3.1 Properties of Rings
Theorem 3.1.1. Let \(R\) be a ring and \(\, a\, , \, b\, , \, c\in R\).
- 1.
- The zero elt of \(R\) is unique.
- 2.
- Each elt of \(R\) has a unique negative.
- 3.
- If \(\, a + b = a + c\), then \(b = c\) (left cancellation)
- 4.
- \(b + a = c + a\, \implies \, b = c\,\) (Right cancellation)
- 5.
- Each of the equations \(a + x = b\) and \(x + a = b \) has a unique solution.
- 6.
- \(-(-a) = a\) and \(-(a + b) = (-a + (-b))\)
- 7.
- If \(m\) and \(n\) are integers, then \((m + n) a = ma + na\), and \(n(a+b) = na + nb\), \(\, \, n(ma) = nma\)
Proof. Each part follows from the distributive law and the uniqueness of additive inverses.
For \(a0 = 0\): since \(0+0=0\), distributivity gives \(a0 = a(0+0) = a0 + a0\). Adding \(-(a0)\) to both sides leaves \(0 = a0\).
For \(a(-b) = -(ab)\): distributivity gives \[ab + a(-b) = a\left (b + (-b)\right ) = a0 = 0,\] so \(a(-b)\) is the additive inverse of \(ab\). The identity \((-a)b = -(ab)\) is the same argument on the other side, and \((-a)(-b) = -\left (a(-b)\right ) = -\left (-(ab)\right ) = ab\).
For \(a(b-c) = ab-ac\): write \(b-c = b+(-c)\) and apply distributivity together with the previous part. □
Theorem 3.1.2. Let \(R\) be a ring, \(0\) the zero of \(R\) and \(a\, , \, b\, , \, c \in R\). Then;
- 1.
- \(0\,a = a\, 0 = 0\)
- 2.
- \(a (-b) = (-a)\, b = - (ab)\)
- 3.
- \((-a)(-b) = ab\)
- 4.
- \(a(b-c) = ab - ac\), and \((b - c)a = ba - ca\)
In the case where \(R\) has a multiplication unity
- 5.
- \((-1)\, a = -a\)
- 6.
- \((-1)(-1) = 1\)
Proof.
- 1.
- \(0\, a + 0\, a = 0\, a = 0\, a + 0\). Now, this has a unique additive inverse and by left cancellation we have \(0\, a = 0\).
Use the same argument to get \(a\, 0 = 0\).
- 2.
- The equation \(\, x + ab = 0\) has \(x = - (ab)\) as solution. But \(a(-b) + (ab) = a(-b + b) = a\, 0 = 0\). Hence \(a(-b)\) is also a solution. By uniqueness, \(-(ab) = a(-b)\). Also \(\, (-a) b + ab = (-a + a) b = 0\).
Hence \((-a)b\) is also a solution so that by uniqueness property we have \(a(-b) = (-a)b = -(ab)\)
Q6 Examples
- \begin {align*} \text {(v)}\hspace {1cm} p (q\cdot r) & = \begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\begin {bmatrix} \begin {pmatrix} w & x\\ y & z\\ \end {pmatrix}\begin {pmatrix} k & l\\ m & n\\ \end {pmatrix} \end {bmatrix}\\\\ & = \begin {bmatrix} awk + axm + byk + bzm & awl + axn + byl + bzn\\ cwk + cxm + dyk + dzm & cwl + cxn + dyl + dzn\\ \end {bmatrix}\\\\ & = \begin {bmatrix} awk + byk + axm + bzm & awl + byl + axn + bzn\\ cwk + dyk + cxm + dzm & cwl + dyl + cxn + dzn\\ \end {bmatrix}\\\\ & = \begin {bmatrix} (aw + by)k + (ax + bz)m & (aw + by)l + (ax + bz)n\\ (cw + dy)k + (cx + dz)m & (cw + dy)l + (cx + dz)n\\ \end {bmatrix}\\\\ & = \begin {bmatrix} aw + by & ax + bz\\ cw + dy & cx + dz\\ \end {bmatrix}\begin {bmatrix} k & l\\ m & n\\ \end {bmatrix} = \begin {bmatrix} \begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\, \begin {pmatrix} w & x\\ y & z\\ \end {pmatrix} \end {bmatrix}\begin {pmatrix} k & l\\ m & n\\ \end {pmatrix}\\\\ & = (p\cdot q)\cdot r\\ \end {align*}
