# Ring Theory

A subset $$S\subseteq R$$ is a subring iff

• $$(S, +)$$ forms an abelian subgroup (so closed under addition and contains inverses)
• $$(S, \cdot)$$ forms a submonoid (so closed under multiplication)

• $$I + J = \left\{{i+j {~\mathrel{\Big\vert}~}i\in I, j\in J}\right\} = \left\langle{I, J}\right\rangle$$ is the smallest ideal containing $$I$$ and $$J$$.
• $$IJ = \left\{{\sum_{k\leq N} x_k y_k {~\mathrel{\Big\vert}~}x_k\in I, y_k \in J, N\in {\mathbf{Z}}^{\geq 0}}\right\}$$ is the ideal generated by all finite sums of products.
• $$I \cap J$$ is an ideal, $$I\cup J$$ is generally not an ideal
• Ideals are comaximal if $$I + J = \left\langle{ 1 }\right\rangle$$.
• If $$I+J = \left\langle{ 1 }\right\rangle$$ then $$I \cap J = IJ$$.

The ideal generated by $$\left\{{a, b}\right\}$$ is defined as \begin{align*} \left\langle{a, b}\right\rangle \coloneqq Ra + Rb \coloneqq\left\{{ r_1 a + r_2 b {~\mathrel{\Big\vert}~}r_i \in R}\right\} .\end{align*}

More generally for a set $$S = \left\{{s_k}\right\}$$, \begin{align*} \left\langle{S}\right\rangle \coloneqq\sum_{k=1}^{{\sharp}S} Rs_k \coloneqq\left\{{ \sum r_k s_k {~\mathrel{\Big\vert}~}r_k\in R, s_k\in S}\right\} .\end{align*}

• $$\left\langle{p, q}\right\rangle = \left\langle{\gcd(p, q)}\right\rangle {~\trianglelefteq~}{\mathbf{Z}}$$.

## Isomorphism Theorems

These are all basically the same for modules.

For any ring morphism $$f:A\to B$$ there is SES of rings \begin{align*} 0 \to \ker f \to A \to \operatorname{im}(f) \to 0 ,\end{align*} and thus $$A/\ker f \cong \operatorname{im}f$$. If $$f$$ is surjective, then $$A/\ker f \cong B$$.

• $$\ker \phi \in \operatorname{Id}(A)$$
• $$\operatorname{im}\phi \leq B$$ is a subring (not necessarily an ideal)
• $$R/\ker \phi \cong \operatorname{im}\phi$$.

Let $$R\in \mathsf{Ring}, S\leq R, I\in \operatorname{Id}(R)$$, then there is an isomorphism: \begin{align*} {S+I \over I} {\xrightarrow{\sim}}{S\over S \cap I} .\end{align*}

Where it’s also true that this statement makes sense:

• $$S+I \leq R$$ is a subring.
• $$S \cap I {~\trianglelefteq~}S$$

For $$I\in \operatorname{Id}(R)$$, the canonical quotient map $$\phi: R \to R/I$$ induces a bijective correspondence: \begin{align*} \left\{{\substack{ J \in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}J\supseteq I }}\right\} &\rightleftharpoons \operatorname{Id}(R/I) \\ J \coloneqq\phi^{-1}(\overline{J}) &\mapsfrom \overline{J} \\ J &\mapsto \overline{J} \coloneqq\phi(J) ,\end{align*} where $$\phi: R\to R/I$$ is the canonical quotient morphism.

• If $$S, I \in \operatorname{Id}(R)$$ with $$S$$ containing $$I$$ then \begin{align*} S/I \leq R/I .\end{align*}

• Every ideal in $$\operatorname{Id}(R/I)$$ is of the form $$\overline{S} \coloneqq S/I$$ for some $$S\in \operatorname{Id}(R)$$ containing $$I$$.

• If $$I, J \in \operatorname{Id}(R)$$ with $$I \subseteq J \subseteq R$$ then there is an isomorphism \begin{align*} {R/I \over J/I} {\xrightarrow{\sim}}{R\over J} .\end{align*}

Moreover, $$A\leq R$$ is a subring containing $$I$$ iff $$A/I \in \operatorname{Id}(R/I)$$.

Show that if $$J\in \operatorname{Id}(R)$$ (with $$J\supseteq I$$) is radical/prime/maximal iff $$\overline{J} \in \operatorname{Id}(R/I)$$ is radical/prime/maximal.

## Important Techniques

$$R \in \mathsf{Field}\iff \operatorname{Id}(R) = \left\{{ 0, R }\right\}$$.

$$\implies$$: If $$0\neq x\in I{~\trianglelefteq~}R$$, using that $$R^{\bullet}= R^{\times}$$, $$x$$ is a unit. So $$x^{-1}\in R$$, and $$xx^{-1}\coloneqq 1 \in I$$ so $$I = R$$.

$$\impliedby$$: Let $$x\in R^{\bullet}$$, then $$Rx = R$$ so $$1\in Rx$$ and $$1=rx$$ for some $$r\in R$$. This forces $$x=r^{-1}$$.

• $$R/{\mathfrak{m}}$$ is a field $$\iff {\mathfrak{m}}\in \operatorname{mSpec}(R)$$ is maximal.
• $$R/{\mathfrak{p}}$$ is an integral domain $$\iff {\mathfrak{p}}\in \operatorname{Spec}(R)$$ is prime.
• $$R/J$$ is reduced $$\iff J$$ is radical.

