Exercises on rings

Exercise 1. Prove that if R is a ring with unit, the unit is unique.

Exercise 2. Let (A,+,) an abelian group. Consider the set

∀f,g ∈ A^A let

Prove that (A^A, +, ∘) in not a ring. Explain which property of ring axiomes is not satisfied.

Considering the set

of the endomorphism of the abelian group A, with the same operations defined above. Prove that End(A, +, ∘) is a ring.

Exercise 3. Prove the ring End(ℤ) with the operation above defined is isomorphic to the ring ℤ.

Exercise 4. Prove the ring End(ℚ) with the operation above defined is isomorphic to the ring ℚ.

Exercise 5. Prove that a ring R such that ∀x ∈ R, x² = x, is such that 2x = 0, ∀x ∈ R and commutative.

Exercise 6. Verify if the following are rings

ℝ is the real set; ⋅ the ordinary multiplication. The other operations are defined as follows

Exercise 7. Prove that the set A of the matrix

with a,b ∈ ℤ₃ is a subring of the ring of all matrix 2x2 with elements in ℤ₃. Prove that A is a field

Exercise 8. Let C[-1, 1] the ring of continuous functions defined in the real interval [-1,1] with real value, with respect to the operations

Prove that C[-1, 1] is not an integral domain.

Exercise 9. Prove that the inverse of an isomorphism φ between R and R' is an isomorphism (between R and R').

Solutions

1. Suppose e and e' are two units. Then ee' = e'e = e (because e' is a unit). It is also true that ee' = e'e = e because e is also unit. Thus e = e'.

2. Consider the distributive properties of ring axiomes

   f∘(g+h) = f∘g + f∘h,   (f+g)∘h = f∘h + g∘h,   ∀f,g ∈ A^A

   f needs to be an additive map to verify the distributive property: f(g + h) = f(g(x)) + f(h(x)), i.e. f(x+y) = f(x) + f(y) (in this case the set A^A is called group homorphism; they'll be covered in the next chapters). This is the case in the case of End(A) by its definition.

   Notice that if f: ℤ → ℤ is a group homorphism then for all n ∈ ℤ, f(n)= nf(1) thus f is entirely determined by the value of f(1). Let φ: End(ℤ) ⟶ ℤ; defined as φ(f) = f(1), then because of what said above ϕ is injective, it is also surjective because for all a ∈ ℤ, ϕ(n→na) = a. It remains to verify that ϕ ring homorphism axioms are satisfied, i.e.

   φ(f+g) = φ(f) + φ(g) = f(1) + g(1);

   Let h = f + g defiend as h(n) = n(f(1) + g(1)) so φ(h) = f(1) + g(1); hence the first axiom is satisfied.

   φ(f∘g) = φ(f) φ(g) = f(1) g(1);

   Let k = f+g defiend as h(n) = f((g(n))) = n⋅g(1)⋅f(1), φ(k) = g(1)⋅f(1), thus φ is indeed a ring isomorphism.

3. Use the same method as in the previous exercise.

4. For each a ∈ R (a + a)² = a + a, writing the square as a² + a² + a² + a² + a² = a + a. Since a² = 1 we have 2a = 0 hence every element has an additive inverse. From (a + b)² = a+b it follows that a² + b² + ab + ba = a + b, and from a² = a and b² = b we have ab + ba = 0. Since any element is equal to its inverse ba + ba = 0 thus ab = ba for all a,b ∈R.

5. (ℝ, ⊕, ⋅) is not a subring because it has not the neuter element. 0 is the neuter element only for positive real numbers. Instead (ℝ, ⊕', ⋅) is a commutative ring.

6. Proving that A is subring boils down to notice that ℤ_n is closed under addition and multiplication. It is a commutative ring since

   $\left(\begin{array}{cc}a& -<!--\; -\; -->b\\ b& a\end{array}\right)\left(\begin{array}{cc}c& -<!--\; -\; -->d\\ d& c\end{array}\right)=\left(\begin{array}{cc}ac-<!--\; -\; -->bd& -<!--\; -\; -->(bc+ad)\\ bc+ad& ac-<!--\; -\; -->bd\end{array}\right)=\left(\begin{array}{cc}c& -<!--\; -\; -->d\\ d& c\end{array}\right)\left(\begin{array}{cc}a& -<!--\; -\; -->b\\ b& a\end{array}\right)$

   and it has unit (1 0; 0 1). Every non-zero element is invertible: we have det (a -b; b a) = a² + b², a, b ∈ ℤ₃. In ℤ₃ the only square elements are [0] and [1] so a² + b² = 0 iff a = b = 0 and only in this case the matrix is null. Thus every non null matrix has determinat different for zero and is invertible.

7. Solution. Consider the functions

   $f_1(x)=\{\begin{array}{ll}-<!--\; -\; -->x& \text{if }-<!--\; -\; -->x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]\\ 0& \text{if }x\in <!--\; \in \; -->[0,+\mathrm{\infty <!--\; \infty \; -->}]\end{array}$        $f_2(x)=\{\begin{array}{ll}0& \text{if }x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]\\ x& \text{if }x\in <!--\; \in \; -->[0,+\mathrm{\infty <!--\; \infty \; -->}]\end{array}$

8. Since x' = φ(x) and y' = φ(y) for one and only one x ∈ R and y ∈ R, then

   φ⁻¹(x' + y') = φ⁻¹(φ(x) + φ(y)) = φ⁻¹(φ(x + y)) = x + y = φ⁻¹(x') + φ⁻¹(y').

   Same argument for the product.

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