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Algebra-II: Companion notes

Abhay Soman

Section 1.6 Examples of subrings and ideals

We give several examples of subrings and ideals.
Let \(R\) be a ring and \(S\) be a nonempty subset of \(R\text{.}\) The centralizer of \(S\) in \(R\) is the set of all those elements of \(R\) that commutes with every element of \(S\text{.}\) It is denoted by \(C_R(S)\text{.}\) Thus,
\begin{equation*} C_R(S)=\{r\in R:rs=sr\;\text{for all}\;s\in S\}. \end{equation*}
Verify that \(C_R(S)\) is a subring of \(R\text{.}\)
The center of a ring \(R\text{,}\) \(\mathcal{Z}(R)\) is the centralizer of \(R\) in \(R\text{.}\) Thus,
\begin{equation*} \mathcal{Z}(R)=\{r\in R:rx=xr\;\text{for all}\;x\in R\}. \end{equation*}
The center of \(R\) is a subring of \(R\text{.}\)
Let \(M_n(R)\) be the ring of \(n\times n\) matrices over a ring \(R\text{.}\) Following subsets of \(M_n(R)\) are subrings.
  1. Upper triangular matrices: \(U_n(R)=\{(a_{ij})\in M_n(R):a_{ij}=0\text{ for }i\gt j\}\)
  2. Lower triangular matrices: \(L_n(R)=\{(a_{ij})\in M_n(R):a_{ij}=\text{ for }i\lt j\}\)
  3. Diagonal matrices: \(D_n=\{(a_{ij})\in M_n(R):a_{ij}=0\text{ for all }i\neq j\}\)
Let \(I\) be a right ideal of a ring \(R\text{.}\) The idealizer of \(I\) is
\begin{equation*} S=\{r\in R:rI\subset I\}. \end{equation*}
Note that the idealizer of \(I\) is a subring containing \(I\) as a two-sided ideal.
Let \(R\) be a ring and let \(r_1,r_2,\ldots,r_n\in R\text{.}\) The ideal generated by \(r_1,\ldots,r_n\) is
\begin{equation*} (r_1,\ldots,r_n)=\left\{\sum_{i=1}^n\sum_{j=1}^{m_i}x_{ij}r_iy_{ij}:x_{ij},y_{ij}\in R\;\text{and}\;n,m_i\in\N\right\}. \end{equation*}
A particular case is the principal ideal generated by \(r\):
\begin{equation*} (r)=RrR=\left\{\sum_{i=1}^nx_iry_i:n\in\mathbb{N}\;\text{and}\;x_i,y_i\in R\right\}. \end{equation*}
Note that \(\Z\) is an integral domain. Hence, every left ideal of \(\Z\) is also a right ideal of \(\Z\text{.}\) In other words, every ideal of \(\Z\) is a two-sided ideal. Let \(I\) be a nonzero ideal of \(\Z\text{.}\) Suppose that \(n\in I\) is the least positive integer in \(I\text{.}\) We claim that \(n\) generates \(I\text{,}\) i.e., \((n)=I\text{.}\) Since \(n\in I\) we have, using same actions defined in (1.2.1) and (1.2.2), \(\underbrace{n+n+\cdots+n}_{m\text{-times}}=mn\in I\) for any \(m\in\Z\text{.}\) Hence \((n)\subseteq I\text{.}\) Suppose that \(s\in I\text{.}\) The division algorithm gives the following.
\begin{equation*} s=nq+r\quad\text{where}\;0\leq r\lt n \end{equation*}
Suppose that \(r\neq 0\text{.}\) Since \(s\) and \(nq\in I\) we get \(r=s-nq\in I\text{.}\) This is a contradiction because \(n\) is the least positive integer in \(I\text{.}\) Therefore, we must have \(r=0\text{,}\) i.e., \(s=nq\in (n)\text{.}\) We thus get \(I=(n)\text{.}\)
Let \(I\) be a nonzero left ideal of a division ring \(D\text{.}\) Let \(d\in I\) be a nonzero element. Hence there exists \(e\in D\) such that \(1=de=ed\text{.}\) In particular, \(1=ed\in I\) and hence for any \(x\in D\) we have \(x=x1\in I\text{.}\) This shows that every nonzero left ideal of \(D\) is \(D\text{.}\) Similarly, every nonzero right ideal of a division ring will be the whole ring.