Use the ideal correspondence theorem: $$\operatorname{Id}(R/{\mathfrak{m}})$$ are ideals of $$R$$ containing $${\mathfrak{m}}$$: \begin{align*} &\quad R/{\mathfrak{m}}\in \mathsf{Field}\\ &\iff \not\exists J/{\mathfrak{m}}\in \operatorname{Id}(R/{\mathfrak{m}})^{\bullet}\text{ such that } J \in \operatorname{Id}(R) \\ &\iff \not\exists {\mathfrak{m}}\subsetneq J \subsetneq R \\ &\iff J\in \operatorname{mSpec}(R) .\end{align*}

$$\impliedby$$: Show $$xy = 0$$ with $$x\neq 0$$ forces $$y=0$$. Let $$x, y\in {\mathfrak{p}}\in \operatorname{Spec}R$$, so $$x = a + I, y = b + I$$ for some $$a,b\in R$$. If $$xy=0 \operatorname{mod}{\mathfrak{p}}$$ with $$y\neq 0\operatorname{mod}{\mathfrak{p}}$$, we can check \begin{align*} xy = (a+{\mathfrak{p}})(b+{\mathfrak{p}}) \coloneqq(ab) + {\mathfrak{p}}= 0 + {\mathfrak{p}}\implies ab\in {\mathfrak{p}} .\end{align*} Since $${\mathfrak{p}}$$ is prime and $$x\neq 0 \implies a\not\in{\mathfrak{p}}$$, so $$b\in {\mathfrak{p}}$$. But then \begin{align*} y \coloneqq b + {\mathfrak{p}}= 0 + {\mathfrak{p}}= 0 \operatorname{mod}{\mathfrak{p}} .\end{align*}

$$\implies$$: Let $$a,b\in R$$ with $$xy\in {\mathfrak{p}}$$, we want to show that if $$x\not \in {\mathfrak{p}}$$ then $$y\in {\mathfrak{p}}$$. Note $$x\not \in {\mathfrak{p}}\iff x \cong 0 \operatorname{mod}{\mathfrak{p}}$$. Setting $$x\coloneqq a + {\mathfrak{p}}, y\coloneqq b + {\mathfrak{p}}$$ yields \begin{align*} xy \coloneqq(a+{\mathfrak{p}})(b+{\mathfrak{p}}) \coloneqq ab + {\mathfrak{p}}= 0 \operatorname{mod}{\mathfrak{p}} .\end{align*} Since $$R/{\mathfrak{p}}$$ is a domain, assuming $$x\neq 0 \operatorname{mod}{\mathfrak{p}}$$ we have $$y = 0 \operatorname{mod}{\mathfrak{p}}$$, so $$y\in {\mathfrak{p}}$$.

Note that this yields a quick proof that $$\operatorname{mSpec}R \subseteq \operatorname{Spec}R$$, using that $$\mathsf{Field}\leq \mathsf{IntDomain}$$: \begin{align*} I \text{ maximal } \iff R/I \in \mathsf{Field}{\color{blue} \implies } R/I \in \mathsf{IntDomain} \iff I \text{ prime} .\end{align*}

If $${\mathfrak{m}}$$ is maximal and $$x \in R\setminus{\mathfrak{m}}$$ then $${\mathfrak{m}}+ Rx = R = \left\langle{ 1}\right\rangle$$.

Notation:

• $$\left\langle{ a }\right\rangle \coloneqq Ra \coloneqq\left\{{ ra {~\mathrel{\Big\vert}~}r \in R }\right\}$$ is the ideal generated by a single element.
• $$R = \left\langle{ 1 }\right\rangle$$ is equivalently the ideal generated by 1.

### Basics

A ring is a triple $$(R, +, \cdot) \in \mathsf{CRing}$$ such that

• $$(R, +)\in {\mathsf{Ab}}{\mathsf{Grp}}$$,
• $$(R, \cdot) \in \mathsf{Mon}$$
• Distributivity: $$a(b+c) = ab + ac$$.

Some of the most important examples of rings:

• The usual suspects: $${\mathbf{Z}}, {\mathbf{Q}}$$
• Their analogs: number fields $$K \coloneqq{\mathbf{Q}}(\zeta)$$, their rings of integers $${\mathbf{Z}}_K$$ or $${\mathcal{O}}_K$$,
• Gaussian integers $${\mathbf{Z}}(i)$$
• Fields $$k = {{ \mathbf{F} }_{p^n}}, {\mathbf{R}}$$
• Fraction fields of rings $$\operatorname{ff}(R)$$, e.g. $$\operatorname{ff}({\mathbf{Z}}) = {\mathbf{Q}}$$.
• Polynomial rings $R { \left[ \scriptstyle {{ {x}_1, {x}2, \cdots, {x}{n}}} \right] }$, particularly for $$R=k$$ a field
• Power series rings $R {\llbracket { {x}_1, {x}2, \cdots, {x}{n}} \rrbracket }$.
• Formal power series rings $R {\llbracket { {x}_1, {x}2, \cdots, {x}{n}} \rrbracket }$.
• $${ {\mathbf{Z}}_{\widehat{p}} }\coloneqq\left\{{a/b {~\mathrel{\Big\vert}~}p \notdivides b }\right\}$$ the ring of $$p{\hbox{-}}$$adic integers
• Rings of germs, e.g. $$C^\infty(X, Y)$$ where $$f\sim g$$ iff there exists some $$U \subseteq X$$ with $${ \left.{{f}} \right|_{{U}} } = { \left.{{g}} \right|_{{U}} }$$.

