Exercises on the fundamental morphism theorems

Exercise 1. Let r and s positive integers such that (r,s)=1.

1. Consider the ring ℤ_r x ℤ_s (with respect to ordinary multiplication and addition, component by component), prove that the application f: ℤ ⟶ ℤ_r x ℤ_s given by a ⟶ (ā_r, ā_s) is an epimorphism of rings.

2. Prove (using the fundamental theorem of isomorphism for rings) that ℤ_rs and ℤ_r x ℤ_s are isomorphic.

Exercise 2. Let I₁ and I₂ two ideals of a ring R such that I₁ + I₂ = R. Prove that

Exercise 3. Using the first isomorphism theorem determine the isomorphism in case that R = ℤ, I = 4ℤ, A = 6ℤ.

Exercise 4. Using the second isomorphism theorem determine the isomorphism for the case R = ℤ, I = 10ℤ, J = 6ℤ.

Exercise 5. Determine all the ideals of ℤ₂₄ by using the correspondence theorem.

Solutions

1. 1. (a+b) maps to ([a+b]_r, [a+b]_s), which is ([a]_r + [b]_r) + ([a]_s + [b]_s).

      (ab) maps to ([ab]_r, [ab]_s), which is ([a]_r [b]_r) ([a]_s [b]_s). So it's a homomorphism.

      To prove surjectivity we must show that for (x,y) in ℤ_r x ℤ_s there's an a ∈ ℤ. Think about what the surjectivity requirement for the function means algebraically: it means for any x,y, the equations [a]_r = x, and [a]_s = y are simultaneously solvable.

      a ≡ x mod r  ⇐⇒  [a]_r = x,   a ≡ y mod s ⇐⇒  [a]_r = y.

   2. We proved that f: ℤ ⟶ ℤ_r x ℤ_s is a ring homomorphism, by the fundamental theorem o of ring homomorphism we know there exists a unique isomorphism f*: ℤ_rs ⟶ ℤ_r x ℤ_s if ker f = rsℤ. We have that ℤ/ℤ_rs = rsℤ. We have that the kernel of φ: ℤ → ℤ/nℤ sending a to [a] is {kn| k ∈ ℤ} Thus ℤ_rs ≃ ℤ_r x ℤ_s

      Alternative proof. The map f: Z_rs ⟶ ℤ_r x ℤ_s, defined by f([x]_rs) = ([x]_r, [x]_s), is a ring isomorphism. By the Chinese remainder theorem the system of congruences

      $\begin{array}{rl}x& \equiv <!--\; \equiv \; -->a(\mathrm{mod}r)\\ x& \equiv <!--\; \equiv \; -->b(\mathrm{mod}s),\end{array}$

      admits exactly one solution modulo rs. This implies the surjectivity and the injectivity of the map.

2. It's sufficient to find an epimorphism from R to R/I₁ ⊕ R/I₂ with kernel I₁ ∩ I₂. We define φ: R ⟶ R/I₁ ⊕ R/I₂ by letting φ(x) = (x +I₁, x + I₂).

   R/I1 ⊕ R/I2 is the cartesian product of those rings so made of the ordered pairs (x+I₁, x+I₂). x ∈ ker(φ) would satisfy, by definition, that x+I₁ = I₁ and x+I₂ = I₂, hence that x is an element of both I₁ and I₂, hence an element of the intersection of these ideals: ker φ = I₂ ∩ I₁.

   To prove surjectivity we must show that given a (r + I₁, s + I₂) ∈ R/I₁ ⊕ R/I₂ there exists a x ∈ R, such that φ(x) = (r +I₁, s + I₂). Since R = I₁ + I₂, r = a₁ + a₂, s = b₁ + b₂ for given a_i,b_i ∈ I_i (i = 1,2). The element x = a₂ + b₁ is such that

   x ≡ a₂ ≡ r (mod I₁),   x ≡ b₁ ≡ r (mod I₂)

   hence φ(x) = (r +I₁, s + I₂).

3. We want to show that

   A/(A ∩ I) ≃ (A + I)/I

   4ℤ ∩ 6ℤ = 12ℤ, since an integer is divisible by both 4 and 6 if and only if it is divisible by 12. An we have 4ℤ + 6ℤ = {4x + 6y: x,y ∈ℤ } = 2ℤ, hence

   6ℤ/12ℤ ≃ 2ℤ/4ℤ

   If π: R ⟶ R/I, let π_|A the restriction of π to A that is

   π_|A: A ⟶ (A+I)/I

   It is a ring homomorpshim whose kernel is A ∩ I. By using the fundamental homomrphsim we have the following isomorphism

   f: A/(A ∩ I) ⟶ (A+I)/I

4. ℤ/I = ℤ/10ℤ, ℤ/J = ℤ/2ℤ, J/I = 2ℤ/10ℤ ≃ [2]₁₀ℤ₁₀.

   Using the second isomorphism theorem, we have the isomorphism f: ℤ/J ⟶ (ℤ/I)/(J/I) is defined as follows f(r+J) := (r + I) + J/I

   f(0) ⟼ 0 + 10ℤ + 2ℤ/10ℤ = [0]₁₀ + [2]₁₀ℤ₁₀ = {[2]₁₀, [4]₁₀, [6]₁₀, [8]₁₀}

   f(1) ⟼ 1 + 10ℤ + 2ℤ/10ℤ = [1]₁₀ + [2]₁₀ℤ₁₀ = {[3]₁₀, [5]₁₀, [7]₁₀, [9]₁₀}

   f(2) ⟼ 2 + 10ℤ + 2ℤ/10ℤ = [2]₁₀ + [2]₁₀ℤ₁₀ = {[4]₁₀, [6]₁₀, [8]₁₀, [10]₁₀}

5. The ideals of ℤ containing 24ℤ are: ℤ, 2ℤ, 3ℤ,4ℤ,6ℤ,8ℤ,12ℤ, 24ℤ, thus the ideals of ℤ₂₄ are ℤ/ℤ₂₄ = ℤ₂₄, 2ℤ/24ℤ, etc.

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