Congruence modulo an ideal and the quotient
Definition (The congruence relation modulo an ideal)
Let \(R\) be a ring and \(I\) an \(R\)-ideal. For \(a, b \in R\), define the relation \(\sim_{I}\) such that
\[a \sim_{I} b \iff a - b \in I.\]Lemma (Congruence modulo an ideal is an equivalence relation)
The relation \(\sim_{I}\) is an equivalence relation on \(R\).
Proof: We verify the three properties:
- Reflexivity: \(a \sim_{I} a\), since \(a - a = 0 \in I\).
- Symmetry: if \(a \sim_{I} b\), then \(a - b \in I\); since \(I\) is closed under additive inverses, \(b - a = -(a - b) \in I\), that is, \(b \sim_{I} a\).
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Transitivity: if \(a \sim_{I} b\) and \(b \sim_{I} c\), then \(a - b \in I\) and \(b - c \in I\). Since \(I\) is closed under sums,
\[a - c = (a - b) + (b - c) \in I,\]that is, \(a \sim_{I} c\).
Note (The equivalence classes are cosets)
Let \(R/I = \{ [a]_{I} :\ a \in R \}\) be the set of equivalence classes of \(\sim_{I}\). Then
\[\begin{aligned} [a]_{I} &= \{ b \in R :\ a \sim_{I} b \} = \{ b \in R :\ a - b \in I \} \\ &= \{ b \in R :\ \exists x \in I\ (x = a - b) \} = \{ b \in R :\ \exists x \in I\ (b = a - x) \} \\ &= \{ a - x :\ x \in I \} = a + I, \end{aligned}\]where the last equality uses that \(I\) is closed under additive inverses, so that \(\{-x : x \in I\} = I\). Notation: \([a]_{I} = a + I\).
Definition (The operations of the quotient ring)
We define \(\tilde{+}\) on \(R/I\) such that
\[(a + I)\ \tilde{+}\ (b + I) = (a + b) + I.\]The ideal \(I\) is the “\(0\)” of \(R/I\) (that is, \([0]_{I}\)). We define \(\tilde{\cdot}\) on \(R/I\) such that
\[(a + I)\ \tilde{\cdot}\ (b + I) = (a \cdot b) + I.\]The class \(1_{R} + I\) is the “\(1\)” of \(R/I\).
Theorem (The quotient is a ring)
\(R/I\) is a ring under these operations. It is called the quotient ring of \(R\) by \(I\).
Proof: The delicate part is to verify that \(\tilde{+}\) and \(\tilde{\cdot}\) are well defined, that is, that they do not depend on the representatives. Suppose that \(a + I = a' + I\) and \(b + I = b' + I\), that is, \(a - a' \in I\) and \(b - b' \in I\). For the sum,
\[(a + b) - (a' + b') = (a - a') + (b - b') \in I,\]so that \((a + b) + I = (a' + b') + I\). For the product, we add and subtract \(a' b\):
\[a b - a' b' = (a - a') b + a' (b - b'),\]and here \((a - a') b \in I\) by absorption on the right and \(a' (b - b') \in I\) by absorption on the left; hence \(a b - a' b' \in I\), that is, \((a b) + I = (a' b') + I\). This is where it is essential that \(I\) is a two-sided ideal.
The ring axioms are inherited from \(R\) by operating with representatives; for example,
\[\big( (a + I)\, \tilde{\cdot}\, (b + I) \big)\, \tilde{\cdot}\, (c + I) = (a b) c + I = a (b c) + I = (a + I)\, \tilde{\cdot}\, \big( (b + I)\, \tilde{\cdot}\, (c + I) \big),\]and likewise the associativity and commutativity of \(\tilde{+}\), distributivity, the identity elements \(I = 0 + I\) and \(1 + I\), and the additive inverse \((-a) + I\).
Definition (The canonical projection)
If \(R\) is a ring and \(I\) is an \(R\)-ideal, let \(\pi_{I} : R \to R/I\) be such that \(a \mapsto a + I\), the canonical projection of \(R\) onto \(R/I\).
Proposition (Properties of the canonical projection)
The canonical projection \(\pi_{I}\) is surjective and \(\operatorname{Ker}(\pi_{I}) = I\).
Proof: Surjectivity is immediate: every class \(a + I \in R/I\) is \(\pi_{I}(a)\). For the kernel, we have the two inclusions:
\[\text{if } x \in I \implies \pi_{I}(x) = x + I = I = [0]_{I} \implies x \in \operatorname{Ker}(\pi_{I}),\]since \(x - 0 = x \in I\); and conversely,
\[\text{if } x \in \operatorname{Ker}(\pi_{I}) \implies \pi_{I}(x) = x + I = I \implies x \in I,\]since \(x = x - 0 \in I\). We conclude that \(\operatorname{Ker}(\pi_{I}) = I\).
Examples of quotients
Example (The integers modulo \(n\) as a quotient)
\(\mathbb{Z}/(n) = \mathbb{Z}_{n}\).
