Algebra - Rings and fields

Tejas N S
Assistant Professor in Mathematics, NIE First Grade College, Mysuru
Email: nstejas@gmail.com Mobile: +91 9845410469

8. Ideal of a ring

  • Definition 8.1: A non empty subset $I$ of a ring $R$ is said to be left ideal, if
    (i) $I$ is a additive subgroup of $R$ i.e., $\forall$ $a,b \in I$ $\Rightarrow$ $a-b \in I$
    (ii) For $a \in I$ and $r\in R$ $\Rightarrow$ $ra \in I$

  • Definition 8.2: A non empty subset $I$ of a ring $R$ is said to be right ideal, if
    (i) $I$ is a additive subgroup of $R$ i.e., $\forall$ $a,b \in I$ $\Rightarrow$ $a-b \in I$
    (ii) For $a \in I$ and $r\in R$ $\Rightarrow$ $ar \in I$

  • Definition 8.3: A non empty subset $I$ of a ring $R$ is said to be ideal or two sided ideal, if
    (i) $I$ is a additive subgroup of $R$ i.e., $\forall$ $a,b \in I$ $\Rightarrow$ $a-b \in I$
    (ii) For $a \in I$ and $r\in R$ $\Rightarrow$ $ar \in I$ and $ra \in I$

  • Theorems on ideals

  • Theorem 8.1: Intersection of two ideals is an ideal.
    Proof: Let $I_1$ and $I_2$ be ideals of a ring $R$.
    To prove: $I_1 \cap I_2$ is an ideal of $R$.
    Since $I_1$ and $I_2$ are ideals of $R$, we have $0 \in I_1$ and $0 \in I_2$ $\Rightarrow$ $0 \in I_1 \cap I_2$
    $\therefore$ $I_1 \cap I_2$ is non empty.
    Again since $I_1$ and $I_2$ are ideals of $R$, we have $I_1 \subseteq R$ and $I_2 \subseteq R$ $\Rightarrow$ $I_1 \cap I_2 \subseteq R$
    Consider $a,b \in I_1 \cap I_2$ then $a \in I_1,I_2$ and $b \in I_1,I_2$
    $\Rightarrow$ $a \in I_1, b \in I_1$ and $a \in I_2, b \in I_2$
    $\Rightarrow$ $a-b \in I_1$ and $a-b \in I_2$ [$\because$ $I_1$ and $I_2$ are ideals of $R$]
    $\Rightarrow$ $a-b \in I_1 \cap I_2$
    $\therefore$ $I_1 \cap I_2$ is an additive subgroup of $R$.
    Now consider $a \in I_1 \cap I_2$ and $r \in R$
    $\Rightarrow$ $a \in I_1$, $a \in I_2$ and $r \in R$
    $\Rightarrow$ $ar, ra \in I_1$ and $ar, ra \in I_2$ [$\because$ $I_1$ and $I_2$ are ideals of $R$]
    $\Rightarrow$ $ar \in I_1 \cap I_2$ and $ra \in I_1 \cap I_2$
    $\therefore$ $I_1 \cap I_2$ is an ideal of $R$. $\blacksquare$

