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Intro to Abstract Algebra Fall 2025

Blake Farman

Chapter 16 Homomorphisms and Factor Rings

Section 16.1 Ideals

Definition 16.1. Ideal.

Assume \(R\) is a ring. We say an additive subgroup \(I \leq R\) is
  1. a left ideal of \(R\) if \(I\) is closed under multiplication from the left; that is, for all \(r \in R\text{,}\)
    \begin{equation*} rI = \left\{rx \;\middle\vert\; x \in I\right\} \subseteq I\text{.} \end{equation*}
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  2. a right ideal of \(R\) if \(I\) is closed under multiplication from the right; that is, for all \(r \in R\text{,}\)
    \begin{equation*} Ir = \left\{xr \;\middle\vert\; x \in I\right\} \subseteq I \end{equation*}
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  3. an ideal of \(R\) if \(I\) is both a left and right ideal.
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Remark 16.2.

If \(R\) is commutative, then
\begin{equation*} rI = \left\{rx \;\middle\vert\; x \in I\right\} = \left\{xr \;\middle\vert\; x \in I\right\} \end{equation*}
implies that every left ideal is also a right ideal and vise-versa. Hence when \(R\) is commutative, these three notions coincide and we typically only consider multiplication on the left.
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Example 16.3.

The ideals of the ring \(\Z\) are precisely the subgroups of the additive group \(\Z\text{.}\) One direction is clear from the definition: if \(I\) is an ideal of \(\Z\text{,}\) then \(I \leq \Z\text{.}\)
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For the converse, assume \(H \leq \Z\text{.}\) By Corollary 7.28, there exists \(n \in \Z\) such that \(H = n\Z\text{.}\) For all \(m \in \Z\text{,}\)
\begin{align*} m\left(n\Z\right) \amp= \left\{m(nx) \;\middle\vert\; x \in \Z\right\}\\ \amp= \left\{n(mx) \;\middle\vert\; x \in \Z\right\}\\ \amp\subseteq n\Z \end{align*}
Hence \(m\left(n\Z\right) \subseteq n\Z\text{.}\) Therefore \(n\Z\) is an ideal of \(\Z\text{.}\)
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Example 16.4.

Assume \(n \in \N\) and \(1 \lt n\text{.}\) The ideals of the ring \(\Z/n\Z\) are precisely the subgroups of the additive group \(\Z/n\Z\text{.}\) One direction is clear from the definition: if \(I\) is an ideal of \(\Z/n\Z\text{,}\) then \(I \leq \Z/n\Z\text{.}\)
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For the converse, assume \(H \leq \Z/n\Z\text{.}\) By Corollary 7.30, there exists \(d \in \Z\) such that \(d \mid n\) and \(H = \langle \overline{n/d} \rangle\text{.}\) Let \(\overline{m} \in \Z/n\Z\) and \(\overline{x} \in H\) be given. Since \(H\) is finite and cyclic, there exists \(k \in \N\) such that
\begin{equation*} \overline{x} = k\overline{\frac{n}{d}} = \underbrace{\overline{\frac{n}{d}} + \cdots + \overline{\frac{n}{d}}}_k\text{.} \end{equation*}
Since multiplication distributes over addition,
\begin{align*} \overline{m}\overline{x} \amp= \underbrace{\overline{m}\overline{\frac{n}{d}} + \cdots + \overline{m}\overline{\frac{n}{d}}}_k\\ \amp= \underbrace{\overline{m(\frac{n}{d})} + \cdots + \overline{m(\frac{n}{d})}}_k\\ \amp= \underbrace{\underbrace{\overline{\frac{n}{d}} + \cdots + \overline{\frac{n}{d}}}_m + \cdots + {\underbrace{\overline{\frac{n}{d}} + \cdots + \overline{\frac{n}{d}}}_m}}_k\\ \amp= (mk) \overline{\frac{n}{d}} \in H \end{align*}
Hence \(mH \subseteq H\text{.}\) Therefore \(H\) is an ideal of \(\Z\text{.}\)
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Section 16.2 Characterizing Ideals

Proof.