A morphism $$f\in \mathsf{CRing}(X, Y)$$ satisfies:

• $$f(1_X) = 1_Y$$
• $$f(a(b+c)) = f(a)f(b) + f(a)f(c)$$

Important notes:

• $$\ker f \coloneqq f^{-1}(\left\{{0}\right\})$$.
• A bijective ring morphisms is automatically an isomorphism in $$\mathsf{CRing}$$.
• $$\ker f {~\trianglelefteq~}X$$ is an ideal, but $$\operatorname{im}f \leq Y$$ is only a subring in general.
• For any ideal $$I{~\trianglelefteq~}R$$ there is a quotient map $$R\to R/I$$, it’s useful to write cosets as $$a+I$$.
• For quotients, $$x \equiv y \operatorname{mod}I \iff x-y \in I$$.

An ideal $$I{~\trianglelefteq~}R$$ is a subset where $$(I, +) \leq (R, +) \in {\mathsf{Grp}}$$ is a subgroup and for $$x\in R, i\in I$$, $$xi \in I$$. Equivalently,

• $$RI \subseteq I$$
• $$I + I \subseteq I$$

Note that $$0$$ is in every ideal.

Using that every ring has a $${}_{{\mathbf{Z}}}{\mathsf{Mod}}$$ structure, the characteristic of a ring $$R$$ is the smallest $$n$$ such that $$n\curvearrowright 1_R = 0_R$$, i.e. $$\sum_{i=1}^n 1_R = 0_R$$.

### Elements

An element $$r\in R$$ is divisible by $$q \in R$$ if and only if there exists some $$c \in R$$ such that $$r = qc$$. In this case, we sometimes write $$q\divides r$$.

An element $$r\in R$$ is a unit if $$r\divides 1$$: there exists an $$s\in R$$ such that $$rs = sr = 1$$. Then $$r^{-1}\coloneqq s$$ is uniquely determined, and the set of units $$(R^{\times}, \cdot) \in {\mathsf{Ab}}{\mathsf{Grp}}$$ forms a group.

An element $$r\in R$$ is irreducible iff \begin{align*} r=ab \implies a \in R^{\times}\text{ or } b\in R^{\times} \end{align*}

An element $$p\in R$$ is prime iff \begin{align*} a,b \in R^{\times}\setminus\left\{{0}\right\}, \quad ab\divides p \implies a\divides p \text{ or } b\divides p .\end{align*}

If $$R$$ is an integral domain, prime $$\implies$$ irreducible. If $$R$$ is a UFD, then irreducible $$\implies$$ prime, so this is an iff.

$$a, b\in R$$ are associates iff there exists a $$u\in R^{\times}$$ such that $$a = ub$$. Equivalently, $$a\divides b$$ and $$b\divides a$$.

### Ideals

• $$\operatorname{Id}({\mathbf{Z}}) = \left\{{ \left\langle{m}\right\rangle {~\mathrel{\Big\vert}~}m \in {\mathbf{Z}}^{\geq 0}}\right\}$$
• $$\operatorname{mSpec}{\mathbf{Z}}= \left\{{ \left\langle{p}\right\rangle {~\mathrel{\Big\vert}~}p\neq 0 \text{ is prime} }\right\}$$
• $$\operatorname{Spec}{\mathbf{Z}}= \operatorname{mSpec}{\mathbf{Z}}\cup\left\{{ \left\langle{0}\right\rangle }\right\}$$.
• For $$k$$ a field and $$f\in k[x_1, \cdots, x_{n}]$$ irreducible, $$\left\langle{f}\right\rangle \in \operatorname{Spec}k[x_1, \cdots, x_{n}]$$.
• $${\mathfrak{m}}\coloneqq\left\{{ f = \sum_I a_I x^I\in k[x_1, \cdots, x_{n}]{~\mathrel{\Big\vert}~}a_0 = 0 }\right\} \in \operatorname{mSpec}k[x_1, \cdots, x_{n}]$$ (i.e. this is the ideal of polynomials with no constant term).

If $$I{~\trianglelefteq~}R$$ is a proper ideal $$\iff I$$ contains no units.

$$r\in R^{\times}\cap I \implies r^{-1}r \in I \implies 1\in I \implies x\cdot 1 \in I \quad \forall x\in R$$.

If $$I_1 \subseteq I_2 \subseteq \cdots$$ are ideals then $$\cup_j I_j$$ is an ideal.

An ideal $$I{~\trianglelefteq~}R$$ is irreducible if it can not be written as the intersection of two larger ideals, i.e. there are not $$J_1, J_2 \supseteq I$$ such that $$J_1 \cap J_2 = I$$.

$${\mathfrak{p}}$$ is a prime ideal $$\iff$$ \begin{align*} ab\in {\mathfrak{p}}\implies a\in {\mathfrak{p}}\text{ or } b\in {\mathfrak{p}} .\end{align*}

In $$R$$ a UFD, an element $$r\in R$$ is prime $$\iff r$$ is irreducible.

For $$R$$ an integral domain, prime $$\implies$$ irreducible, but generally not the converse. Take $$R \coloneqq k[x, y]/\left\langle{x^2-y^3}\right\rangle \cong k[x^2, y^3]$$, which is a domain, But here $$[x^2] = [y^3]$$ as equivalence classes where $$[y^3]$$ is irreducible since every element in $$r\in R$$ has $$\deg_y(r) = 0,3,6,\cdots$$. But $$[y^3]$$ is not prime since it divides $$[x^2]$$ but doesn’t divide $$[x]$$.

The prime spectrum (or just the spectrum) of $$R$$ is defined as \begin{align*} \operatorname{Spec}(R) = \left\{{{\mathfrak{p}}{~\trianglelefteq~}R {~\mathrel{\Big\vert}~}{\mathfrak{p}}\text{ is prime}}\right\} .\end{align*}

An ideal $${\mathfrak{m}}$$ is maximal iff whenever $$I{~\trianglelefteq~}R$$ with $${\mathfrak{m}}\subsetneq I$$ a proper containment then $$I = R$$.

