KindiakMath

Ring Isomorphisms

Let R, S be rings.

Recall the first isomorphism theorem for rings.

Theorem 1. Let \phi : R \to S be a ring homomorphism. Then \ker(\phi) is an ideal and R/{\ker(\phi)} \cong \phi(R) as rings.

Problem 1. Suppose K \subseteq R as a sub-ring and I \subseteq R as an ideal. Show that:

  • K + I \subseteq R as a sub-ring,
  • K \cap I \subseteq K as an ideal,
  • K \cap (K+I) = K.

Deduce that

(K+I)/I \cong (K \cap (K+I))/(K \cap I) = K/(K \cap I).

This result is known as the second isomorphism theorem for rings.

(Click for Solution)

Solution. We first verify the three properties.

  • Sum of sub-rings remain as sub-rings.
  • For the ideal claim, fix x \in K and z \in K \cap I. Since I \subseteq R as an ideal, xz \in I. Since K is a sub-ring, xz \in K. Therefore, xz \in K \cap I. Since I is also a right-ideal, we have zx \in K \cap I. Therefore, K \cap I \subseteq K as an ideal.
  • For the final claim, the direction (\subseteq) is trivial, while the direction (\supseteq) holds via x = x+ 0 \in K + I for any x \in K.

Hence, define the map \phi : K \to (K + I)/I by x \mapsto x + I, which is a surjective group homomorphism under addition with kernel K \cap I. Since I \subseteq K +I as an ideal, we also obtain a ring structure: for x,y \in K \subseteq K+I,

\phi(x \cdot y) = x \cdot y + I = (x+I) \cdot (y+I) = \phi(x) \cdot \phi(y).

By the first isomorphism theorem,

K/(K \cap I) = K/{\ker(\phi)} \cong \phi(K) = (K+I)/I.

Problem 2. Suppose I \subseteq J \subseteq R as ideals. Show that:

  • J/I \subseteq R/I as an ideal,
  • (R/I)/(J/I) \cong R/J.

This result is known as the third isomorphism theorem for rings.

(Click for Solution)

Solution. Check that the map \phi : R/I \to R/J defined by r + I \mapsto r + J is a surjective ring homomorphism with kernel J/I and conclude using the first isomorphism theorem.

Remark 1. Let I \subseteq R be an ideal and \phi : R \to R/I denote the canonical quotient ring homomorphism r \mapsto r + I, so that \ker(\phi) = I. Define the collections

\begin{aligned} \Sigma_1 &:= \{ K \subseteq R\ \text{sub-ring} : I \subseteq K\}, \\ \Sigma_2 &:= \{ J \subseteq R/I\ \text{sub-ring}\}. \end{aligned}

Then the map \psi : \Sigma_1 \to \Sigma_2 defined by S \mapsto S/I is a well-defined bijection. This result is known as the correspondence theorem for rings.

—Joel Kindiak, 22 Jan 26, 1833H

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