  • Theorem 8.2: Sum of two ideals is an ideal.
    Proof: Let $I_1$ and $I_2$ be ideals of a ring $R$.
    To prove: $I_1+I_2=\{x+y/x\in I_1$ and $y \in I_2\}$ is an ideal of $R$
    Since $I_1$ and $I_2$ are ideals of $R$, we have $0 \in I_1$ and $0 \in I_2$ $\Rightarrow$ $0+0 \in I_1 + I_2$
    $\therefore$ $I_1 + I_2$ is non empty.
    Again since $I_1$ and $I_2$ are ideals of $R$, we have $I_1 \subseteq R$ and $I_2 \subseteq R$ $\Rightarrow$ $I_1 + I_2 \subseteq R$
    Let $a,b \in I_1+I_2$ $\Rightarrow$ $a=x_1+y_1$ and $b=x_2+y_2$ where $x_1,x_2 \in I_1$; $y_1,y_2 \in I_2$
    $x_1,x_2 \in I_1$ $\Rightarrow$ $x_1-x_2 \in I_1$ and $y_1,y_2 \in I_2$ $\Rightarrow$ $y_1-y_2 \in I_2$ [$\because$ $I_1$ and $I_2$ are ideals of $R$]
    Hence $a-b=(x_1+y_1)-(x_2+y_2)=(x_1-x_2)+(y_1-y_2) \in I_1 + I_2$
    $\therefore$ $I_1 + I_2$ is a additive subroup of $R$.
    Now consider $a \in I_1 + I_2$ and $r \in R$
    $a \in I_1 + I_2$ $\Rightarrow$ $a=x_1+y_1$ where $x_1 \in I_1$ and $y_1 \in I_2$
    $\Rightarrow$ $x_1r, rx_1 \in I_1$ and $y_1r, ry_1 \in I_2$ [$\because$ $I_1$ and $I_2$ are ideals of $R$]
    Hence, $ar=(x_1+y_1)r=x_1r+y_1r \in I_1 + I_2$ and $ra=r(x_1+y_1)=rx_1+ry_1 \in I_1 + I_2$
    $\therefore$ $I_1 + I_2$ is an ideal of $R$. $\blacksquare$

  • Theorem 8.3: The only ideals of a field are $\{0\}$ and itself. (trivial ideals)
    Proof: Let $F$ be a field and $I$ be the ideal of $F$.
    To prove: $I=\{0\}$ or $I=F$
    If $I=\{0\}$ nothing to prove. Suppose $I\neq \{0\}$
    Let $a \in I$ then $a \neq 0$
    $\Rightarrow$ $a \in F$ [$\because$ $I\subseteq F$]
    $\Rightarrow$ $a^{-1} \in F$ [$\because$ $F$ is a field]
    Now $a \in I$ and $a^{-1} \in F$ $\Rightarrow$ $aa^{-1} \in I$ [$\because$ $I$ is an ideal of $F$] $\Rightarrow$ $1 \in I$
    Now for $1 \in I$ and $a\in F$ $\Rightarrow$ $1.a \in I$ [$\because$ $I$ is an ideal of $F$] $\Rightarrow$ $a \in I$
    $\therefore$ $F\subseteq I$........$(1)$
    Since $I$ is an ideal of $F$, we have $I\subseteq F$........$(2)$
    From $(1)$ and $(2)$ we have $I=F$. $\blacksquare$

  • Theorem 8.4: Every ideal of a ring $R$ is a subring of $R$. Converse may not be true.
    Proof: Let $I$ be an ideal of ring $R$.
    To prove: $I$ is a subring of $R$
    Since $I$ is an ideal of ring $R$, it is a non empty subset of $R$
    Let $a,b \in I$ $\Rightarrow$ $a-b \in I$ [$\because$ $I$ is an ideal of $R$]
    Since $I$ is a ideal of $R$, we have $I\subseteq R$, So $a\in I$ $\Rightarrow$ $a \in R$
    Now $a \in R$ and $b \in I$ $\Rightarrow$ $ab \in I$ [$\because$ $I$ is an ideal of $R$]
    $\therefore$ $I$ is a subring of $R$.
    Converse of the above theorem may not be true.
    Example: $I=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$ is neither a right ideal or left ideal of a ring of $2 \times 2$ real matrices but a subring of a ring of $2 \times 2$ real matrices. (Really!) $\blacksquare$

Problems 8

  1. Show that the set of all multiples of an integer $n$, where $n$ is fixed is an ideal of $\mathbb{Z}$.
    Solution: Set of integers, $\mathbb{Z}=\{0,\pm 1,\pm 2,\pm 3,...\}$
    Let $n \in \mathbb{Z}$ be fixed. Let $I=\{nx/ x\in \mathbb{Z}\} $.
    To prove: $I$ is an ideal of ring $\mathbb{Z}$. Clearly $I \subseteq \mathbb{Z}$ and $I$ is non empty
    Let $a,b \in I$ $\Rightarrow$ $a=nx_1$ and $b=nx_2$ where $x_1,x_2 \in \mathbb{Z}$
    Consider $a-b=nx_1-nx_2=n(x_1-x_2) \in I$
    $\therefore$ $I$ is a additive subgroup of $\mathbb{Z}$.
    Let $a \in I$ and $r \in \mathbb{Z}$
    Consider $ar=(nx_1)r=n(x_1r) \in I$ and $ra=r(nx_1)=n(rx_1) \in I$
    $\therefore$ $I$ is an ideal of $\mathbb{Z}$ $\spadesuit$