Some examples. Reminder: maximal always implies prime, and for PIDs, prime and nonzero implies maximal. Maximals quotient to fields, primes to domains.

• Prime and maximal:
• $$p{\mathbf{Z}}\in \operatorname{Id}({\mathbf{Z}})$$. Maximal (and thus prime) since $${\mathbf{Z}}/p$$ is a field and a domain.
• $$\left\langle{2, x}\right\rangle \in \operatorname{Id}({\mathbf{Z}}[x])$$. Maximal (and thus prime) $${\mathbf{Z}}[x]/\left\langle{2, x}\right\rangle \cong {\mathbf{Z}}/2$$ is a field and a domain.
• Prime but not maximal:
• $$\left\langle{0}\right\rangle \in \operatorname{Id}({\mathbf{Z}})$$, since $$m{\mathbf{Z}}\supseteq\left\langle{0}\right\rangle$$ for any $$m$$.
• $$\left\langle{x}\right\rangle \in R[x]$$ over any integral domain since $$R[x]/\left\langle{x}\right\rangle \cong R$$ is a domain (making it maximal), but $$R$$ can be chosen not to be a field (making it non-prime).
• Not prime, not maximal:
• $$m{\mathbf{Z}}\in \operatorname{Id}({\mathbf{Z}})$$, since $$m$$ composite implies $${\mathbf{Z}}/m$$ is not a domain since it has nonzero zero divisors. For example, in $${\mathbf{Z}}/6$$, $$[3]$$ is a zero divisors since $$[2][3] = 0$$.
• Useful examples:
• $$\operatorname{mSpec}{\mathbf{Z}}= \left\{{p{\mathbf{Z}}}\right\}$$ and $$\operatorname{Spec}{\mathbf{Z}}= \left\{{p{\mathbf{Z}}}\right\} \cup\left\langle{0}\right\rangle$$.
• $$\operatorname{mSpec}{\mathbf{C}}[x] = \left\{{x-a {~\mathrel{\Big\vert}~}a\in {\mathbf{C}}}\right\}$$, since over a PID $$\left\langle{\alpha}\right\rangle$$ is maximal iff $$\alpha$$ is irreducible, and over $${\mathbf{C}}$$ irreducibles are degree 1.
• $$\operatorname{mSpec}k[x_1, \cdots, x_{n}] = \left\{{\left\langle{x-a_1, x-a_2, \cdots, x-a_n}\right\rangle {~\mathrel{\Big\vert}~}a_k \in k}\right\}$$.
• A ring with no maximal ideals: the Prüfer $$p{\hbox{-}}$$group $${\mathbf{Z}}(p^\infty) = \left\{{\zeta_{p^k}}\right\}_{k=1}^{\infty}$$ with the trivial ring structure $$xy = 0$$. The subgroups are $$H_k \coloneqq\left\{{\zeta_{p^k}}\right\}$$, which form an increasing chain that doesn’t stabilize.

The max spectrum of $$R$$ is defined as \begin{align*} \operatorname{mSpec}(R) = \left\{{{\mathfrak{m}}{~\trianglelefteq~}R {~\mathrel{\Big\vert}~}{\mathfrak{m}}\text{ is maximal}}\right\} .\end{align*}

$$3\in {\mathbf{Z}}[\sqrt{-5}]$$. Check norm to see irreducibility, but $$3 \divides 9 = (2+\sqrt{-5})(2-\sqrt{-5})$$ and doesn’t divide either factor.

Maximal ideals of $$R[x]$$ are of the form $$I = (x - a_i)$$ for some $$a_i \in R$$.

## Types of Rings

A division ring is any (potentially noncommutative) ring $$R$$ for which $$R\setminus\left\{{0}\right\}\subset {\mathbf{R}}^{\times}$$, i.e. every nonzero element is a unit and thus has a multiplicative inverse.

An element $$r\in R$$ is a zero-divisor iff there exists an $$a\in R\setminus\left\{{0}\right\}$$ such that $$ar = ra = 0$$, i.e. $$r\divides 0$$. Equivalently, the map \begin{align*} r\cdot: R &\to R \\ x &\mapsto rx \end{align*} fails to be injective.

A ring is an integral domain if and only if it has no nonzero zero divisors: \begin{align*} a, b\in R\setminus\left\{{0}\right\}, ab = 0 \implies a = 0 \txt{ or } b = 0 .\end{align*}

Examples of integral domains: ${\mathbf{Z}}, k { \left[ \scriptstyle {{ {x}_1, {x}2, \cdots, {x}{n}}} \right] }$. Non-examples: $${\mathbf{Z}}/6, \operatorname{Mat}(2\times 2; k)$$

A field is a commutative division ring, i.e. every nonzero element is a uni, i.e. every nonzero element is a unit

Show that TFAE:

• $$A\in \mathsf{Field}$$
• $$A$$ is a simple ring, so $$\operatorname{Id}(A) = \left\{{ 0, A }\right\}$$.
• If $$B\in \mathsf{Field}$$ is nonzero then every ring morphism $$A\to B$$ is injective.

Every field is an integral domain, but e.g. $${\mathbf{Z}}$$ is an integral domain that is not a field.

### The Big Ones

An ideal $$I {~\trianglelefteq~}R$$ if principal if there exists an $$a\in R$$ such that $$I = \left\langle{a}\right\rangle$$, i.e. $$I = Ra$$.

