Definitions and first examples
Definition (Ring homomorphism)
Let \(R, S\) be rings. Then \(\varphi : R \to S\) is a ring homomorphism if:
- \(\varphi(a +_{R} b) = \varphi(a) +_{S} \varphi(b)\) for all \(a, b \in R\);
- \(\varphi(a \cdot_{R} b) = \varphi(a) \cdot_{S} \varphi(b)\) for all \(a, b \in R\);
- \(\varphi(1_{R}) = 1_{S}\).
Note (Homomorphisms preserve zero)
The definition implies that \(\varphi(0_{R}) = 0_{S}\), since
\[\varphi(0_{R}) = \varphi(0_{R} + 0_{R}) = \varphi(0_{R}) +_{S} \varphi(0_{R}) \implies 0_{S} = \varphi(0_{R}),\]cancelling \(\varphi(0_{R})\) in the group \((S, +)\).
Definition (Monomorphisms, isomorphisms and automorphisms)
Let \(\varphi : R \to S\) be a ring homomorphism.
- If \(\varphi\) is injective, then \(\varphi\) is a monomorphism.
- If \(\varphi\) is bijective, then \(\varphi\) is an isomorphism.
- If \(\varphi\) is bijective and \(R = S\), then \(\varphi\) is an automorphism.
If \(R\) and \(S\) are rings such that there exists an isomorphism \(\varphi : R \to S\), we say that \(R\) and \(S\) are isomorphic and write \(R \cong S\).
Example (\(\mathbb{R}[x]/(x^{2}+1)\) is isomorphic to \(\mathbb{C}\))
We have already seen that \(\mathbb{R}[x]/(x^{2} + 1) = \{ [a + b x] :\ a, b \in \mathbb{R} \}\) (the normal form argument is the same as for \(\mathbb{Z}[x]/(x^{2})\)). The map
\[\varphi : \mathbb{C} \longrightarrow \mathbb{R}[x]/(x^{2} + 1), \qquad a + i b \longmapsto [a + b x],\]is an isomorphism (exercise). In particular, \(\mathbb{R}[x]/(x^{2}+1) \cong \mathbb{C}\).
Lemma (Homomorphisms preserve units)
Let \(R, S\) be rings and \(\varphi : R \to S\) a homomorphism. If \(a \in R^{\times}\), then \(\varphi(a) \in S^{\times}\) and \(\varphi(a^{-1}) = (\varphi(a))^{-1}\).
Proof: Let \(a \in R^{\times}\) with inverse \(a^{-1} \in R\), so that \(a \cdot a^{-1} = 1_{R} = a^{-1} \cdot a\). Applying \(\varphi\),
\[\varphi(a \cdot a^{-1}) = \varphi(1_{R}) \implies \varphi(a) \cdot_{S} \varphi(a^{-1}) = 1_{S} = \varphi(a^{-1}) \cdot_{S} \varphi(a),\]where the second equality is obtained by applying \(\varphi\) to \(a^{-1} \cdot a = 1_{R}\). Hence \(\varphi(a^{-1})\) is a multiplicative inverse of \(\varphi(a)\), that is, \(\varphi(a^{-1}) = (\varphi(a))^{-1}\) and \(\varphi(a) \in S^{\times}\).
Example (The inclusion of a subring and reduction modulo \(n\))
Two basic examples of homomorphisms:
- Let \(S\) be a subring of \(R\). The map \(\varphi : S \to R\) such that \(x \mapsto x\) is a monomorphism.
- The map \(\varphi : \mathbb{Z} \to \mathbb{Z}_{n}\) such that \(x \mapsto [x]\) is a homomorphism.
Kernel and image
Definition (Kernel and image of a homomorphism)
Let \(R, S\) be rings and \(\varphi : R \to S\) a homomorphism.
- \(\operatorname{Ker}(\varphi) = \{ x \in R :\ \varphi(x) = 0_{S} \} \subseteq R\);
- \(\operatorname{Im}(\varphi) = \{ \varphi(x) :\ x \in R \} = \varphi(R) \subseteq S\).
