3.3 Characteristics of Ring
Note that each \(\, x\in \mathbb {Z}_3[i]\,\), we have \(3x = 0\)
Similarly in the sub ring \(\, \{0\, , \, 2\, , \, 4\}\, \) of \(\mathbb {Z}_6\,, \, \, 3x = 0\) for any \(x\in \{0\, , \, 2\, , \, 4\}\).
Definition 3.3.1. The characteristic of a Ring \(R\) denoted as Char\(\,R\) is the least \(+ve\) integer \(n\) such that
\(nx = 0\) for all \(x\in R\). If no such integer exists, then Char \(R = 0\).
- 1.
- \(\mathbb {Z}\,\) has characteristic.
- 2.
- \(\mathbb {Z}_n\) has characteristic.
- 3.
- Consider \(\, \mathbb {Z}_2[x]\, , \, p(x) = a_0 + a_1 x + a_2 x^2 + \cdots \cdots + a_nx^n\,\) where \(\, a_i\in \{0\, , \, 2\}\). \[\textit {Char}\, \mathbb {Z}_2 [x] = 2\]
- 4.
- The ring \(\, \big (\zeta (s)\, , \, + \, , \times \big )\,\) where \(\zeta (s)\) is the power set for a non-empty set \(s\) where for \(A\, , \, B\in \zeta (s)\, , \, \, A + B = \big (A\cup B\big )/\big (A\cap B\big )\,\) and \(\, A\cdot B = A\cap B\).
Thus, \(\, \) Char\(\big (\zeta (s)\big ) = 2\).
Example
\(s = \{a\, , \, b\} \, \implies \zeta (s) = \big \{\emptyset \, , \{a\}\, , \, \{b\}\, , \, s\big \}\)
\(0 = \emptyset \, ,\, \hspace {0.2cm} 1 = s\,\) and \(\, A+ A = A\cup A/ A\cap A = A/A = 0\, \implies 2A = 0\)
\(\implies \, \) Char \(\zeta (s) = 2\)
Theorem 3.3.3 (Characteristic of Ring with unity). Let \(R\) be a ring with unity 1. If 1 has
infinite order under addition, then \(\textit {Char}R = \). If 1 has order \(n\), then Char\(R = n\).
Proof. If \(\,\begin {vmatrix} 1\\ \end {vmatrix} = \infty \,\), then there is not \(+ve\) integer such that \(n\cdot 1 = 0\). So Char\(R = 0\). Suppose \(\begin {vmatrix} 1\\ \end {vmatrix} = n\). Then \(n\cdot 1 - 0\), so for any \(x\in R,\hspace {0.3cm} nx = n(1\, x) = (n\cdot 1)\, x = 0\cdot x = 0\). Thus
Char\(R = n\).
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Note. When the ring is an integral domain, the possibilities for the characteristic are,
constrained.
Theorem 3.3.4 (Characteristic of an integral domain). If \(D\) is an integral domain, then Char\(RD\) is
either \(0\) or prime.
Proof. By the previous theorem it suffices to show that if the additive order of 1 is finite, then
\(\begin {vmatrix} 1\\ \end {vmatrix} = \)prime.
Suppose that \(\begin {vmatrix} 1\\ \end {vmatrix} = n\) and \(n = st\) where \(1\leq s\, , \, t\leq n\). Then, \(0 = n\cdot 1 = (st)1 = (s\cdot 1)(t\cdot 1)\), so either \(s\cdot 1 = 0\) or \(t\cdot 1 = 0\) since \(n\) is the least positive integer with this
property that \(n\cdot 1 = 0\). We must have \(s=n\) or \(t=n\). Thus \(n = n\cdot 1\) or \(n = 1\cdot n\). Thus \(n\) is prime.
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Theorem 3.3.5. If \(D\) an integral domain of characteristic 0, then \(D\) contains a subring isomorphic
to \(\mathbb {Z}\).
Proof. Let 1 denote the unity of \(D\). Then define \(\,Q :\mathbb {Z} \longrightarrow D\,\) by \(Q(n) = n\cdot 1\) for each \(n\in \mathbb {Z}\). We show that \(Q\) is one-to-one and
preserves the ring operations and that it is an isomorphism. If \(Q(m) = Q n\,\) for \(m\, ,\, n\in \mathbb {Z}\)
\[m\cdot 1 = n\cdot 1\, \implies m\cdot 1 - n\cdot 1 = 0\]
\[\implies \, (m - n)1 = 0\, \implies m - n = 0\implies m = n\]
Therefore \(Q\) is one-to-one.
\[Q(m + n) = (m + n)\cdot 1 = m\cdot 1 + n\cdot 1 = Q(m) + Q(n)\]
\[Q(mn) = mn\cdot 1 = (m\cdot 1)(n\cdot 1) = Q(m)\, Q(n)\]
Thus the image of \(Q\), which is \(Q(\mathbb {Z})\) in \(D\) is isomorphic to \(\mathbb {Z}\) since we have constructed an isomorphism
for \(\mathbb {Z}\) onto \(Q(\mathbb {Z}) < D\).
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Corollary 3.3.6. If \(D\) is an integral domain of prime Char\( = p\). Then \(D\) contains a subring isomorphic
to \(\mathbb {Z}_p\).
Proof. Similar to the previous result. Define a map \(Q :\mathbb {Z}_p \longrightarrow D\) by \([R] \longrightarrow R\cdot 1\, , \,\,\)
\( R\in \{0\, , \, 1\, , \, 2\, , \, \cdots \cdots ,\, p - 1\}\). We just need to show that this is indeed an isomorphism i.e \(\mathbb {Z}_p \, \cong \, Q(\mathbb {Z}_p\subset D\).
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Check (Exercise)
- 1.
- \(Q\) is well defined (i.e \(Q\) is a ring homomorphism).
- 2.
- \(Q\) is one-to-one and onto.
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