\(\mathbf{(1) \implies (2):}\) Assume \(I\) is an ideal. First, we prove that the multiplication \((a + I)(b + I)\) is well-defined. Let \(a,b,c,d \in R\) be given. Assume that \(a + I = c + I\) and \(b + I = d + I\text{.}\) We must prove that \(ab + I = cd + I\text{.}\) By definition, there exist \(x,y \in I\) such that
\begin{equation*} a - c = x \iff a = x + c\quad\text{and}\quad b - d = y \iff b = d + y\text{.} \end{equation*}
Hence
\begin{align*} ab \amp= (x + c)(d + y)\\ \amp= xd + cy + cd \end{align*}
implies \(ab - cd = xd + cy \in I\) because \(I\) is an ideal. Therefore \(ab + I = cd + I\) and multiplication is well-defined.
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We leave the proof that this multiplication is associative and distributive as an exercise for the reader.
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\(\mathbf{(2) \implies (3):}\) Assume \(R/I\) is a group. Define the function
\begin{align*} \pi \colon R \amp\to R/I\\ x \amp\mapsto x + I \end{align*}
The function \(\pi\) is a morphism of groups by Theorem 11.2, and given \(a,b \in R\text{,}\)
\begin{equation*} \pi\left(ab\right) = (ab) + I = (a + I)(b + I)\text{.} \end{equation*}
Hence \(\pi\) is a morphism of rings. Moreover, by Theorem 11.2, \(\ker\pi = I\text{.}\) Therefore if \(R/I\) is a ring, then \(I\) is the kernel of the morphism of rings \(\pi \colon R \to R/I\text{.}\)
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\(\mathbf{(3) \implies (1):}\) Assume \(f \colon R \to S\) is a morphism of rings and \(I = \ker{f}\text{.}\) Let \(r \in R\) be given. For all \(x \in I\text{,}\)
\begin{equation*} f(rx) = f(r)f(x) = f(r)0 = 0 = 0f(r) = f(x)f(r) = f(xr) \end{equation*}
implies \(rx \in \ker{f} = I\) and \(xr \in \ker{f} = I\text{.}\) Hence \(rI \subseteq I\) and \(Ir \subseteq I\text{.}\) Therefore if \(I\) is the kernel of a morphism of rings, then \(I\) is an ideal.
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Definition 16.6.

Assume \(R\) is a ring and \(I\) is an ideal of \(R\text{.}\) The ring \(R/I\) is called the quotient of \(R\) by \(I\). One usually reads \(R/I\) as “\(R\) modulo \(I\)” or, simply, “\(R\) mod \(I\)”.
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Remark 16.7.

If the ring \(R\) is unital, then so is \(R/I\text{,}\) with multiplicative identity \(1 + I\) because for all \(r \in R\text{,}\)
\begin{equation*} (r + I)(1 + I) = (r1) + I = r + I = (1r) + I = (1 + I)(r + I)\text{.} \end{equation*}
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Section 16.3 Mapping Properties of the Quotient

Proof.

Regard \(f \colon R \to S\) as a morphism of abelian rings and apply Theorem 11.13 to produce the morphism of abelian rings \(\overline{f} \colon R/I \to S\) defined by the property
\begin{equation*} \overline{f} \circ \pi(x) = \overline{f}(x + I) = f(x)\text{.} \end{equation*}
First, we check this is indeed a morphism of rings. Let \(x_1, x_2 \in R/I\) be given. Then
\begin{equation*} \overline{f}\left((x_1 + I)(x_2 + I)\right) = \overline{f}((x_1x_2) + I) = f(x_1x_2) = f(x_1)f(x_2) = \overline{f}(x_1 + I)\overline{f}(x_2 + I)\text{.} \end{equation*}
Hence \(\overline{f}\) is a morphism of rings.
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Assume \(g \colon R/I \to S\) is another morphism of rings that satisfies the condition
\begin{equation*} f = g \circ \pi\text{.} \end{equation*}
For all \(x + I \in R/I\text{,}\)
\begin{equation*} g(x + I) = g \circ \pi(x) = f(x) = \overline{f} \circ \pi(x) = \overline{f}(x + I)\text{.} \end{equation*}
Therefore the induced morphism \(\overline{f} \colon R/I \to S\) is the unique morphism with this property.
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Proof.

Assume \(f \colon R \to S\) is a morphism of rings and \(K = \ker{f}\text{.}\) For all \(k \in K\text{,}\) \(f(k) = 0\text{.}\) Hence by the universal mapping property, there exists a unique morphism \(\overline{f} \colon R/K \to S\) such that \(\overline{f} \circ \pi = f\text{.}\)
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Observe that \(\overline{f}\) is injective because
\begin{equation*} \overline{f}(x + K) = f(x) = 0 \iff x \in K \iff x + K = 0 + K \end{equation*}
and satisfies
\begin{equation*} \overline{f}(R/K) = \overline{f}\left(\pi(R)\right) = f(R)\text{.} \end{equation*}
Hence \(\overline{f}\) is a surjection onto \(f(R)\text{.}\) Therefore \(R/K \cong f(R)\text{.}\) The final statement follows from the fact that \(f\) is surjective if and only if \(f(R) = S\text{.}\)
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Exercises 16.4 Exercises for Undergrads & Grads

1.

Assume \(f \colon R \to S\) is a morphism of rings. Prove that \(f\) is injective if and only if \(\ker{f} = \{0\}\text{.}\)
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Solution.
Regard \(f \colon R \to S\) as a morphism of groups. Then the underlying function between sets is injective if and only if the kernel of \(f\) is the trivial group, \(\{0\}\text{.}\)
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2.