Note (The kernel is an ideal and detects injectivity)
\(\operatorname{Ker}(\varphi)\) is an \(R\)-ideal. Moreover, \(\operatorname{Ker}(\varphi) = \{0_{R}\}\) if and only if \(\varphi\) is injective.
Note (Composition of homomorphisms)
The composition of homomorphisms is a homomorphism, and the inverse of an isomorphism is an isomorphism. Moreover,
\[\varphi_{1} \circ (\varphi_{2} \circ \varphi_{3}) = (\varphi_{1} \circ \varphi_{2}) \circ \varphi_{3}.\]Example (Reduction of coefficients modulo \(n\) and its kernel)
Let \(r_{n} : \mathbb{Z}[x] \longrightarrow \mathbb{Z}_{n}[x]\) be such that
\[r_{n}(a_{0} + a_{1} x + \dots + a_{m} x^{m}) = [a_{0}] + [a_{1}] x + \dots + [a_{m}] x^{m}.\]The map \(r_{n}\) is a homomorphism (exercise) and is surjective. Moreover, \(\operatorname{Ker}(r_{n}) = (n\mathbb{Z})[x]\), since
\[\begin{aligned} a_{0} + a_{1} x + \dots + a_{m} x^{m} \in \operatorname{Ker}(r_{n}) &\iff [a_{i}] = [0] \quad \forall i \in \{0, 1, \dots, m\} \\ &\iff n \mid a_{i} \quad \forall i \in \{0, 1, \dots, m\}. \end{aligned}\]Example (The evaluation homomorphism and its kernel)
Let \(a \in \mathbb{R}\) and let \(\varphi_{a} : \mathbb{R}[x] \to \mathbb{R}\) be such that \(\varphi_{a}(p(x)) = p(a)\). Then \(\varphi_{a}\) is a homomorphism (exercise). Its kernel is
\[\begin{aligned} \operatorname{Ker}(\varphi_{a}) &= \{ p(x) \in \mathbb{R}[x] :\ p(a) = 0 \} \\ &= \{ p(x) \in \mathbb{R}[x] :\ \exists q(x) \in \mathbb{R}[x]\ \big( p(x) = (x - a) \cdot q(x) \big) \} \\ &= \{ (x - a) \cdot q(x) :\ q(x) \in \mathbb{R}[x] \} \\ &= (x - a) \cdot \mathbb{R}[x] \neq \{ 0_{\mathbb{R}[x]} \}, \end{aligned}\]where the second equality is the root factorisation theorem.
Lemma (Properties invariant under isomorphisms)
Let \(R, S\) be rings and \(\varphi : R \to S\) an isomorphism. Then:
- \(R\) is commutative \(\iff\) \(S\) is commutative;
- \(x \in R^{\times} \iff \varphi(x) \in S^{\times}\);
- \(R\) is a field \(\iff\) \(S\) is a field;
- \(x \in R\) is a zero divisor \(\iff\) \(\varphi(x) \in S\) is a zero divisor.
Proof: Exercise.
Example (\(\mathbb{R} \times \mathbb{R}\) is not isomorphic to \(\mathbb{C}\) as a ring)
Let \(\varphi : \mathbb{R} \times \mathbb{R} \to \mathbb{C}\) be such that \((a, b) \mapsto a + i b\). It is an isomorphism of abelian groups, but not of rings, since \(\mathbb{C}\) is an integral domain and \(\mathbb{R} \times \mathbb{R}\) is not (it has zero divisors).
Example (\(\mathbb{Q}(i)\) is not isomorphic to \(\mathbb{Q}\))
Let \(\mathbb{Q}(i) = \{ a + i b :\ a, b \in \mathbb{Q} \}\). Then \(\mathbb{Q}(i)\) is a field and \(\mathbb{Q}(i) \not\cong \mathbb{Q}\). Indeed, there exists \(\alpha \in \mathbb{Q}(i)\) such that \(\alpha^{2} = -1\) (namely, \(\alpha = i\)), that is, such that \(\alpha^{2} + 1 = 0\). If \(f : \mathbb{Q}(i) \to \mathbb{Q}\) were an isomorphism, we would have that
\[f(\alpha^{2} + 1) = f(0) = 0 \implies (f(\alpha))^{2} + f(1) = 0 \implies (f(\alpha))^{2} + 1 = 0.\]This would imply that in \(\mathbb{Q}\) the equation \(x^{2} + 1 = 0\) has a solution, which is impossible, since \(x^{2} \geq 0\) for every \(x \in \mathbb{Q}\). We conclude that no such isomorphism exists.