  2. Show that $S=\Bigg\{ \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$ is a left ideal of a ring of $2 \times 2$ matrices over $\mathbb{Z}$. Is it a right ideal?
    Solution: Consider, $M=\Bigg\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Big/ a,b,c,d \in \mathbb{Z} \Bigg\}$ and $S=\Bigg\{ \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$
    Clearly $S \subseteq M$ and since $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$, $S$ is non empty.
    Let $A,B \in S$ $\Rightarrow$ $A=\begin{pmatrix} a_1 & 0 \\ b_1 & 0 \end{pmatrix}$ and $B=\begin{pmatrix} a_2 & 0 \\ b_2 & 0 \end{pmatrix}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$
    Consider $A-B=\begin{pmatrix} a_1 & 0 \\ b_1 & 0 \end{pmatrix}-\begin{pmatrix} a_2 & 0 \\ b_2 & 0 \end{pmatrix}=\begin{pmatrix} a_1-a_2 & 0 \\ b_1-b_2 & 0 \end{pmatrix} \in S$
    $\therefore$ $S$ is an additive subgroup of $M$.
    Let $X\in M$ $\Rightarrow$ $X=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a,b,c,d \in \mathbb{Z}$
    Consider $XA=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} a_1 & 0 \\ b_1 & 0 \end{pmatrix}=\begin{pmatrix} aa_1+bb_1 & 0 \\ ca_1+db_1 & 0 \end{pmatrix} \in S$
    $\therefore$ $S$ is an left ideal of $M$.
    Consider $AX=\begin{pmatrix} a_1 & 0 \\ b_1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\begin{pmatrix} a_1a & a_1b \\ b_1a & b_1b \end{pmatrix} \notin S$
    $\therefore$ $S$ is not an right ideal of $M$. $\spadesuit$

  3. Show that $S=\Bigg\{ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$ is a right ideal of a ring of $2 \times 2$ matrices over $\mathbb{Z}$. Show that $S$ is not a left ideal.
    Solution: Consider, $M=\Bigg\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Big/ a,b,c,d \in \mathbb{Z} \Bigg\}$ and $S=\Bigg\{ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$
    Clearly $S \subseteq M$ and since $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$, $S$ is non empty.
    Let $A,B \in S$ $\Rightarrow$ $A=\begin{pmatrix} a_1 & b_1 \\ 0& 0 \end{pmatrix}$ and $B=\begin{pmatrix} a_2 & b_2 \\ 0 & 0 \end{pmatrix}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$
    Consider $A-B=\begin{pmatrix} a_1 & b_1 \\ 0 & 0 \end{pmatrix}-\begin{pmatrix} a_2 & b_2 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} a_1-a_2 & b_1-b_2 \\ 0 & 0 \end{pmatrix} \in S$
    $\therefore$ $S$ is an additive subgroup of $M$.
    Let $X \in \mathbb{M}$ $\Rightarrow$ $X=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a,b,c,d \in \mathbb{Z}$
    Consider $AX=\begin{pmatrix} a_1 & b_1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\begin{pmatrix} a_1a+b_1c & a_1b+b_1d \\ 0 & 0 \end{pmatrix} \in S$
    $\therefore$ $S$ is an right ideal of $M$.
    Consider $XA=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} a_1 & b_1 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} aa_1 & ab_1 \\ ca_1 & db_1 \end{pmatrix} \notin S$
    $\therefore$ $S$ is not an left ideal of $M$. $\spadesuit$