Example (The quotient \(\mathbb{Z}[x]/(x^{2})\))
If \(p(x) + (x^{2}) \in \mathbb{Z}[x]/(x^{2})\), then there exist unique \(a, b \in \mathbb{Z}\) such that
\[[p(x)]_{(x^{2})} = [a + b x]_{(x^{2})}.\]For existence, write \(p(x) = a + b x + x^{2} q(x)\) with \(q(x) \in \mathbb{Z}[x]\); then \(p(x) - (a + bx) = x^{2} q(x) \in (x^{2})\). We now prove uniqueness. Let \(a_{0}, a_{1}, b_{0}, b_{1} \in \mathbb{Z}\) be such that
\[[a_{0} + a_{1} x]_{(x^{2})} = [b_{0} + b_{1} x]_{(x^{2})}.\]Then
\[(a_{0} - b_{0}) + (a_{1} - b_{1}) x \in (x^{2}).\]But every non-zero element of \((x^{2}) = x^{2} \cdot \mathbb{Z}[x]\) has degree at least \(2\), that is, its coefficients of degrees \(0\) and \(1\) vanish; hence \(a_{0} = b_{0}\) and \(a_{1} = b_{1}\). We conclude that
\[\mathbb{Z}[x]/(x^{2}) = \{ [a + b x]_{(x^{2})} :\ a, b \in \mathbb{Z} \}.\]Note moreover that
\[[3x]_{(x^{2})} \cdot [5x]_{(x^{2})} = [15 x^{2}]_{(x^{2})} = [0]_{(x^{2})},\]so that \(\mathbb{Z}[x]/(x^{2})\) has zero divisors.
Exercise (Zero divisors and units of \(\mathbb{Z}[x]/(x^{2})\))
Show that the zero divisors of \(\mathbb{Z}[x]/(x^{2})\) are \(\{ [b x] :\ b \neq 0 \}\) and that the units are \(\{ [\pm 1 + b x] :\ b \in \mathbb{Z} \}\).
Example (A quotient of the Gaussian integers)
Let \(\mathbb{Z}[i] = \{ a + b i :\ a, b \in \mathbb{Z} \}\). Then \(\mathbb{Z}[i]/(1 + i) = \{ [0], [1] \}\) (exercise).
Example (The quotient \(\mathbb{R}[x]/(x^{2} + 1)\))
Let \(R = \mathbb{R}[x]/(x^{2} + 1)\) and \(\alpha = [x]\). Then
\[\alpha^{2} = [x^{2}] = [-1],\]since \(x^{2} - (-1) = x^{2} + 1 \in (x^{2} + 1)\). That is, in \(R\) the equation “\(t^{2} + 1 = 0\)” has a solution. We shall see that \(R \cong \mathbb{C}\).
The correspondence theorem
Definition (Quotient of a subset)
Let \(R\) be a ring, \(I\) an \(R\)-ideal and \(S \subseteq R\) (merely a subset). We define
\[S/I := \{ s + I :\ s \in S \} \subseteq R/I.\]Theorem (Correspondence between ideals and subrings of the quotient)
The map \(S \mapsto S/I\) induces inclusion-preserving bijections between the following sets:
- \(\{ J :\ J \text{ ideal of } R \text{ such that } J \supseteq I \} \to \{ \text{ideals of } R/I \}\);
- \(\{ \text{subrings of } R \text{ that contain } I \} \to \{ \text{subrings of } R/I \}\).
In both cases the inverse function is \(A \mapsto \{ r \in R :\ r + I \in A \}\).
Proof: We prove the case of ideals; that of subrings is identical, using the subring criterion instead of the definition of an ideal. Write \(\Phi(J) = J/I\) and \(\Psi(A) = \{ r \in R :\ r + I \in A \}\).
\(\Phi\) lands where it should. If \(J \supseteq I\) is an ideal of \(R\), then \(J/I\) is an ideal of \(R/I\): it is non-empty; it is closed under sums, since \((j + I) + (j' + I) = (j + j') + I\) with \(j + j' \in J\); and it absorbs products, since \((r + I)(j + I) = r j + I\) with \(r j \in J\) by the absorption of \(J\) in \(R\).
\(\Psi\) lands where it should. If \(A\) is an ideal of \(R/I\), then \(\Psi(A)\) is an ideal of \(R\): if \(r, r' \in \Psi(A)\), then \((r + r') + I = (r + I) + (r' + I) \in A\), so \(r + r' \in \Psi(A)\); and if \(s \in R\), then \((s r) + I = (s + I)(r + I) \in A\) by the absorption of \(A\). Moreover \(I \subseteq \Psi(A)\): if \(x \in I\), then \(x + I = I = [0] \in A\), since every ideal contains zero.
They are mutually inverse. First, \(\Psi(\Phi(J)) = \{ r :\ r + I \in J/I \} = J\): the inclusion “\(\supseteq\)” is clear, and if \(r + I = j + I\) for some \(j \in J\), then \(r - j \in I \subseteq J\), whence \(r = j + (r - j) \in J\). Second, \(\Phi(\Psi(A)) = \{ r + I :\ r + I \in A \} = A\), since every element of \(A\) is a class \(r + I\) for some \(r \in R\).
They preserve inclusion. If \(J \subseteq J'\), then \(J/I \subseteq J'/I\) directly; and if \(A \subseteq A'\), then \(\Psi(A) \subseteq \Psi(A')\) directly. (In the case of subrings, \(\Psi(B) \supseteq I\) holds because \([0] \in B\), and the computation of the inverses uses \(I \subseteq S\) in the same way.)