A ring $$R$$ is a principal ideal domain iff every ideal is principal.

Let $$R$$ be a PID.

• Show primes are maximal, so $$\operatorname{Spec}R \subseteq \operatorname{mSpec}R$$ and nonzero ideals are prime iff maximal.
• Show that $$R$$ is Noetherian.
• Show that every element is a finite product of irreducibles.
• Show that in a PID, every maximal ideal is generated by an irreducible element.
• Show that not $${\mathbf{Z}}$$ is Noetherian but not Artinian.
• Hint: take a chain $$n{\mathbf{Z}}\supseteq n^2{\mathbf{Z}}\supseteq\cdots$$.

A ring $$R$$ is a unique factorization domain iff $$R$$ is an integral domain and every $$r\in R\setminus\left\{{0}\right\}$$ admits a decomposition \begin{align*} r = u \prod_{i=1}^n p_i \end{align*} where $$u\in R^{\times}$$ and the $$p_i$$ irreducible, which is unique up to associates.

An integral domain $$R$$ is Euclidean if $$R$$ admits a degree function $$d:R\to {\mathbf{Z}}_{\geq 0}$$ such that for all $$x,y\in R$$ there exist $$q,r\in R$$ with $$x = qy + r$$ and either $$f(r) < f(y)$$ or $$r=0$$.

### Others

A ring $$R$$ is Noetherian if the ACC holds: every ascending chain of ideals $$I_1 \leq I_2 \cdots$$ stabilizes in the sense that there exists some $$N$$ such that $$I_N = I_{N+1} = \cdots$$.

A ring $$R$$ is reduced if $$R$$ contains no nonzero nilpotent elements.

A ring $$R$$ is local iff it contains a unique maximal ideal $${\mathfrak{m}}$$, so $$\operatorname{mSpec}R = \left\{{ 0, {\mathfrak{m}}}\right\}$$. As a consequence, there is a uniquely associated residue field $$\kappa \coloneqq R/{\mathfrak{m}}$$.

Show that if $$R$$ is a nonzero ring where every element is either a unit or nilpotent, then $$R$$ is local.

Show that if $$p\in \operatorname{Spec}R$$ then $R \left[ { \scriptstyle { {p}^{-1}} } \right]$ is local.

Suppose $${\mathfrak{m}}\in \operatorname{mSpec}R$$ is a proper maximal ideal. Show that under either of the following two conditions, $$R$$ is local:

• $$R\setminus{\mathfrak{m}}\subseteq R^{\times}$$, so every element of $$R\setminus{\mathfrak{m}}$$ is a unit, or
• $$1 + {\mathfrak{m}}\subseteq R^{\times}$$
• Sketch: $${\mathfrak{m}}$$ must contain every non-unit.
• If $$I \neq R$$ then $$I$$ contains no units, so $$I\subseteq N \coloneqq R\setminus R^{\times}$$, i.e. $$I$$ is contained in the non-units. But $$N \subseteq {\mathfrak{m}}$$ since no element of $${\mathfrak{m}}$$ is a unit and no element of $$R\setminus{\mathfrak{m}}$$ is a non-unit.
• Sketch: show that every $$r\in R\setminus{\mathfrak{m}}$$ is a unit and apply the first part.
• If $$r\in R\setminus{\mathfrak{m}}$$ then $$\left\langle{r, {\mathfrak{m}}}\right\rangle = R = \left\langle{ 1 }\right\rangle$$ so $$rt + m = 1$$ for some $$t\in R, m\in {\mathfrak{m}}$$, so $$rt = 1-m \in 1 + {\mathfrak{m}}\subseteq R^{\times}$$ by assumption. Now apply (1).

A Dedekind domain is an integral domain for which the monoid $$\operatorname{Id}(R)$$ of nonzero ideals of $$R$$ satisfies unique factorization: every ideal can be decomposed uniquely into a product of prime ideals.

Show that a Dedekind domain $$R$$ is a PID iff $$R$$ is a UFD.

A valuation ring is an integral domain $$R$$ such that for every $$x\in \operatorname{ff}(R)$$, $$x\in R$$ or $$x^{-1}\in R$$.

A discrete valuation ring or DVR is a local PID with a unique maximal ideal.

A commutative ring $$R$$ is regular if $$R$$ is Noetherian and for every $$p\in \operatorname{Spec}R$$ the localization $R \left[ { \scriptstyle { {p}^{-1}} } \right]$ is a regular local ring: it has a maximal ideal $${\mathfrak{m}}$$ which admits a minimal generating set of $$n$$ elements where $$n$$ is the Krull dimension of $R \left[ { \scriptstyle { {p}^{-1}} } \right]$.

Motivation: if $$R = {\mathcal{O}}_{X, x}$$ is the ring of germs at $$x$$ of an algebraic variety $$X$$, then $$R$$ is regular iff $$X$$ is nonsingular at $$x$$.

## Comparing and Transporting Ring Types

• $$R$$ a commutative division ring $$\implies R$$ is a field
• $$R$$ a finite integral domain $$\implies R$$ is a field.
• $${ \mathbf{F} }$$ a field $$\iff { \mathbf{F} }[x]$$ is a PID.
• $${ \mathbf{F} }$$ is a field $$\iff { \mathbf{F} }$$ is a commutative simple ring.
• $$R$$ is a UFD $$\iff R[x]$$ is a UFD.
• $$R$$ a PID $$\implies R[x]$$ is a UFD
• $$R$$ a PID $$\implies R$$ Noetherian

Show that $$R[x]$$ a PID $$\iff R$$ is a field.

