Today we shall look and proof the isomorphism theorems for rings today.

Let’s begin.

First isomorphism theorem

Let \(R\) and \(S\) be rings and let \(\phi: R \to S\) be a ring homomorphism. Then,

  1. \(ker(\phi)\) is an ideal of \(R\)
  2. The image of \(R\) is a subring of \(S\)
  3. The image of \(\phi\) is isomorphic to the quotient ring \(R/ker(\phi)\), which we write \(R/ker(\phi) \cong \phi(R)\)

In particular, if \(\phi\) is surjective, then \(S\) is isomorphic to \(R/ker(\phi)\)

Proof. We will only consider the case where a subset of \(R\) is a left ideal and leave the case of right ideal and two-sided idea to the reader.

(1) \(ker(\phi)\) is a subring: for \(k_1, k_2 \in ker(\phi)\)

\[\phi(k_1 - k_2) = \phi(k_1) + \phi(-k_2) = 0_S\] \[\phi(k_1 k_2) = \phi(k_1) \cdot \phi(-k_2) = 0_S\]

Next, \(rk \in ker(\phi)\) for \(k \in ker(\phi)\), \(r \in R\), since

\[\phi(rk) = \phi(r)\cdot\phi(k) = \phi(r) \cdot 0_{S} = 0_{S}\]

(2) Let \(r_1, r_2 \in R\), so \(\phi(r_1), \phi(r_2) \in \phi(R)\),

\[\phi(r_1) - \phi(r_2) = \phi(r_1 - r_2) \in \phi(R)\] \[\phi(r_1)\phi(r_2) = \phi(r_1 r_2) \in \phi(R)\]

(3) Let \(\psi : R/ker(\phi) \to \phi(R)\) be defined by \(\psi(r + ker(\phi)) = \phi(r)\), and it is clear that \(\psi\) is surjective. It suffices to show that \(\psi\) is well-defined and injective. Let \(r_1, r_2 \in R\) be given

\[r_1 + ker(\phi) = r_2 + ker(\phi)\Leftrightarrow r_1 - r_2 \in ker(\phi) \Leftrightarrow \phi(r_1 - r_2) = 0_S\] \[\phi(r_1 - r_2) = 0_S \Leftrightarrow \phi(r_1) = \phi(r_2) \Leftrightarrow \psi(r_1 + ker(\phi)) = \psi(r_2+ker(\phi))\]

The forward direction shows well-defined: any two equivalent different expressions of the equivalence class in \(R/ker(\phi)\) will evaluate to the same value on \(\psi\). The converse shows \(\psi\) is injective.

Second isomorphism theorem

Let \(R\) be a ring, \(S\) be a subring of \(R\) and \(I\) be an ideal of \(R\).

  1. The sum \(S+I:=\{s+i \mid s \in S, i \in I\}\) is a subring of \(R\)
  2. The intersection \(S \cap I\) is an ideal of \(S\)
  3. The quotient rings \((S+I)/I\) and \(S/(S \cap I)\) are isomorphic

Proof.

(1) Let \(s_1+ i_1, s_2+i_2 \in S + I\) where \(s_k \in S, i_K \in I\) for \(i = 1,2\). It suffices to show that \(S+I\) is closed under \(-\) and \(\cdot\).

\[(s_1+ i_1) - (s_2+i_2) = (s_1 - s_2) + (i_1 - i_2) \in S + I\] \[(s_1+ i_1) \cdot (s_2+i_2) = s_1s_2 + (s_1i_2 + i_1s_2 + i_1i_2) \in S + I\]

since \(s_1s_2 \in S\) and \(s_1i_2 + i_1s_2 + i_1i_2 \in I\)

(2) Let \(x_1, x_2 \in S \cap I\). First, \(S \cap I\) is an subring: By considering the membership of \(x_1, x_2\) in \(S\) and \(I\) respective we see that \(x_1 - x_2\) and \(x_1 \cdot x_2\) are contained in \(S \cap I\), thus \(S \cap I\) is a subring. Next we consider the left ideal case and leave the right to the reader. For \(x \in S \cap I\) and \(s \in S\), we have \(s \cdot x \in S\) since \(x \in S\). \(s \cdot x \in I\) since \(I\) is an ideal of \(R\) and \(s \in S \subseteq R\). Thus \(S \cap I\) is an ideal of \(S\).