- \begin {align*} \text {(vi)}\hspace {1cm} p (q + r) & = \begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\, \begin {bmatrix} \begin {pmatrix} w & x\\ y & z\\ \end {pmatrix} + \begin {pmatrix} k & l\\ m & m\\ \end {pmatrix} \end {bmatrix} = \begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\, \begin {bmatrix} w + k & x + l\\ y + m & z + n\\ \end {bmatrix}\\\\ & = \begin {bmatrix} aw + ak + by + bm & ax + al + bz + bn\\ cw + ck + dy + dm & cx + cl + dz + dn\\ \end {bmatrix}\\\\ & = \begin {bmatrix} aw + by + ak + bm & ax + bz + al + bn\\ cw + dy + ck + dm & cx + dz + cl + dn\\ \end {bmatrix}\\\\ & = \begin {bmatrix} aw + by & ax + bz\\ cw + dy & cx + dz\\ \end {bmatrix} + \begin {bmatrix} ak + bm & al + bn\\ ck + dm & cl + dn\\ \end {bmatrix}\\\\ & = pq + pr\\\\ \end {align*}
- 3.
- We have \(\, (-a)(-b) = (- (-a))(b) = ab\).
- 4.
- We write \(\, a\, (b - c) = a\, (b + (-c)) = ab + a\, (-c)\) \begin {align*} & = ab + \big (- (ac)\big )\\ & = ab - ac \end {align*}
Show also for \((a - b)\,c = ac - bc\)
We write \(\, (a - b)\, c = \big ( a + (-b)\big ) \, c = ac + (-b)c\) \begin {align*} & = ac + (-(bc))\\ & = ac - bc\\\\ \end {align*}
Exercise
For the given examples of rings, verify that the ring properties are satisfied.
□
Theorem 3.1.3. If a ring \(R\) has unity, then it is unique. If a ring elt has an inverse , then it is
unique.
Proof. Suppose \(R\) has no unity, let \(1\, , \, 1'\in R\) be unity in \(R\). \(\, a\, 1 = a\, 1'\) for \(a\neq 0\,\)
\(\implies \, a\, 1 - a\, 1' = 0\, \implies a (1 - 1') = 0\). Now, since \(a\neq 0\), then \(1 - 1' = 0\, \implies 1 = 1'\).
The proof for uniqueness of inverse is similar (check).
Suppose all elt \(a\in R\) has atleast two inverses \(b\) and \(c\). Then we have that \(a + b = 0\) and \(a + c = 0\). Now, we have \(a + b = a + c\, \implies a + b - a - c = a- a + b - c = 0 \implies b - c = 0\implies b = c\). As
required.
Suppose \(a\in R\) has no unique inverse such that \(b\, , \, c\in R\) are its inverses. Then we have \(a\, b = 1 = a\, c\implies ab = ac\implies ab - ac = 0\implies a(b-c) = 0\implies b - c = 0\implies b=c \) since \(a\neq 0\).
□
Definition 3.1.4 (Sub ring). A subset \(S\) of a ring \(R\) is a sub ring of \(R\) if \(S\) is itself a ring with respect
to the operations in \(R\).
Similar to subgroups there is a simple test for for sub rings.
Theorem 3.1.5 (Sub ring Test). A non-empty set \(S\) of a ring \(R\) is a subgroup if \(S\) is closed under
subtraction and multiplication. i.e \(\, \forall a\, , \, b\in S\, \, \, a- b \in S\) and \(ab\in S\).
Proof. Addition is commutative in \(R\) and \(S\). Now, suppose \(S\) is closed under subtraction. Then
by one-step subgroup test, \(\, \forall \, a\, , \, b\in S\,\,\, a-b\in S\). Also, multiplication is associative and distributive over addition.
Finally closure of multiplication shows that indeed multiplication is a binary operation on \(S\).
□
- 1.
- \(\{0\}\) and \(R\) are subrings of any ring \(R\). \(\{0\}\) is called the trivial subring of \(R\).
- 2.