Hint: take $$r\in R$$, then $$\left\langle{r, x}\right\rangle = \left\langle{f}\right\rangle$$ for some $$f$$. Write $$r = fp$$ and $$x = fq$$ for $$p, q\in R[x]$$, show $$\deg f = 0$$ and $$\deg q = 1$$. Write $$f = c$$ a constant, $$q(x) = ax + b$$ to get $$c(ax+b)=x \implies ca=1 \implies c\in R^{\times}\implies \left\langle{f}\right\rangle = R[x]$$. Conclude by writing $$1= ar_1(x) + xr_2(x)$$, evaluate at $$x=0$$ to get $$a^{-1}= r_1(0)$$.

A polynomial ring over a PID is not necessarily a PID: take $$\left\langle{2, x}\right\rangle {~\trianglelefteq~}{\mathbf{Z}}[x]$$.

Fields $$\subset$$ Euclidean domains $$\subset$$ PIDs $$\subset$$ UFDs $$\subset$$ Integral Domains $$\subset$$ Rings

Sketch proofs of the inclusions:

• Field $$\implies$$ Euclidean: given $$x,y$$ we need to write $$x=qy+r$$, so just take $$q=y^{-1}$$ and $$r=0$$.

• Euclidean $$\implies$$ PID: to divide is to contain, and the Euclidean algorithm terminates to yield a gcd. Alternatively, pick an element $$a\in I$$ of minimal degree. If $$I\neq Ra$$ pick $$b\in Ra$$ that $$a$$ doesn’t divide and write $$b = aq + r$$ with $$d(r) < d(a)$$. Then $$r = b-aq \in I$$. $$\contradiction$$

• PID $$\implies$$ UFD:

• To get existence, use that PIDS are Noetherian, maximals are generated by irreducibles, and irreducibles are prime. Write $$a =a_1 b_1$$ a proper factorization to get a proper containment $$\left\langle{a}\right\rangle \subset \left\langle{a_1}\right\rangle$$ that eventually stabilizes to yield an irreducible factor $$a_r$$. Use the same idea to write $$a$$ as finitely many irreducible factors.
• To get uniqueness, write $$a = \prod p_i = \prod q_i$$ as primes and divide everything over.

• A Euclidean Domain that is not a field: $$k[x]$$ for $$k$$ a field.
• Proof: Use that $$k$$ a field implies $$k[x]$$ is a PID, and PID implies UFD. But this is not a field since the element $$x$$ is not invertible.
• A PID that is not a Euclidean Domain: $${\mathbf{Z}}\left[\frac{1 + \sqrt{-19}}{2}\right]$$.
• Proof: complicated.
• A UFD that is not a PID: $${\mathbf{Z}}[x]$$.
• Proof: $${\mathbf{Z}}$$ a UFD implies $${\mathbf{Z}}[x]$$ is a UFD, but $$\left\langle{2, x}\right\rangle = 2{\mathbf{Z}}[x] + x{\mathbf{Z}}[x] = \left\{{\sum r_ix^i {~\mathrel{\Big\vert}~}r_0 \in 2{\mathbf{Z}}}\right\}$$ is not principal. Why: if $$\left\langle{2, x}\right\rangle = \left\langle{f}\right\rangle$$ and $$f$$ is constant, then every polynomial in this ideal has even coefficients and thus misses $$g(x) \coloneqq x$$. Otherwise, $$\deg f \geq 1$$ and we miss 2, which has degree zero.
• An integral domain that is not a UFD: $${\mathbf{Z}}[\sqrt{-5}]$$
• Proof: $$(2+\sqrt{-5})(2-\sqrt{-5})=9=3\cdot 3$$, where all factors are irreducible (check norm).
• A ring that is not an integral domain: $${\mathbf{Z}}/4$$
• Proof: $$[2]_4$$ is a zero divisor since $$[2]_4 [2]_4 = [0]_4$$.

For an ideal $$I{~\trianglelefteq~}R$$, the radical \begin{align*} \sqrt{I} \coloneqq\left\{{r\in R{~\mathrel{\Big\vert}~}r^n\in I\text{ for some } n\geq 0}\right\} ,\end{align*} so $$x^n \in I \implies x\in \sqrt{I}$$.

An ideal is radical iff $$\sqrt{I} = I$$.

In general, “radical” refers to “bad elements” of some type to be quotiented out, not necessarily $$\sqrt{{-}}$$.

An element $$r\in R$$ is nilpotent if $$r^n = 0$$ for some $$n \in {\mathbf{Z}}^{\geq 0}$$.

The binomial expansion works in any ring: \begin{align*} (a+b)^n = \sum_{k\leq n} {n\choose k} a^k b^{n-k} .\end{align*}

This is useful when considering nilpotents or radicals.

Notation: let $$N$$ or $$N(R)$$ be the set of nilpotents in $$R$$. Let $$ZD$$ or $$ZD(R)$$ be the set of zero divisors. Let $$U, U(R), R^{\times}$$ be the units of $$R$$.