Let \(V\) be a finite-dimensional vector space over a field \(F\) and let \(U\) be a subspace of \(V\text{.}\) We consider an \(F\)-linear map \(T\colon V\to U\text{,}\) by composing it with the inclusion \(i\colon U\to V\text{,}\) to be an \(F\)-endomorphism. In this way, we consider \(\Hom_F(V,U)\) as a subset of \(\End_F(V)\text{.}\) Moreover, \(\Hom_F(V,U)\) is a subgroup of \(\End_F(V)\text{.}\) Let \(f\in\End_F(V)\) and \(T\in\Hom_F(V,U)\text{.}\) Thus,
\begin{equation*} V\xrightarrow{f}V\xrightarrow{T}U \end{equation*}
shows that \(T\circ f\in\Hom_F(V,U)\text{.}\) Therefore, \(\Hom_F(V,U)\) is a right ideal of \(\End_F(V)\text{.}\) In fact, there is a one-one correspondence between subspaces \(U\) of \(V\) and right ideals of \(\End_F(V)\) given by \(U\mapsto\Hom_F(V,U)\text{.}\)
Indeed, suppose that \(I\) is a right ideal of \(\End_F(V)\text{.}\) Consider the following subspace.
\begin{equation*} U=\langle{\rm Im}(f):f\in I\rangle \end{equation*}
Thus \(I\subseteq\Hom_F(V,U)\text{.}\) Let \(\{u_1,\ldots,u_r\}\) be a basis for \(U\) and suppose that \(w_i\in V\) are such that \(f_i(w_i)=u_i\) for some \(f_i\in I\) 1  Note that we may have \(f_i=f_j\) for \(i\neq j\text{.}\) Let \(\{v_1,\ldots,v_n\}\) be a basis of \(V\) and let \(f\in\Hom_F(V,U)\text{.}\) Assume that an \(F\)-linear map \(f\) is given as follows.
\begin{equation*} f(v_k)=\sum_{\ell=1}^{r}a_{\ell k}u_\ell \end{equation*}
We consider, for each \(\ell\in\{1,2,\ldots,r\}\text{,}\) \(E_\ell\in\Hom_F(V,U)\) as follows:
\begin{equation*} E_\ell(v_k)=a_{\ell k}w_{\ell},\text{ where }k\in\{1,2,\ldots,n\} \end{equation*}
Hence for every \(\ell\in\{1,2,\ldots,r\}\) we get
\begin{equation*} f_{\ell}\circ E_\ell(v_k)=a_{\ell k}u_\ell \end{equation*}
Since \(f_\ell\) belongs to a right ideal \(I\text{,}\) we must have \(f_\ell\circ E_\ell\in I\text{.}\) Therefore,
\begin{align*} f(v_k)\amp=\sum_{\ell=1}^{r}a_{\ell k}u_\ell\\ \amp=\sum_{\ell=1}^{r}f_\ell\circ E_\ell(v_k)\in I \end{align*}
We thus get \(\Hom_F(V,U)\subseteq I\text{.}\)
Similarly, there is a one-one correspondence between subspaces \(U\) of \(V\) and left ideals of \(\End_F(V)\) given by \(U\mapsto\Hom_F(V/U,V)\text{.}\)
Let \(R\) be a ring and let \(M_n(R)\) be the ring of matrices over \(R\text{.}\) If \(S\) is a subset of \(R\) we denote by \(M_n(S)\) the set of all \(n\times n\) matrices with entries from \(S\text{.}\)
We show that every ideal, i.e., every two-sided ideal of \(M_n(R)\) is of the form \(M_n(I)\) for some ideal \(I\) of \(R\text{.}\)
We leave it to the reader to verify that if \(I\) is an ideal of \(R\) then \(M_n(I)\) is an ideal of \(M_n(R)\text{.}\)
Suppose that \(\mathcal{J}\) be an ideal of \(M_n(R)\text{.}\) Consider the following subset of \(R\text{.}\)
\begin{equation*} I=\{a\in R:a\text{ is the }(1,1)\text{-th entry of }A\in\mathcal{J}\} \end{equation*}
The set \(I\) is an ideal of \(R\text{.}\) Let \(E_{ij}\) be a matrix such that \((i,j)\)-th entry is \(1\) and all other entries are zero. Then for any \(A\in M_n(R)\) we have the following.