- \(\{0\,, \, 2\, , \, 4\}\) is a subring of the ring \(\mathbb {Z}_6\) of integers modulo 6.
\(a - b \in \{0\, , \, 2 \, , \, 4\}\hspace {0.4cm}\forall \, a\, , \, b\in \{0\, , \, 2 \, , \, 4\}\)\(+\) \(-0 = 0\) \(-2 = 4\) \(-4 = 2\) 0 0 4 2 2 2 0 4 4 4 2 0 \(ab\in \{0\, , \, 2 \, , \, 4\}\hspace {0.4cm}\forall \, a\, , \, b \in \{0\, , \, 2 \, , \, 4\}\)\(\times \) 0 2 4 0 0 0 0 2 0 4 2 4 0 2 4 \(\therefore \,\, \{0\, , \, 2 \, , \, 4\}\,\) is a sub ring of \(\mathbb {Z}_6\)
- 3.
- The set of even integers \(\, 2\mathbb {Z} = \big \{0\, , \, \pm 2\, , \, \pm 4\, , \, \pm 6\, , \cdots \cdots \}\) is a subring of the integers \(\mathbb {Z}\).
Check \(\hspace {0.4cm} a\, , \, b\in 2\mathbb {Z} \, , \, a-b\in 2\mathbb {Z}\)
Let \(\, a = 2k\hspace {0.3cm} b = 2m\). Then \(a - b = 2k -2m = 2(k - m) \in 2\mathbb {Z}\)
also \(\, ab = 2k\cdot 2m = 4km = 2(2km) \in 2\mathbb {Z}\hspace {0.2cm} \forall \, k\, , \, m\mathbb {Z}\).
- 4.
- The set of Gaussian integers. \(\mathbb {Z}[i] = a + bi\, , \, a\, , b\in \mathbb {Z}\) is a sub ring of the complex numbers \(\mathbb {C}\).
Check as Exercise
Let \(\, a = x + iy\, \) and \(\, b = m + i n\in \mathbb {Z}[i]\). Then we have- (a)
- \(a + b = x + i y + m + ni = x + m + i(y+n) \hspace {0.3cm} \forall x\, , \, y\, , m\, , \, n\in \mathbb {Z}\) \(\implies \, a + b = p + i q\in \mathbb {Z}[i]\,\,\) where \(p = x + m\, , \, q = y + n\in \mathbb {Z}\)
- (b)
- \(ab = (x + iy)(m + in) = xm - yn + ixn + iym = x m - yn + i(xn + ym) = R + iS\in \mathbb {Z}[i]\)
where \(R = xm - yn\) and \(S = xn + ym\implies R\, , \, S\in \mathbb {Z}\).
Thus a subring of \(\mathbb {C}\) indeed.
Some Non Examples
- 1.
- \(\mathbb {N}\) is not a subring of \(\mathbb {Z}\) since \(0\in \mathbb {N}\), for each \(0\in \mathbb {N}\, , \, -a\not \in \mathbb {N}\).
- 2.
- \(\mathbb {Q}^+\) is not a subring of \(\mathbb {Q}\) since \(0\in \mathbb {Q}^+\), for all \(a\, , \, b\in \mathbb {Q}\, , \,\, -a/b\not \in \mathbb {Q}\).
Definition 3.1.7. An element \(a\neq 0\) in a commutative ring \(R\) is called a zero divisor in \(\,\exists \,\) an element \(b\neq 0\)
in \(R\) such that \(ab = 0\).
- 1.
- \(\mathbb {Z}\), the ring of integers has no zero divisors since \(\big (a\cdot \big (-a\big )\big ) = 0\not \in \mathbb {Z}\).
- 2.
- Integers 2 and 3 are zero divisors of \(\mathbb {Z}_6\).
- 3.
- Integers \(2\, , \, 3\, , \, 6\, , \, 9\) are zero divisors in \(\mathbb {Z}_{18}\).
Definition 3.1.9. A commutative ring with unity \(1\neq 0\) and no zero divisors is called an integral
domain.
- 1.
- The ring of integers \(\mathbb {Z}\) is an integral domain. (Since there are no zero divisors).
- 2.
- The ring of \(\mathbb {Q}\) of rational numbers is an integral domain.
- 3.