• Show that every nilpotent is either zero or a zero divisor.
• Solution: $$a^m=0$$ with $$a\neq 0$$ and $$m>1$$, then $$x x^{m-1} = 0$$, so $$x^{m-1}$$ is a nontrivial element annihilating $$x$$.
• Show that $$R$$ commutative and unital and $$x$$ nilpotent implies $$1+x$$ is a unit, and moreover $$N + R^{\times}= R^{\times}$$ (the sum of a nilpotent and unit is a unit).
• Solution: expand $$1/(1+x) = \sum_{k=0}^\infty (-x)^k = \sum_{k=0}^n (-x)^k \coloneqq f(x)$$, so $$(1+x)f(x) =1$$. Now use that $$RN = N$$ since $$x^n=0$$ implies $$(rx)^n = rxrx\cdots rx = r^n x^n = 0$$. Taking $$n + u\in N + R^{\times}$$, then $$u+n = u^{-1}(1 + u^{-1}n) \in R^{\times}R^{\times}$$ since $$u^{-1}n\in N$$ and $$1+u^{-1}\in R^{\times}$$ by the first part.
• Show that $$f(x) = \sum a_k x^k \in R[x]$$ iff $$f\in R[x]^{\times}\iff a_0\in R^{\times}, a_{k>1}\in N$$.
• Solution: use that if $$a_k$$ is nilpotent, $$a_k x^k$$ is nilpotent. Then $$a_0$$ a unit at $$a_1 x$$ nilpotent implies $$a_0 + a_1 x$$ is a unit, and inductively $$f$$ is a unit. If $$f$$ is a unit, take $$fg=1$$ with $$f = \sum_{k=0}^na_k x^k$$ and $$g = \sum_{k=0}^m a_k x^k$$. Write $$fg(x) = \sum_{k=0}^{n+m} c_k x^k$$ where $$c_k = \sum_{j=0}^k a_j b_{k-j}$$. Using $$fg=1$$, $$c_{0} = a_0 b_0 = 1$$ so $$a_0, b_0$$ are units, and proceed inductively by descending coefficients, checking that $$a_n b_m$$ is the $$r=0$$ case.
• Show that $$f(x) \in N(R[x]) \iff a_k \in N(R)$$ for all $$k$$.
• Solution: $$f$$ nilpotent with $$f(x) = \sum a_k x^k$$ implies $$f^m=0$$, and check the leading term $$a_n^m x^{nm}$$. Induct down: $$f, a_nx^n$$ nilpotent implies $$f - a_n x^n$$ nilpotent. Conversely, if $$a_i^{n_i} = 0$$, use that $$N(R) {~\trianglelefteq~}R$$ form an ideal.
• Show that $$f\in ZD(R[x]) \iff f\neq 0$$ and $$rf(x) = 0$$ for some $$r\in R$$.

The nilradical of $$R \in \mathsf{CRing}$$ is \begin{align*} {\sqrt{0_{R}} } \coloneqq\left\{{x\in R {~\mathrel{\Big\vert}~}x \text{ is nilpotent}}\right\} .\end{align*}

Show $${\sqrt{0_{R}} } {~\trianglelefteq~}R$$ is an ideal and $$A/{\sqrt{0_{R}} }$$ is reduced.

• $$R{\sqrt{0_{R}} } \subseteq R$$: For $$r$$ nilpotent of order $$n$$ and $$x\in R$$, $$xr$$ is nilpotent since \begin{align*} (xr)^n = (xr)(xr)\cdots(xr) = x^n r^n = x^n 0 = 0 .\end{align*}

• $$R^2 \subseteq R$$, for $$r,s\in {\sqrt{0_{R}} }$$ write $$r^{n} = s^m = 0$$, then \begin{align*} (r+s)^{N} = \sum_{k\geq 0}{N \choose k} r^k s^{N-k} ,\end{align*} so just choose $$N$$ large enough so that either $$k>n$$ or $$N-k> m$$ always holds, e.g. $$N\coloneqq n+m-1$$.

• $$R/{\sqrt{0_{R}} }$$ has no nonzero nilpotents: Take $$\overline{r} \in R/{\sqrt{0_{R}} }$$ for some $$r\in R$$, then $$\phi(r^n) = \phi(r)^n = \overline{r}^n$$. So \begin{align*} \overline{r}^n=0 \operatorname{mod}{\sqrt{0_{R}} } \iff \overline{r^n} \equiv 0 \operatorname{mod}{\sqrt{0_{R}} } \iff r^n \in {\sqrt{0_{R}} } \iff r\in 0_R .\end{align*}

Show that the nilradical is the intersection of all prime ideals.

See A&M 1.8

Write $$P$$ as the intersection of all prime ideals of $$R$$.\

$${\sqrt{0_{R}} } \subseteq P$$: Suppose $$r\in {\sqrt{0_{R}} }$$ so $$r^n = 0$$ and let $${\mathfrak{p}}\in \operatorname{Spec}R$$. Then use that $$0\in I$$ for any ideal: $$r^n = 0 \in {\mathfrak{p}}\implies r\in {\mathfrak{p}}$$ since $${\mathfrak{p}}$$ is prime.\

$${\sqrt{0_{R}} }^c \subseteq P^c$$: Fix $$f$$ non-nilpotent, we want to show $$f$$ is not in any prime ideal. set $$S \subseteq R$$ to be all ideals $$I$$ such that $$f^{>0} \not \in I$$. Apply Zorn’s lemma: $$S\neq \emptyset$$ since $$0\in S$$, so after ordering $$I$$ by inclusions $$S$$ contains a maximal $${\mathfrak{p}}$$ which we claim is prime. If $$a,b \in {\mathfrak{p}}^c$$ then $${\mathfrak{p}}+ \left\langle{ a }\right\rangle$$ and $${\mathfrak{p}}+ \left\langle{b}\right\rangle supset {\mathfrak{p}}$$ strictly, and by maximality they aren’t in $$S$$. So there exist $$m,n$$ such that $$f^m\in {\mathfrak{p}}+ \left\langle{ a }\right\rangle$$ and $$f^n \in {\mathfrak{p}}+ \left\langle{b}\right\rangle$$. Then $$f^{m+n} \in {\mathfrak{p}}+ \left\langle{ab}\right\rangle$$, so $${\mathfrak{p}}+ \left\langle{ab}\right\rangle$$ is not in $$S$$. Thus $$ab\not \in {\mathfrak{p}}$$ so $$f\not\in {\mathfrak{p}}$$. Letting $${\mathfrak{p}}$$ be arbitrary yields $$f\not \in P$$.