Example (The only automorphism of \(\mathbb{Z}\) is the identity)
\(\operatorname{Aut}(\mathbb{Z}) = \{\mathrm{id}_{\mathbb{Z}}\}\). Indeed, let \(f \in \operatorname{Aut}(\mathbb{Z})\). Then
\[\begin{aligned} f(0) &= 0, \\ f(1) &= 1, \\ f(2) &= f(1) + f(1) = 2, \\ &\ \,\vdots \\ f(n) &= n \quad \text{if } n \in \mathbb{N}, \end{aligned}\]where each step uses additivity: \(f(n + 1) = f(n) + f(1)\). In general, \(f(-n) = -f(n) = -n\) if \(n > 0\), since homomorphisms preserve additive inverses. We conclude that \(f = \mathrm{id}_{\mathbb{Z}}\).
Example (Conjugation is an automorphism of matrices)
Let \(U \in GL_{n}(\mathbb{R})\). The map \(\varphi : M_{n \times n}(\mathbb{R}) \longrightarrow M_{n \times n}(\mathbb{R})\) given by \(A \longmapsto U \cdot A \cdot U^{-1}\) is an automorphism.
Extension and contraction of ideals
Lemma (The inverse image of an ideal is an ideal)
Let \(R, S\) be rings, \(\varphi : R \to S\) a homomorphism and \(J \subseteq S\) an \(S\)-ideal. Then \(\varphi^{-1}(J)\) is an \(R\)-ideal.
Proof: Exercise.
Note (The direct image of an ideal is not necessarily an ideal)
If \(I \subseteq R\) is an \(R\)-ideal, \(\varphi(I)\) need not be an ideal of \(S\). Why?
Definition (Extension and contraction of ideals)
Let \(\varphi : R \to S\) be a ring homomorphism.
- Let \(I\) be an \(R\)-ideal. Define the extension of \(I\), denoted \(I^{e}\), to be the smallest ideal of \(S\) containing \(\varphi(I)\).
- If \(J \subseteq S\) is an \(S\)-ideal, the contraction of \(J\), denoted \(J^{c}\), is \(\varphi^{-1}(J)\).
Proposition (Iterated contractions and extensions)
Let \(\varphi : R \to S\) be a ring homomorphism, \(I \subseteq R\) an \(R\)-ideal and \(J \subseteq S\) an \(S\)-ideal. Then:
- \(I \subseteq (I^{e})^{c}\) and \(J \supseteq (J^{c})^{e}\);
- \(I^{e} = I^{ece}\);
- \(J^{c} = J^{cec}\).
Proof: We shall use two monotonicity properties. If \(I_{1} \subseteq I_{2}\) are \(R\)-ideals, then \(I_{1}^{e} \subseteq I_{2}^{e}\): indeed, \(\varphi(I_{1}) \subseteq \varphi(I_{2}) \subseteq I_{2}^{e}\), and since \(I_{1}^{e}\) is the smallest ideal containing \(\varphi(I_{1})\), we have that \(I_{1}^{e} \subseteq I_{2}^{e}\). If \(J_{1} \subseteq J_{2}\) are \(S\)-ideals, then \(J_{1}^{c} = \varphi^{-1}(J_{1}) \subseteq \varphi^{-1}(J_{2}) = J_{2}^{c}\), directly.