  4. Show that $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$ is a subring of a ring of $2 \times 2$ real matrices but not an ideal of a ring of $2 \times 2$ real matrices.
    Solution: Let $M=\Bigg\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Big/ a,b,c,d \in \mathbb{R} \Bigg\}$ and given $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$
    Clearly $S \subseteq M$ and Since $0=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$ we have $S$ is non empty
    Let $A,B \in S$ $\Rightarrow$ $A=\begin{pmatrix} a_1 & 0 \\ 0 & b_1 \end{pmatrix}$, $B=\begin{pmatrix} a_2 & 0 \\ 0 & b_2 \end{pmatrix}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$.
    Consider $A-B$ $=\begin{pmatrix} a_1 & 0 \\ 0 & b_1 \end{pmatrix}-\begin{pmatrix} a_2 & 0 \\ 0 & b_2 \end{pmatrix}$ $=\begin{pmatrix} a_1-a_2 & 0 \\ 0 & b_1-b_2 \end{pmatrix} \in S$
    Consider $AB$ $=\begin{pmatrix} a_1 & 0 \\ 0 & b_1 \end{pmatrix}\begin{pmatrix} a_2 & 0 \\ 0 & b_2 \end{pmatrix}$ $=\begin{pmatrix} a_1a_2 & 0 \\ 0 & b_1b_2 \end{pmatrix} \in S$
    $\therefore$ $S$ is a subring of $M$.
    Let $X \in M$ $\Rightarrow$ $X=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a,b,c,d \in \mathbb{R}$
    Consider $XA=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} a_1 & 0 \\ 0 & b_1 \end{pmatrix}=\begin{pmatrix} aa_1 & bb_1 \\ ca_1 & db_1 \end{pmatrix} \notin S$
    $\therefore$ $S$ is not an left ideal of $M$.
    Consider $AX=\begin{pmatrix} a_1 & 0 \\ 0 & b_1 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\begin{pmatrix} a_1a & a_1b \\ b_1c & b_1d \end{pmatrix} \notin S$
    $\therefore$ $S$ is not an right ideal of $M$. $\spadesuit$

  5. Let $R$ be a ring and $a \in R$. If $Ra=\{x\in R / ax=0 \}$ then prove that $Ra$ is a right ideal of $R$.
    Solution: Clearly $Ra \subseteq R$ and $Ra$ is non empty.
    Let $x,y \in Ra$ $\Rightarrow$ $ax=0$ and $ay=0$ where $a \in R$
    Consider $a(x-y)=ax-ay=0$ $\Rightarrow$ $x-y \in Ra$
    Let $r \in R$
    Consider $a(xr)=(ax)r=0$ $\Rightarrow$ $xr \in Ra$
    $\therefore$ $Ra$ is an right ideal of $R$. $\spadesuit$

  6. Let $U$ be an ideal of ring $R$. If $R(U)=\{x \in R/xu=0$ $\forall$ $u \in U\}$ then prove that $R(U)$ is an ideal of $R$.
    Solution: Clearly $R(U) \subseteq R$ and $R(U)$ is non empty.
    Let $x,y \in R(U)$ $\Rightarrow$ $xu=0$ and $yu=0$ $\forall$ $u \in U$
    Consider, $(x-y)u=xu-yu=0$ $\Rightarrow$ $x-y \in R(U)$
    Let $r \in R$
    Consider, $(xr)u=(xu)r=0$ $\Rightarrow$ $xr \in R(U)$ and $(rx)u=r(xu)=0$ $\Rightarrow$ $rx \in R(U)$
    $\therefore$ $R(U)$ is an ideal of $R$ $\spadesuit$

  7. If $A$ and $B$ are ideals of a ring $R$ such that $A \cap B = \{0\}$. Then show that $ab=0$ $\forall$ $a \in A$ and $b \in B$.
    Solution: $a \in A$ and $b \in B$ $\Rightarrow$ $a \in A$ and $b \in R$ $\Rightarrow$ $ab \in A$ [$\because$ $A$ is an ideal of $R$]
    Also $a \in A$ and $b \in B$ $\Rightarrow$ $a \in R$ and $b \in B$ $\Rightarrow$ $ab \in B$ [$\because$ $B$ is an ideal of $R$]
    Hence $ab \in A \cap B$
    Since $A \cap B=\{0\}$ $\Rightarrow$ $ab=0$ $\spadesuit$

