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Algebra-II: Companion notes

Abhay Soman

Section 1.4 Types of Rings

We define certain types of rings.

Definition 1.4.1. (Commutative Ring).

A ring \(R\) with unity is said to be a commutative ring if for every \(r\in R\) and every \(s\in R\) we have \(rs=sr\text{.}\)
Thus \(R\) is a commutative monoid under multiplication.

Definition 1.4.2. (Integral Domain).

A nonzero ring \(R\) is said to be an integral domain if \(R\) is a commutative ring satisfying the following condition.
\begin{equation*} \text{If}\; rs=0\;\text{then either}\;r=0\;\text{or}\;s=0. \end{equation*}
The ring of integers (see Example 1.2.2), the complex, real and rational numbers (see Example 1.2.3), the ring of Gaussian integers (see Example 1.2.4), polynomial ring over an integral domain (see Example 1.2.8), the opposite ring of an integral domain (see Example 1.2.9) are some of the examples of integral domains and hence commutative rings.
If \(R_i\) are commutative rings then their product (see Example 1.2.10) is also commutative. Finite cyclic groups with ring structures defined in Example 1.2.5 are commutative rings.

Observation 1.4.4. (Cancellative Monoid).

Suppose that \(R\) is an integral domain. Let \(r,s\) and \(t\) be elements of \(R\text{.}\) We have the following cancellation property.
\begin{equation*} \text{Left cancellation:}\quad\text{If}\;r\neq 0\;\text{and}\;rs=rt\;\text{ then }\; s=t. \end{equation*}
Since \(rs=rt\text{,}\) by a distributive law, \(r(s-t)=0\text{.}\) As \(R\) is assumed to be an integral domain and \(r\neq 0\) we have\(s=t\text{.}\)
Note that, by similar arguments, we also have the right cancellation.
\begin{equation*} \text{Right cancellation:}\quad\text{If}\;r\neq 0\;\text{and}\;sr=tr\;\text{ then }\; s=t. \end{equation*}

Definition 1.4.5. (Reduced Ring).

A ring is said to be reduced if \(r^2=0\) then \(r=0\) in \(R\text{.}\)

Definition 1.4.6. (Division Ring).

A nonzero ring is said to be a division ring if every nonzero element is invertible, i.e., if \(r\in R\) is nonzero then there exists a unique \(s\in R\) such that \(rs=sr=1\text{.}\)

Definition 1.4.7. (Field).

A commutative division ring is called a field.

Definition 1.4.8. (Associative Algebra over a Field).

Let \(F\) be a field and \(A\) be a ring as well as a vector space over \(F\text{.}\) We say that \(A\) is an associative algebra over a field \(F\) if following conditions are satisfied.
  1. The underlying set, the addition, and the additive identity \(0\) of \(A\) as a ring and \(A\) as a vector space are the same. In other words, the underlying abelian group structure \((A,+)\) is the same for \(A\) as a ring and for \(A\) as a vector space.
  2. For \(\alpha\in F\text{,}\) \(x,y\in A\) we have
    \begin{equation} \alpha\cdot(xy)=(\alpha\cdot x)y=x(\alpha\cdot y).\tag{1.4.1} \end{equation}
    Note that \(\alpha\cdot(xy)\) (resp., \(\alpha\cdot x\) and \(\alpha\cdot y\)) is scalar multiplication inherited from the vector space structure. While \((\alpha\cdot x)y\) (resp. \(x(\alpha\cdot y)\)) is the multiplication of ring elements \(\alpha\cdot x\) and \(y\) (resp., \(x\) and \(\alpha\cdot y\)).
An associative algebra over a field \(F\) is said to be finite-dimensional if it is finite-dimensional as a vector space over \(F\text{.}\)

Observation 1.4.9.

Let \(A\) be an \(F\)-algebra. For any \(\alpha\in F\) and every \(x\in A\text{,}\) using (1.4.1), we have the following.
\begin{equation} \alpha\cdot x=\alpha\cdot(1 x)=(\alpha\cdot 1)x\tag{1.4.2} \end{equation}
Similarly, using (1.4.1), we have the following.
\begin{equation} x(\alpha\cdot 1)=\alpha\cdot(x 1)=\alpha\cdot x\tag{1.4.3} \end{equation}
Therefore, the scalar multiplication of \(\alpha\) and \(x\in A\) coincides with the ring multiplication of \(\alpha\cdot 1\) and \(x\text{.}\) Furthermore, by (1.4.2) and (1.4.3) we get that \(\alpha\cdot 1\) commutes with every element of \(A\text{,}\) i.e., \(\alpha\cdot 1\) belongs to the center of \(A\text{,}\) \(\mathcal{Z}(A)\text{.}\)
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