The Jacobson radical $${J ({R}) }$$ is the intersection of all maximal ideals, i.e. \begin{align*} {J ({R}) } = \bigcap_{{\mathfrak{m}}\in \operatorname{mSpec}R} {\mathfrak{m}} .\end{align*}

Show $$x\in {J ({R}) } \iff 1-xR \subseteq R^{\times}$$.

## Structure Theorems

A module $$M$$ is simple iff every submodule $$M' \leq M$$ is either $$0$$ or $$M$$. A ring $$R$$ is simple if and only if it is simple as an $$R{\hbox{-}}$$module, i.e. there are no nontrivial proper ideals.

A module $$M$$ is semisimple if and only if it admits a decomposition \begin{align*} M = \bigoplus_{j\in J} M_j \end{align*} with each $$M_j$$ simple.

If $$R$$ is a nonzero, unital, semisimple ring then \begin{align*} R \cong \bigoplus_{i=1}^m \mathrm{Mat}(n_i, D_i) ,\end{align*} a finite sum of $$n_i\times n_i$$ matrix rings over division rings $$D_i$$.

If $$M$$ is a simple ring over $$R$$ a division ring, the $$M$$ is isomorphic to a matrix ring.

Every finite division ring is a field, i.e. finite division rings must be commutative.

## Zorn’s Lemma

In a poset, a chain is a totally ordered subset. An upper bound on a subset $$S$$ of a poset $$X$$ is any $$x\in X$$ such that $$s\leq x$$ for all $$s\in S$$.

If $$P$$ is a poset in which every chain has an upper bound, then $$P$$ has a maximal element.

You can always form a subset poset, and restrict with any sub-collection thereof with a set predicate. To use Zorn’s lemma, you need to take an arbitrary chain in your poset $$X$$, produce an upper bound $$U$$ (e.g. by taking a union), and showing that $$U$$ is still in $$X$$ (i.e. it still satisfies the right predicate).

Every ring has a proper maximal ideal, and any proper ideal is contained in a maximal ideal.

Every proper ideal is contained in a maximal ideal.

Let $$0 < I < R$$ be a proper ideal, and consider the set \begin{align*} S = \left\{{J {~\mathrel{\Big\vert}~}I \subseteq J < R}\right\} .\end{align*}

Note $$I\in S$$, so $$S$$ is nonempty. The claim is that $$S$$ contains a maximal element $$M$$.

$$S$$ is a poset, ordered by set inclusion, so if we can show that every chain has an upper bound, we can apply Zorn’s lemma to produce $$M$$.

Let $$C \subseteq S$$ be a chain in $$S$$, so $$C = \left\{{C_1 \subseteq C_2 \subseteq \cdots}\right\}$$ and define $$\widehat{C} = \cup_i C_i$$.

$$\widehat{C}$$ is an upper bound for $$C$$: This follows because every $$C_i \subseteq \widehat{C}$$.

$$\widehat{C}$$ is in $$S$$: Use the fact that $$I \subseteq C_i < R$$ for every $$C_i$$ and since no $$C_i$$ contains a unit, $$\widehat{C}$$ doesn’t contain a unit, and is thus proper.

Show that every non-unit of $$R$$ is contained in a maximal ideal.

This follows because if $$x\in R\setminus R^{\times}$$, then $$Rx {~\trianglelefteq~}R$$ and $$Rx\neq R$$ implies $$R/Rx \neq 0$$. Then there exists some $$\overline{{\mathfrak{m}}}\in \operatorname{mSpec}R/Rx$$, and by the correspondence theorem this lifts to some $${\mathfrak{m}}\in \operatorname{mSpec}R$$ containing $$Rx$$.

## Unsorted

Division algorithm for Euclidean domains.

\todo[inline]{todo}

For $$R\in \mathsf{CRing}$$ an integral domain, the field of fractions of $$R$$ can be constructed as \begin{align*} \operatorname{ff}(R) \coloneqq\qty{R \times R^{\bullet}}/\sim && (a,s)\sim (b, t) \iff at-bs = 0_R .\end{align*}

Checking transitivity requires having no nonzero zero divisors.

For $$R\in \mathsf{CRing}$$ and $$S \subseteq R$$ a multiplicatively closed subset, so $$RS \subseteq S$$ and $$1_R\in S$$, the localization of $$R$$ at $$S$$ can be constructed as \begin{align*} R \left[ { \scriptstyle { {S}^{-1}} } \right] \coloneqq\qty{R\times S} / \sim && (a, s)\sim (b, t) \iff \exists u\in S\quad (at-bs)u = 0_R .\end{align*}

Why the $$u$$: use in proof of transitivity.

\todo[inline]{Universal property.}

There is a canonical ring morphism \begin{align*} R &\to R \left[ { \scriptstyle { {S}^{-1}} } \right] \\ x &\mapsto {x\over 1} ,\end{align*} but this may not be injective.

For integral domains $$R$$, \begin{align*} \operatorname{ff}(R) \cong R \left[ { \scriptstyle { { (R^{\bullet}) }^{-1}} } \right] .\end{align*}

\todo[inline]{todo}

An ideal $$I{~\trianglelefteq~}R$$ is primary iff whenever $$pq\in I$$, $$p\in I$$ and $$q^n\in I$$ for some $$n$$.