  8. If $I$ is an ideal of ring $R$ and $1 \in I$, then prove that $I=R$.
    Solution: $I$ is an ideal of $R$ $\Rightarrow$ $I \subseteq R$.......$(1)$
    Let $x \in R$
    Since $1 \in I$ and $x \in R$ $\Rightarrow$ $1.x \in I$ $x \in I$ [$\because$ $I$ is an ideal of $R$]
    Hence $R \subseteq I$......$(2)$
    From $(1)$ and $(2)$, we have $I=R$ $\spadesuit$

Exercise 8

  1. Let $I_1$ and $I_2$ be ideals of ring $R$. Let $I_1I_2$ be the set of all elements that can be written as a finite sum of elements of the form $i_1i_2$ where $i_1 \in I_1$ and $i_2 \in I_2$. Then show that $I_1I_2$ is an ideal of $R$.

  2. Show that $2\mathbb{Z}$ is an ideal of $\mathbb{Z}$.

  3. Show that the set of all nilpotent elements of ring $R$ is an ideal of $R$.

  4. Let $R$ be a commuatative ring with identity whose ideals are $\{0\}$ and $R$. Then prove that $R$ is a field.

8.1 Principal ideals


  • Definition 8.1.1: An ideal $I$ of a commutative ring $R$ is said to be prinicpal ideal, if there exists an element $a \in R$ such that $I=\{xa/ x \in R\}$. We denote this ideal $I$ by $<a>$ or $Ra$ or $(a)$.

  • Theorem 8.1.1: If $p$ is an integer then $p\mathbb{Z}$ is a maximal ideal of $\mathbb{Z}$ if and only if $p$ is prime.

  • Problems 8.1

  1. Find the principal ideals of $\mathbb{Z}_4$.
    Solution: $\mathbb{Z}_4=\{0,1,2,3\}$
    $<0>=\{0,0,0,0\}=\{0\}$
    $<1>=\{0,1,2,3\}=\mathbb{Z}_4$
    $<2>=\{0,2,0,2\}=\{0,2\}$
    $<3>=\{0,3,2,1\}=\mathbb{Z}_4$
    $\therefore$ Principal ideals of $\mathbb{Z}_4$ are $\{0\}$, $\{0,2\}$ and $\mathbb{Z}_4$. $\spadesuit$

  2. Find the principal ideals of $\mathbb{Z}_{11}$.
    Solution: $\mathbb{Z}_{11}$ is a field. (Why?)
    We know that the ideals of field are trivial ideals. $\therefore$ $\{0\}$ and $\mathbb{Z}_{11}$ are ideals of $\mathbb{Z}_{11}$
    The principal ideals are $\{0\}$ and $\mathbb{Z}_{11}$. $\spadesuit$

  3. Find the principal ideal generated by $4$ in $\mathbb{Z}$.
    Solution: $<4>=\{4x/x \in \mathbb{Z}\}=\{0,\pm 4, \pm 8, \pm 12, \pm 16,......\}$ $\spadesuit$

Exercise 8.1

  1. Show that set of all multiples of a fixed integer $n$ is a principal ideal in $\mathbb{Z}$.

  2. Find all the principal ideals of $\mathbb{Z}_6$.

  3. Find all the principal ideals of $\mathbb{Z}_3,\mathbb{Q},\mathbb{R},\mathbb{C}$.

8.2 Maximal ideals and prime ideals


  • Definition 8.2.1: An ideal $M$ of a ring $R$ is said to be maximal ideal if when $I$ is an ideal of $R$ such that $M \subseteq I \subseteq R$ then $M=R$ or $I=R$.

  • Definition 8.2.2: An ideal $P$ of a ring $R$ is said to be prime ideal if $ab \in P$ $\Rightarrow$ $a \in P$ or $b \in P$.

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