Assume \(R\) is a ring and \(I\) is an ideal of \(R\text{.}\) Prove that the multiplication of cosets
\begin{equation*} (a + I)(b + I) = (ab) + I \end{equation*}
is associative.
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Solution.
Let \(a,b,c \in R\) be given. Since multiplication on \(R\) is associative,
\begin{align*} (a + I)\left[(b + I)(c + I)\right] \amp= (a + I)(bc + I)\\ \amp= \left(a(bc)\right) + I\\ \amp= \left((ab)c\right) + I\\ \amp= \left((ab) + I\right)(c + I)\\ \amp= \left[(a + I)(b + I)\right](c + I) \end{align*}
Hence the multiplication on \(R/I\) is associative.
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3.

Assume \(R\) is a ring and \(I\) is an ideal of \(R\text{.}\) Prove that the multiplication of cosets
\begin{equation*} (a + I)(b + I) = (ab) + I \end{equation*}
is distributive.
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Solution.
Let \(a,b,c \in R\) be given. Since multiplication on \(R\) is distributive,
\begin{align*} (a + I)\left[(b + I) + (c + I)\right] \amp= (a + I)((b + c) + I)\\ \amp= a(b + c) + I\\ \amp= (ab + ac) + I\\ \amp= (ab + I) + (ac + I)\\ \amp= (a + I)(b + I) + (a + I)(c + I) \end{align*}
and
\begin{align*} \left[(a + I) + (b + I)\right](c + I) \amp= ((a + b) + I)(c + I)\\ \amp= (a + b)c + I\\ \amp= (ac + bc) + I\\ \amp= (ac + I) + (bc + I)\\ \amp= (a + I)(c + I) + (b + I)(c + I) \end{align*}
Therefore if \(I\) is an ideal, then \(R/I\) is a ring.
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4.

Assume \(f \colon R \to S\) is a morphism of rings. Prove that if \(R^\prime \subseteq R\) is a subring of \(R\text{,}\) then
\begin{equation*} f(R^\prime) = \left\{f(x) \;\middle\vert\; x \in R^\prime\right\} \subseteq S \end{equation*}
is a subring of \(S\text{.}\)
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Solution.
Since \(f\) is a morphism of groups and \(R^\prime \leq R\text{,}\) \(f(R^\prime) \leq S\text{.}\) Let \(s_1, s_2 \in f(R^\prime)\) be given. There exist \(r_1, r_2 \in R^\prime\) such that \(f(r_1) = s_1\) and \(f(r_2) = s_2\text{.}\) Hence
\begin{equation*} s_1s_2 = f(r_1)f(r_2) = f(r_1r_2) \in f(R^\prime) \end{equation*}
implies that \(f(R^\prime)\) is closed under the multiplication on \(S\text{.}\) As such, the multiplication on \(f(R^\prime)\) is associative and distributive. Therefore \(f(R^\prime)\) is a subring of \(S\text{.}\)
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5.

Prove that if \(f \colon R \to S\) is a morphism of rings, and \(I\) is an ideal of \(R\text{,}\) then \(f(I)\) is an ideal of \(f(R)\text{.}\)
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Solution.
Since \(f\) is a morphism of abelian groups, \(f(I) \leq f(R)\text{.}\) Let \(s \in f(R)\) and \(y \in f(I)\) be given. There exists \(r \in R\) such that \(s = f(r)\) and there exists \(x \in I\) such that \(f(x) = y\text{.}\) Then \(rx, xr \in I\) imply
\begin{equation*} sy = f(r)f(x) = f(rx) \in f(I) \quad\text{and}\quad ys = f(x)f(r) = f(xr) \in f(I)\text{.} \end{equation*}
Therefore \(f(I)\) is an ideal of \(f(R)\text{.}\)
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6.

Prove that if \(f \colon R \to S\) is a morphism of rings, and \(J\) is an ideal of \(S\text{,}\) then \(f^{-1}(J)\) is an ideal of \(R\text{.}\)
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Solution.
Since \(f\) is a morphism of groups, \(f^{-1}(J) \leq R\text{.}\) Let \(r \in R\) and \(x \in f^{-1}(J)\) be given. Since \(J\) is an ideal, \(f(r) \in S\text{,}\) and \(f(x) \in J\text{,}\)
\begin{equation*} f(rx) = f(r)f(x) \in J \quad\text{and}\quad f(xr) = f(x)f(r) \in J\text{.} \end{equation*}
Hence \(rx, xr \in f^{-1}(J)\text{.}\) Therefore \(f^{-1}(J)\) is an ideal of \(R\text{.}\)
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