For (a): if \(x \in I\), then \(\varphi(x) \in \varphi(I) \subseteq I^{e}\), so \(x \in \varphi^{-1}(I^{e}) = (I^{e})^{c}\); that is, \(I \subseteq I^{ec}\). On the other hand, \(\varphi(J^{c}) = \varphi(\varphi^{-1}(J)) \subseteq J\), and since \(J\) is an ideal containing \(\varphi(J^{c})\), the smallest ideal with that property satisfies \((J^{c})^{e} \subseteq J\).
For (b): applying the first part of (a) to the ideal \(I\) and then the monotonicity of the extension, \(I \subseteq I^{ec}\) implies \(I^{e} \subseteq I^{ece}\). And applying the second part of (a) with \(J = I^{e}\), we obtain \(I^{ece} = (I^{ec})^{e} = ((I^{e})^{c})^{e} \subseteq I^{e}\). Both inclusions give \(I^{e} = I^{ece}\).
For (c): applying the first part of (a) with \(I = J^{c}\), we obtain \(J^{c} \subseteq (J^{c})^{ec} = J^{cec}\). And applying the monotonicity of the contraction to \((J^{c})^{e} \subseteq J\), we obtain \(J^{cec} = ((J^{c})^{e})^{c} \subseteq J^{c}\). Both inclusions give \(J^{c} = J^{cec}\).
Example (An extension that grows: \(n\mathbb{Z}\) inside \(\mathbb{Q}\))
Let \(\varphi : \mathbb{Z} \to \mathbb{Q}\) be such that \(n \mapsto n\), and let \(n \neq 0, 1\). Then \(\varphi(n) = n \in \mathbb{Q}^{\times}\), and since the only ideal of \(\mathbb{Q}\) containing a unit is all of \(\mathbb{Q}\), we have that
\[\langle \varphi((n)) \rangle = \mathbb{Q} \implies (n\mathbb{Z})^{e} = \mathbb{Q} \implies ((n\mathbb{Z})^{e})^{c} = \mathbb{Q}^{c} = \varphi^{-1}(\mathbb{Q}) = \mathbb{Z}.\]That is, \(I = n\mathbb{Z} \subsetneq I^{ec} = \mathbb{Z}\): the contraction of the extension may be strictly larger than the original ideal.
The isomorphism theorems
Theorem (First isomorphism theorem)
Let \(R, S\) be rings and \(f : R \to S\) a ring homomorphism. Let \(F : R/\operatorname{Ker}(f) \to \operatorname{Im}(f)\) be such that \([a] \longmapsto f(a)\). Then \(F\) is an isomorphism and, in particular,
\[R/\operatorname{Ker}(f) \cong \operatorname{Im}(f).\]Proof: \(F\) is well defined. If \([a] = [b]\), then \(a - b \in \operatorname{Ker}(f)\), so that
\[f(a - b) = 0 \implies f(a) = f(b) \implies F([a]) = F([b]).\]\(F\) is a homomorphism. Using that \(f\) is one,
\[\begin{aligned} F([a] + [b]) &= F([a + b]) = f(a + b) = f(a) + f(b) = F([a]) + F([b]), \\ F([a] \cdot [b]) &= F([a \cdot b]) = f(a \cdot b) = f(a) \cdot f(b) = F([a]) \cdot F([b]), \\ F([1]) &= f(1) = 1. \end{aligned}\]\(F\) is surjective. Every element of \(\operatorname{Im}(f)\) is of the form \(f(a) = F([a])\) for some \(a \in R\).
\(F\) is injective. We compute the kernel:
\[F([a]) = 0_{S} \implies f(a) = 0_{S} \implies a \in \operatorname{Ker}(f) \implies [a] = [0],\]that is, \(\operatorname{Ker}(F) = \{ 0_{R/\operatorname{Ker}(f)} \} = \{ [0] \}\), and therefore \(F\) is injective. We conclude that \(F\) is an isomorphism.
