2.1 Review of Rings

Convention

In our course, all rings have identity.

Definition

A ring is a non-empty set $R$ with $+$ and $\cdot$ such that

• $(R,+)$ is an abelian group;
• $a(bc)=(ab)c$;
• $a(b+c)=ab+ac$;
• there exists $1_R\in R$ such that $1_{R}a=a{1}_R=a$ for all $a\in R$.

Definition

$f:R\to S$ is called a homomorphism of rings if $f(a+b)=f(a)+f(b)$, $f(ab)=f(a)f(b)$ and $f(1_R)=1_S$.

Definition

Let $R$ be a ring. We say $0\ne a\in R$ is a left (rep. right) zero-divisor if there exists $b\ne 0$ such that $ab=0$ (rep. $ba=0$). We say $a$ is a zero-divisor if there exist $b,c\in R$ such that $ab=0=ca$ (possibly $b\ne c$).

Definition

If there exists a smallest $n\in {\mathbb{Z}}_{⩾0}$ such that $na=0$ for all $a\in R$, then we say $R$ has characteristic $n$, denote $\mathrm{char}R=n$. If such $n$ does not exist, then we say $\mathrm{char}R=0$.

Theorem

Suppose $\mathrm{char}R=n>0$.

• $n$ is the least integer such that $n{1}_R=0$;
• if $R$ has no zero-divisor, then $n$ is a prime.

\begin{proof} i) If $k\cdot 1_R=0$, then $ka=k\cdot 1_R\cdot a=0$.

ii) If $n=mk$, then $n\cdot 1_R=(m\cdot 1_R)(k\cdot 1_R)=0$. Since $R$ has no zero-divisor, we have one of $m,k=1$ and so $n$ is a prime. \end{proof}

Convention

From now on all rings are commutative and have identity. Assume that ring homomorphisms map the identity $1$ to $1$.

Definition

Let $R$ be a ring, and let $X\subseteq R$ be a subset. Let $(X)$ be the ideal generated by $X$, then

$$(a_1,\cdots ,a_n)=\left\{\sum _{i=1}^n{a}_i{r}_i:r_i\in R\right\}.$$

In particular, if $X=\{a\}$, then $(a)=Ra$ is called a principal ideal.

Theorem

Let $I\subseteq R$ be an ideal. Then $R/I$ is a ring with $(a+I)(b+I)=ab+I$.

Theorem

The follows hold:

• if $f:R\to S$ is a homomorphism, then $\ker f$ is an ideal;
• if $I\subseteq R$ is an ideal, then $\pi :R\to R/I$ is a project and $\ker \pi =I$.

Theorem

If $f:R\to S$ is a homomorphism, then $R/\ker f\simeq \mathrm{Im}f$ is an isomorphism as rings.

Theorem

Let $I,J\subseteq R$ be ideals and $I\subseteq J$. Then $J/I\subseteq R/I$ is an ideal, and $(R/I)/(J/I)\simeq R/J$.

Definition

Let $I,J\subseteq R$ be ideals. Then:

• $I+J=\{a+b:a\in I,b\in J\}$ is an ideal;
• $IJ$ is the ideal generated by $\{ab:a\in I,b\in J\}$ and $IJ=\left\{{\sum }_{k=1}^n{a}_k{b}_k,a_k\in I,b_k\in J\right\}$.

Definition

We say an ideal $P⊊R$ is a prime ideal if for any $A,B\subseteq R$ ideals such that $AB\subseteq P$, then either $A\subseteq P$ or $B\subseteq P$.

Theorem

For an ideal $P⊊R$, $P$ is prime iff $\forall a,b\in R$ such that $ab\in P$, then $a\in P$ or $b\in P$.

\begin{proof} One direction is easy. If $P$ is prime, then for any $ab\in P$, $(ab)\subseteq P$ and so one of $(a),(b)$ is contained in $P$.

For another direction, assume that $AB\subseteq P$. WLOG $A⊈P$, then there exists $a\in A$ such that $a\notin P$. Then for any $b\in B$, $ab\in P$ and so $b\in P$. Therefore, $B\subseteq P$. \end{proof}

Theorem

Let $P⊊R$ be an ideal. Then $P$ is prime iff $R/P$ is an integral domain.

\begin{proof} $P$ is prime $⟺$ $ab\in P$ yields $a\in P$ or $b\in P$ $⟺$ $ab+P\in P$ yields $a+P\in P$ or $b+P\in P$ $⟺$ $\overline{ab}=0$ yields $\overline{a}=0$ or $\overline{b}=0$ $⟺$ $R/P$ is an integral domain. \end{proof}

Definition

We say an ideal $M⊊R$ is a maximal ideal if there does not exist another ideal $N$ such that $M⊊N⊊R$.

Fact. For any ring, maximal ideal exists. In fact, for any ideal $I\subseteq R$ is contained in some maximal ideal by Zorn’s lemma. (It states that a partially ordered set containing upper bounds for every chain (that is, every totally ordered subset) necessarily contains at least one maximal element.)

Theorem

A maximal ideal is a prime ideal.

\begin{proof} Suppose $M⊊R$ is an maximal ideal, and suppose $ab\in M$ with $a\notin M,b\notin M$. Since $M+(a)$ is an ideal, there is $M+(a)=R$. Similarly, we have that $M+(a)=M+(b)$. Also, there exists $m_1,m_2\in M$ such that $m_1+a{x}_1=m_2+b{x}_2=1$ and so $(m_1+a{x}_1)(m_2+b{x}_2)=1\in M$, which is impossible. \end{proof}

Theorem

Let $M⊊R$ be an ideal. Then $M$ is a maximal ideal iff $R/M$ is a field.

\begin{proof} Assume that $M$ is maximal, then $M$ is prime and so $R/M$ is an integral domain. Consider $\overline{a}\in R/M$ with $\overline{a}\ne \overline{0}$, that is $a\notin M$. Therefore, $M+(a)=R$ and there exists $b\in R$ such that $1=m+ab$ and so $\overline{b}=(\overline{a}{)}^{-1}$. It yields that $R/M$ is a field.

