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) 

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

    

  

    
    
      
        
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          \(\left\langle{p, q}\right\rangle = \left\langle{\gcd(p, q)}\right\rangle {~\trianglelefteq~}{\mathbf{Z}}\).
        
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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 \(A/\ker f \cong B\).
    

  
    

      More traditionally stated:
    

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

  
    
    
      
        
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          \(S+I \leq R\) is a subring.
        
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          \(S \cap I {~\trianglelefteq~}S\)
        
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      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.
    

  
    

      More traditionally:
    

  
    
    
      
        
  • 
          
    
    
      If \(S, I \in \operatorname{Id}(R)\) with \(S\) containing \(I\) then 

\begin{align*}
S/I \leq R/I
.\end{align*}

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

    
    
  
        
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      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 \(x=r^{-1}\).
    

  

    
    
      
        
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          \(R/{\mathfrak{m}}\) is a field \(\iff {\mathfrak{m}}\in \operatorname{mSpec}(R)\) is maximal.
        
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          \(R/{\mathfrak{p}}\) is an integral domain \(\iff {\mathfrak{p}}\in \operatorname{Spec}(R)\) is prime.
        
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          \(R/J\) is reduced \(\iff J\) is radical.
        
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      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 \({\mathfrak{m}}+ Rx = R = \left\langle{ 1}\right\rangle\).
    

  
Undergrad Review

    

      Notation:
    

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

    

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

  
    
    
      
        
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          \((R, +)\in {\mathsf{Ab}}{\mathsf{Grp}}\),
        
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          \((R, \cdot) \in \mathsf{Mon}\)
        
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          Distributivity: \(a(b+c) = ab + ac\).
        
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      Some of the most important examples of rings:
    

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

  
    
    
      
        
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          \(f(1_X) = 1_Y\)
        
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          \(f(a(b+c)) = f(a)f(b) + f(a)f(c)\)
        
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      Important notes:
    

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

  
    
    
      
        
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          \(RI \subseteq I\)
        
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          \(I + I \subseteq I\)
        
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      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

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

  
    
    
      
        
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          Prime and maximal:
    
      
      
        
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          \(p{\mathbf{Z}}\in \operatorname{Id}({\mathbf{Z}})\). Maximal (and thus prime) since \({\mathbf{Z}}/p\) is a field and a domain.
        
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          \(\left\langle{2, x}\right\rangle \in \operatorname{Id}({\mathbf{Z}}[x])\). Maximal (and thus prime) \({\mathbf{Z}}[x]/\left\langle{2, x}\right\rangle \cong {\mathbf{Z}}/2\) is a field and a domain.
        
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          Prime but not maximal:
    
      
      
        
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          \(\left\langle{0}\right\rangle \in \operatorname{Id}({\mathbf{Z}})\), since \(m{\mathbf{Z}}\supseteq\left\langle{0}\right\rangle\) for any \(m\).
        
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          \(\left\langle{x}\right\rangle \in R[x]\) 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).
        
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          Not prime, not maximal:
    
      
      
        
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          \(m{\mathbf{Z}}\in \operatorname{Id}({\mathbf{Z}})\), 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 \([2][3] = 0\).
        
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          Useful examples:
    
      
      
        
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          \(\operatorname{mSpec}{\mathbf{Z}}= \left\{{p{\mathbf{Z}}}\right\}\) and \(\operatorname{Spec}{\mathbf{Z}}= \left\{{p{\mathbf{Z}}}\right\} \cup\left\langle{0}\right\rangle\).
        
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          \(\operatorname{mSpec}{\mathbf{C}}[x] = \left\{{x-a {~\mathrel{\Big\vert}~}a\in {\mathbf{C}}}\right\}\), since over a PID \(\left\langle{\alpha}\right\rangle\) is maximal iff \(\alpha\) is irreducible, and over \({\mathbf{C}}\) irreducibles are degree 1.
        
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          \(\operatorname{mSpec}k[x_1, \cdots, x_{n}] = \left\{{\left\langle{x-a_1, x-a_2, \cdots, x-a_n}\right\rangle {~\mathrel{\Big\vert}~}a_k \in k}\right\}\).
        
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          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.
        
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      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:
    

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

  
    
    
      
        
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          Show primes are maximal, so \(\operatorname{Spec}R \subseteq \operatorname{mSpec}R\) and nonzero ideals are prime iff maximal.
        
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          Show that \(R\) is Noetherian.
        
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          Show that every element is a finite product of irreducibles.
        
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          Show that in a PID, every maximal ideal is generated by an irreducible element.
        
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          Show that not \({\mathbf{Z}}\) is Noetherian but not Artinian.
    
      
      
        
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          Hint: take a chain \(n{\mathbf{Z}}\supseteq n^2{\mathbf{Z}}\supseteq\cdots\).
        
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      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).
        
      • 
      
    
    
  
        
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      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.
        
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    • 
      
        
  • 
          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).
        
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  • 
          A ring that is not an integral domain: \({\mathbf{Z}}/4\)
    
      
      
        
    • 
          Proof: \([2]_4\) is a zero divisor since \([2]_4 [2]_4 = [0]_4\).
        
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Radicals

    

      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 \(n\).