Example (The quotient of a product by a factor)
Let \(R, S\) be rings and \(\pi : R \times S \to S\) such that \((x, s) \mapsto s\), which is a surjective homomorphism. Its kernel is
\[\operatorname{Ker}(\pi) = R \times \{0\},\]since \(\pi(x, s) = 0_{S}\) if and only if \(s = 0_{S}\). Hence, by the first isomorphism theorem,
\[(R \times S)/(R \times \{0\}) \cong S.\]Example (The quotient of a product by a product of ideals)
Let \(R, S\) be rings, \(I\) an \(R\)-ideal and \(J\) an \(S\)-ideal. Then
\[(R \times S)/(I \times J) \cong R/I \times S/J,\]using the homomorphism \(\varphi : R \times S \to R/I \times S/J\) such that \((r, s) \mapsto (r + I, s + J)\) (exercise).
Theorem (Second isomorphism theorem)
Let \(R\) be a ring, \(S \leq R\) a subring and \(I\) an \(R\)-ideal. Then:
- \(S \cap I\) is an \(S\)-ideal;
- \(S + I\) is a subring of \(R\);
- \(I\) is an \((S + I)\)-ideal;
- \(S/(S \cap I) \cong (S + I)/I\).
Proof: For (1): \(0 \in S \cap I\), and \(S \cap I\) is closed under sums because \(S\) and \(I\) are. For absorption, let \(s \in S\) and \(x \in S \cap I\); then \(s x \in S\), since \(S\) is closed under products, and \(s x \in I\), by the absorption of \(I\) in \(R\) (note that \(s \in S \subseteq R\)); that is, \(s x \in S \cap I\), and similarly \(x s \in S \cap I\).
For (2): we use the subring criterion. First, \(1 = 1 + 0 \in S + I\). If \(s + x, s' + x' \in S + I\), then
\[(s + x) - (s' + x') = (s - s') + (x - x') \in S + I,\]and for the product,
\[(s + x)(s' + x') = s s' + (s x' + x s' + x x'),\]where \(s s' \in S\) and the three terms in brackets lie in \(I\) by absorption; hence the product lies in \(S + I\).
For (3): \(I \subseteq S + I\) (since \(x = 0 + x\)), \(I\) is non-empty and closed under sums, and absorption by elements of \(S + I \subseteq R\) is a particular case of the absorption of \(I\) in \(R\).
For (4): consider \(\varphi : S \to (S + I)/I\) given by \(\varphi(s) = s + I\), the restriction to \(S\) of the canonical projection. It is a homomorphism, since the operations of \((S+I)/I\) are computed with representatives. It is surjective: every class of \((S + I)/I\) is of the form \((s + x) + I\) with \(s \in S\), \(x \in I\), and \((s + x) + I = s + I = \varphi(s)\), since \(x \in I\). Its kernel is
\[\operatorname{Ker}(\varphi) = \{ s \in S :\ s + I = I \} = \{ s \in S :\ s \in I \} = S \cap I.\]By the first isomorphism theorem, \(S/(S \cap I) \cong \operatorname{Im}(\varphi) = (S + I)/I\).
Theorem (Third isomorphism theorem)
Let \(R\) be a ring and let \(I \subseteq J \subseteq R\) be ideals. Then
\[(R/I)\big/(J/I) \cong R/J.\]Proof: Consider \(f : R/I \to R/J\) given by \(f([x]_{I}) = [x]_{J}\). It is well defined: if \([x]_{I} = [y]_{I}\), then \(x - y \in I \subseteq J\), so that \([x]_{J} = [y]_{J}\). It is a homomorphism, since the operations in both quotients are computed with representatives:
\[f([x]_{I} + [y]_{I}) = f([x + y]_{I}) = [x + y]_{J} = [x]_{J} + [y]_{J},\]and similarly for the product; moreover \(f([1]_{I}) = [1]_{J}\). It is surjective: every class \([x]_{J}\) is \(f([x]_{I})\). Its kernel is
\[\operatorname{Ker}(f) = \{ [x]_{I} :\ [x]_{J} = [0]_{J} \} = \{ [x]_{I} :\ x \in J \} = J/I.\]By the first isomorphism theorem,
\[(R/I)\big/(J/I) = (R/I)\big/\operatorname{Ker}(f) \cong \operatorname{Im}(f) = R/J.\]