- \(2\mathbb {Z}\) the ring of even integers is not an integral domain. (since \(1\not \in 2\mathbb {Z}\)).
- 4.
- The ring \(\mathbb {Z}_6\) is not an integral domain. (since \(\mathbb {Z}_6\) has zero divisors).
- 5.
- The ring \(\, M(2,\mathbb {Z})\, , \, M(2,\mathbb {Z}) = \Big \{\begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\, \Big |\, a\, , \, b\, , \, c\, , \, d\in \mathbb {Z}\Big \}\) is not an integral domain. (since it is not commutative).
- 6.
- \(\mathbb {Z}\oplus \mathbb {Z} = \big \{(a\, , \, b)\, \big |\, a\, , \, b\in \mathbb {Z}\big \}\hspace {0.5cm} (a\, , \, b) = (1\, , \, 1) = \) unity. However we need a situation \(\, (a_1\, , \, b_1) \neq 0\) and \(\, (a_2\, , \, b_2)\neq 0\, \hspace {0.3cm} (a_1\, , \, b_1)(a_2\,, \, b_2) = (a_1a_2\, , \, b_1b_2) = (0\, , \, 0)\) it has zero divisors i.e \((a\, , \, 0) \neq 0\) and \((0\, , \, b) \neq 0\) and \((a\, ,\, 0)(0\, , \, b) = (0\, , \, 0)\). Thus \((a\, , \, 0)\) and \((0\,, \, b)\)
are zero divisors in \(\mathbb {Z}\oplus \mathbb {Z}\).
Theorem 3.1.11 (Cancellation). Let \(a\, , \, b \) and \(c\) belong to an integral domain. If \(a\neq 0\) and \(ab = ac\), then \(b = c\).
Proof. From \(ab = ac\), we have \(ab - ac = 0 \implies a (b - c) = 0\) since an integral domain has no zero divisors, \(a\neq 0\implies b - c = 0 \implies b = c\).
In some cases, one may use the cancellation property to define an integral domain.
□
Definition 3.1.12 (Ring Homomorphism). Let \(R\) and \(S\) be rings. A mapping \(Q : R\longrightarrow S\,\) is a ring homomorphism if ;
- \(Q(a + b) = Q(a) + Q(b)\)
- \(Q(ab) = Q(a)\cdot Q(b).\hspace {0.5cm}\forall a\, , \, b\in R\).
Note.
- •
- A Ring Homomorphism that is bijective is called a ring isomorphism.
- •
- By the first condition \(Q(a+b) = Q(a) + Q(b)\,\) a ring homomorphism is purely an additive group homomorphism.
- 1.
- \(Q : \mathbb {Z} \longrightarrow \mathbb {Z}_n\hspace {0.5cm} Q(a) = [a]_n\) is a ring homomorphism. Since for \(a\, , \, b\in \mathbb {Z}\, \) \begin {align*} Q(a + b) & = [a + b]_n = [a]_n + [b]_n\\ & = Q(a) + Q(b) \end {align*}
e.g mod\(_9\hspace {0.5cm} [ 11 + 13]_9 = [24]_9 = [6]_9\) \[[11]_9 + [13]_9 = [2]_9 + [4]_9 = [6]_9\] And \(\hspace {0.5cm} Q(ab) = [ab]_n = [a]_n\, [b]_n = Q(a)\cdot Q(b)\).
- 2.
- \(\phi : \mathbb {Z} \longrightarrow M_2(\mathbb {Z})\hspace {0.5cm} \phi (a) \longrightarrow \begin {pmatrix} 0 & 0\\ a & a\\ \end {pmatrix}\)
Check!!
Let \(\, a\, , \, b\in \mathbb {Z}\,\) \begin {align*} \phi (a + b) & = \begin {pmatrix} 0 & 0\\ a + b & a + b\\ \end {pmatrix} = \begin {pmatrix} 0 & 0\\ a & a\\ \end {pmatrix} + \begin {pmatrix} 0 & 0\\ b & b\\ \end {pmatrix}\\\\ & = \phi (a) + \phi (b) \end {align*}\[\phi (ab) = \begin {pmatrix} 0 & 0\\ ab & ab\\ \end {pmatrix} = \begin {pmatrix} 0 & 0\\ a & a\\ \end {pmatrix}\, \begin {pmatrix} 0 & 0\\ b & b\\ \end {pmatrix} = \phi (a)\cdot \phi (b)\]
\(\phi \, \) is indeed a ring homomorphism.