Now assume that $R/M$ is a field. If $M$ is not maximal, then there exists an ideal $N$ such that $M⊊N⊊R$ and $a\in N\setminus M$. Hence, $\overline{a}\ne 0$ in $R/M$ and so there exists $\overline{b}$ such that $ab+M=1+M$. It follows that $ab+m=1\in N$, which is a contradiction. \end{proof}

Example. All prime ideals of $\mathbb{Z}$ are $(0),(p)$ for all prime $p$, and all maximal ideals are $(p)$. (Recall: $\mathrm{Spec}\mathbb{Z}=\{(0),(p):p\text{mbox}isaprime\}$.)

Corollary

The followings are equivalent:

• $R$ is a field;
• $R$ has no proper ideal;
• $(0)$ is a maximal ideal;
• for any nonzero homomorphism $f:R\to S$, $f$ is a monomorphism.

\begin{proof} Easy. \end{proof}

Hilbert’s Nullstellensatz

Suppose $f_1,\cdots ,f_m\in \mathbb{C}[x_1,\cdots ,x_n]:=R$ such that $(f_1,\cdots ,f_m)\ne R$, then the system $f_i=0(i=1,\cdots ,m)$ has a solution.

(Another form) Let $R=\mathbb{C}[x_1,\cdots ,x_n]$, then $I\subseteq R$ is a maximal ideal iff $I=(x-a_1,\cdots ,x-a_n)$ for some $a_i\in \mathbb{C}$.

Remark. We do not give the proof here.

• second form → first form: if $(f_1,\cdots ,f_m)\ne R$, then $(f_1,\cdots ,f_m)\subseteq M=(x-a_1,\cdots ,x-a_n)$ and so $(a_1,\cdots ,a_n)$ is a solution of the system.
• for the second form, one direction is easy. Define $R/I\to \mathbb{C}$ by $f\mapsto f(a_1,\cdots ,a_n)$.

Recall that for a family of rings $\{R_i{\}}_{i\in S}$, the (external) direct product $\prod R_i=\{(a_i{)}_{i\in S}:a_i\in R_i\}$.

Lemma

Let $\{R_i{\}}_{i\in S}$ be a family of rings.

• If $I_i\subseteq R_i$ is an ideal, then $\prod I_i\subseteq \prod R_i$ is an ideal;
• conversely, if $I\subseteq \prod R_i$ is an ideal, then it is NOT necessarily of form $I=\prod I_i$;
• but (2) is true if $|S|<\infty$.

\begin{proof} Suppose $I\subseteq {\prod }_{i=1}^n{R}_i$. Let $I_i=\{x\in R_i:(a_1,\cdots ,a_{i-1},x_i,a_{i+1},\cdots ,a_n)\in I\}$. It is easy to check that $I_i\subseteq R_i$ is an ideal. Also clearly $I\subseteq {\prod }_{i=1}^n{I}_i$. Conversely, for given $x_i\in I_i$, we aim to show $(x_1,\cdots ,x_n)\in I$. Note that

$$(x_1,\cdots ,x_n)=(x_1,0,\cdots ,0)+(0,x_2,0,\cdots ,0)+\cdots +(0,\cdots ,0,x_n)$$

and $(0,\cdots ,0,x_i,0,\cdots ,0)\in I$ because $(a_1,\cdots ,a_{i-1},x_i,a_{i+1},\cdots ,a_n)(0,\cdots ,0,1,0,\cdots ,0)\in I$. \end{proof}

Remark. When $|S|=\infty$, we can construct the following example such that ii) fails. For example, see here.

Lemma

Let $\alpha ,\beta \subseteq R$ be ideals such that $\alpha +\beta =R$. Then $\alpha \cap \beta =\alpha \beta$ and the homomorphism $\phi :R\to R/\alpha \times R/\beta$ induces an isomorphism $\theta :R/\alpha \beta \to R/\alpha \times R/\beta$. (We say $\alpha ,\beta$ are coprime if $\alpha +\beta =R$.)

\begin{proof} Note that $\alpha \beta \subseteq \alpha \cap \beta$ for any ideals $\alpha ,\beta$. Let $x\in \alpha \cap \beta$. Since $\alpha +\beta =R$, there exists $a\in \alpha ,b\in \beta$ such that $a+b=1$ and so $ax+bx=x\in \alpha +\beta$. Therefore, $\alpha \cap \beta =\alpha \beta$. Note that $\ker \phi =\alpha \cap \beta$, then it suffices to show $\phi$ is surjective. For any $\overline{c}\in R/\alpha$ and $\overline{d}\in R/\beta$, we have that $\phi (cb+da)=(\overline{c},\overline{d})$ and now we finish the proof. \end{proof}

Example. $(6)+(35)=\mathbb{Z}$ and so $\mathbb{Z}/210\mathbb{Z}\simeq \mathbb{Z}/35\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$.

Lemma

Let ${\alpha }_1,\cdots ,{\alpha }_n\subseteq R$ be ideals such that ${\alpha }_i+{\alpha }_j=R$ for any $i\ne j$. Then $R/({\alpha }_1\cdots {\alpha }_n)\simeq {\prod }_{i=1}^{n}R/{\alpha }_i$.