(3) With the earlier parts proven, it is then meaningful to consider \((S+I)/I\) and \(S/(S \cap I)\).

Let \(\phi : R \to R/I\) and the inclusion map \(\iota : S \to R\), then define \(\psi = \phi \circ \iota : S \to (S+I)/I\) by \(s \mapsto s + I\). The \(ker(\psi) = \{s \in S \mid s \in I\} = S \cap I\) and \(\psi(S) = (S+I)/I\). Note that \(\psi (S) \neq S/I\) since it is not necessary that \(I \subseteq S\). For \(s_1, s_2 \in S\),

\[\psi(s_1 + s_2) = (s_1 + s_2) + I = (s_1 + I) + (s_2 + I) = \psi(s_1) + \psi(s_2)\] \[\psi(s_1s_2) = (s_1s_2) + I = (s_1 + I) \cdot (s_2 + I) = \psi(s_1) \cdot \psi(s_2)\]

thus \(\psi\) is a ring homomorphism, and the first isomorphism theorem says that \(S/(S \cap I)\cong (S+I)/I\).

Third isomorphism theorem

Let \(R\) be a ring, and \(I\) an ideal of \(R\). Then if \(J\) is an ideal of \(R\) such that \(I \subseteq J \subseteq R\), then the quotient ring \((R/I)/(J/I)\) is isomorphic to \(R/J\)

Proof. Let \(\phi : R/I \to R/J\) defined by \(r + I \to r + J\) for \(r \in R\). Then \(ker(\phi) = \{r + I \mid r \in J\}\) which is \(J/I\). The image \(\phi(R/I) = R/J\). We now verify that \(\phi\) is well-defined and a ring homomorphism. For \(r_1, r_2 \in R\),

\[r_1 + I = r_2 +I \Leftrightarrow r_1 - r_2 \in I \subseteq J \Leftrightarrow r_1 + J = r_2 + J \Leftrightarrow \phi(r_1 + I) = \phi(r_2 + I)\]

therefore \(\phi\) is well-defined. Next,

\[\phi(r_1 + r_2) = (r_1 + r_2) + J = (r_1 + J) + (r_2 + J) = \phi(r_1) + \phi(r_2)\] \[\phi(r_1r_2) = (r_1r_2) + J = (r_1 + J) \cdot (r_2 + J) = \phi(r_1) \cdot \phi(r_2)\]

thus by the first isomorphism theorem we get \((R/I)/(J/I)\cong R/J\).

Fourth isomorphism theorem (Lattice isomorphism theorem for rings)

Let \(I\) be an ideal of \(R\). The correspondence \(S \leftrightarrow S/I\) is an inclusion preserving bijection between the set of subrings \(S\) of \(R\) that contain \(I\) and the set of subrings of \(R/I\). Furthermore, \(S\) (a subring containing \(I\)) is an ideal of \(R\) if and only if \(S/I\) is an ideal of \(R/I\).

Proof. Take the natural quotient map, \(p : R \to R/I\) defined by \(r \mapsto r + I\). It is easy to see that \(p\) is a bijective ring homomorphism. Let \(s_1, s_2 \in S \supseteq I\) be a given subring, then observe

\[s_1 - s_2 \in S \Leftrightarrow p(s_1) - p(s_2) \in S/I\] \[s_1s_2 \in S \Leftrightarrow p(s_1)p(s_2) \in S/I\]

which shows that \(S/I\) is a subring of \(R/I\). If \(S \supseteq I\) is an ideal, for \(r \in R\)

\[rs \in S \Leftrightarrow p(rs) \in S/I\]

Lastly to show the bijection between the set of subrings(ideals) of \(R\) containing \(I\) and the set of subrings(ideals) of \(R/I\), let \(S_1, S_2 \supseteq I\) be subrings(ideals) such that \(p(S_1) = p(S_2)\). Then there exists \(s_1 \in S_1, s_2 \in S_2\) such that for \(i=1,2\)

\[\phi(s_i) = \phi(s_{3-i}) \Leftrightarrow s_i - s_{3-i} \in I \subseteq S_{3-i} \Leftrightarrow s_i \in S_{3-i}\]

Therefore \(S_1 = S_2\). Now for surjectivity of \(p\), let \(S/I\) be a subring(ideal), then \(p^{-1}(S/I) = S\supseteq I\), since \(0_{S/I} \in S/I\) and \(ker(p) = I\). Thus the inclusion preserving bijection is shown.