Check if \(1-1\, / \) onto. If \(\phi (a) = \phi (b)\). Then \(\begin {pmatrix} 0 & 0\\ a & a\\ \end {pmatrix} = \begin {pmatrix} 0 & 0\\ b & b\\ \end {pmatrix}\implies a = b\). Thus \(\phi \) is injection and \(\hspace {0.3cm} \forall \, \begin {pmatrix} 0 & 0\\ a & a\\ \end {pmatrix} \in M_2(\mathbb {Z})\, \exists \, a\in \mathbb {Z}\ni \phi (a) = \begin {pmatrix} 0 & 0\\ a & a\\ \end {pmatrix}\). Thus onto.
- 3.
- \(\phi : \mathbb {Z}[x]\longrightarrow \mathbb {Z}\hspace {0.5cm}\) by \(\hspace {0.2cm} \phi (f(x))\longrightarrow f(0) = a_0\,\) from \(f(x) = a_0 + a_1 x + a_2 x^2 + \cdots \cdots + a_nx^n\)
Note. Let \(\, f(x)\, , \, g(x)\in \mathbb {Z}[x]\)
\begin {align*} f(x) & = a_0 + a_1x + a_2 x^2 + \cdots \cdots + a_n x^n\\ g(x) & = b_0 + b_1x + b_2 x^2 + \cdots \cdots + b_n x^n\hspace {0.8cm} a_i\, , \, b_i\in \mathbb {Z} \end {align*}
\begin {align*} \phi (f(x) + g(x)) & = \phi \big [(a_0 + b_0) + (a_1x + b_1x) + (a_2x^2 + b_2x^2) + \cdots \cdots + (a_nx^n + b_nx^n)\big ]\\ & = \big [a_0 + b_0\big ]\hspace {.7cm} \text {since}\hspace {0.3cm} x = 0\\ & = (a_0) + (b_0)\\ & = f(0) + g(0)\\ & = \phi (f(x)) + \phi (g(x)) \end {align*}
\begin {align*} \phi \big (f(x)\,g(x)\big ) & = \phi \big [(a_0 + a_1x + a_2 x^2 + \cdots \cdots + a_n x^n)(b_0 + b_1x + b_2 x^2 + \cdots \cdots + b_n x^n)\big ]\\ & = \phi \big (a_0b_0 + a_0b_1x + a_0b_2x^2 + \cdots \cdots + b_0a_2x + b_0a_2x^2 + \cdots \cdots \big )\\ & = (a_0\cdot b_0)\hspace {1cm} \boxed {\text {after}\hspace {0.2cm} \phi \hspace {0.2cm}\text {works on}\hspace {0.2cm} f(x)\, g(x)\hspace {0.3cm} x = 0.\, \, \text {yielding}\hspace {0.2cm} a_0\cdot b_0}\\ & = (a_0)\cdot (b_0)\\ & = f(0)\cdot g(0)\\ & = \phi (f(x))\cdot \phi (g(x))\\ \end {align*}
- 4.
- \(\phi : \mathbb {Z}[i] \longrightarrow \mathbb {Z}[i]\hspace {0.5cm} a + bi \longrightarrow a - bi\)
Note. Let \(\, a + b i\, \, , \, \, c + di \in \mathbb {Z}[i]\)
\begin {align*} \phi \big ((a + bi) + (c + di)\big ) & = \phi \big ((a + c) + (b + d)i\big )\\ & = (a + c) - (b+d)i = (a - bi) + (c - di)\\ & = \phi (a + bi) + \phi (c + di) \end {align*}
\begin {align*} \phi \big ((a+bi)(c+di\big ) & = \phi \big ((ac-bd) + (ad + bc)i\big )\\ & = \big ((ac - bd) - (ad + bc)i\big )\\ & = ac - adi + bc i - bd\\ & = a(c - di) - b(c - di) = (a-bi)(c-di)\\ & = \phi (a + bi)\cdot \phi (c + di)\\\\ \end {align*}
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.