\begin{equation*} E_{ij}AE_{k\ell}=a_{jk}E_{i\ell} \end{equation*}
Therefore, for \(A\in\mathcal{J}\) we have
\begin{equation*} E_{1j}AE_{k1}=a_{jk}E_{11}\in\mathcal{J}. \end{equation*}
By the construction of \(I\) we get that \(a_{jk}\in I\) for any \(j,k\in\{1,2,\ldots,n\}\text{.}\) In particular, \(\mathcal{J}\subseteq M_n(I)\text{.}\)
Conversely, suppose \(B=(b_{ij})\in M_n(I)\text{.}\) By the construction of \(I\text{,}\) \(b_{ij}\) is the \((1,1)\)-th entry of some \(B^\prime\in\mathcal{J}\) (here \(B^\prime\) may depend on \(b_{ij}\)). Thus, \(E_{i1}B^\prime E_{1j}=b_{ij}E_{ij}\in\mathcal{J}\text{.}\) Hence, \(\sum b_{ij}E_{ij}=B^\prime\in\mathcal{J}\) and therefore, \(M_{n}(I)\subseteq\mathcal{J}\text{.}\)
Let \(R\) be a ring and \(I\) be a right ideal of \(R\text{.}\) The annihilator of \(I\) is defined as follows.
\begin{equation*} I^0=\{r\in R:rx=0\;\text{for all}\;x\in I\} \end{equation*}
The annihilator of a right ideal is a left ideal of \(R\text{.}\)
Similarly, if \(J\) is a left ideal of \(R\) we define the annihilator of \(J\) as follows.
\begin{equation*} J^0=\{r\in R:xr=0\;\text{for all}\;x\in J\} \end{equation*}
The annihilator \(J^0\) is a right ideal of \(R\text{.}\)
Let \(F\) be a field and let \(F[X]\) be the ring of polynomials over \(F\) in one variable. Fix \(\alpha\in F\text{.}\) We define the evaluation map at \(\alpha\in F\) as follows.
\begin{equation*} \ev_{\alpha}\colon F[X]\to F\quad\text{given by}\quad\sum_{i=1}^na_iX^i\mapsto\sum_{i=1}^na_i\alpha^i \end{equation*}
Verify that the following set is an ideal of \(F[X]\text{:}\)
\begin{equation*} \{f\in F[X]:\ev_{\alpha}(f)=0\} \end{equation*}
Let \(\C[X_1,X_2,\ldots,X_n]\) be the polynomial ring in \(n\) variables. Fix \(a_1,a_2,\ldots,a_n\in\C\text{.}\) The following set is a two-sided ideal of \(\C[X_1,X_2,\ldots,X_n]\text{.}\)
\begin{equation*} \mathfrak{m}=(X_1-a_1,X_2-a_2,\ldots,X_n-a_n) \end{equation*}
Let \(C(I)\) be the ring of continuous real-valued functions on a nonempty interval \(I\subseteq\R\) (see Example 1.2.12). Fix \(a_0\in I\text{.}\) Consider the set of all continuous functions on \(I\) vanishing at \(a_0\) i.e.,
\begin{equation*} \mathfrak{m}=\{f\in C(I):f(a_0)=0\}. \end{equation*}
If \(f,g\in\mathfrak{m}\) then \(f-g\in\mathfrak{m}\text{.}\) For any \(g,h\in C(I)\) and \(f\in\mathfrak{m}\) we have \((f\cdot g)(a_0)=f(a_0)g(a_0)=0\) and \((h\cdot f)(a_0)=h(a_0)f(a_0)=0\text{.}\) Thus, \(\mathfrak{m}\) is a two-sided ideal of \(C(I)\text{.}\)
Let \(R\) and \(S\) be two rings and consider the product ring \(R\times S\text{.}\) If \(I\subseteq R\) and \(J\subseteq S\) are ideals then \(I\times J\) is an ideal of \(R\times S\text{.}\)
That is, we are assuming that \(\{u_i\}\subset\bigcup_{f\in I}{\rm Im}(f)\text{.}\) Indeed, we may consider \(\{u_i\}\) to be the maximal linearly independent subset of \(\bigcup_{f\in I}{\rm Im}(f)\text{.}\)
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