathfrak{a}_{1}^{\mathfrak{e}} \mathfrak{a}_{2}^{\mathfrak{e}},} & {\left(\mathfrak{b}_{1} \mathfrak{b}_{2}\right)^{\mathfrak{c}} \supseteq \mathfrak{b}_{1}^{\mathfrak{c}} \mathfrak{b}_{2}^{\mathfrak{c}}} \\ {\left(\mathfrak{a}_{1}: \mathfrak{a}_{2}\right)^{\mathfrak{e}} \subseteq\left(\mathfrak{a}_{1}^{\mathfrak{e}}: \mathfrak{a}_{2}^{\mathfrak{e}}\right),} & {\left(\mathfrak{b}_{1}: \mathfrak{b}_{2}\right)^{\mathfrak{c}} \subseteq\left(\mathfrak{b}_{1}^{\mathfrak{c}}: \mathfrak{b}_{2}^{\mathfrak{c}}\right)} \\ {r(\mathfrak{a})^{e} \subseteq r\left(\mathfrak{a}^{e}\right),} & {r(\mathfrak{b})^{c}=r\left(\mathfrak{b}^{c}\right)}\end{array}

Ch 1 Notes

Chapter 1