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Chapter 3: Problem 6

Let \(R\) be a ring such that \(x^{2}=x\) for all \(x \in R\). Show that \(R\) is commutative.

Short Answer

Expert verified
In order to prove that the ring R is commutative, we show that for any elements x and y in R, \(xy = yx\). Given the condition \(x^2 = x\) for all x in R, we consider the expression \((x + y)^2\) and expand it using the binomial theorem. By substituting our given condition and simplifying the resulting equation, we obtain the relationship \(xy + yx = 0\), or equivalently, \(xy = -yx\). Since \(x^2 = x\), we have \(yx = -yx\), which implies that \(xy = yx\). Thus, R is commutative.

Step by step solution

01

Analyze the given condition

We are given that for any element x in the ring R, \(x^2 = x\). This means that for any element x, \(x * x = x\).
02

Start with x + y

Let x and y be two elements in R. Consider the expression \((x + y)^2\). Since \(x^2 = x\) and \(y^2 = y\), we just need to simplify this expression using these identities.
03

Expand (x + y)^2

Using the binomial theorem, we have: \[(x + y)^2 = (x + y)(x + y) = x(x + y) + y(x + y)\]
04

Distribute the product

Distribute the product with the associative and distributive laws, we get: \[(x + y)^2 = xx + xy + yx + yy\]
05

Replace x^2 and y^2

Now, we can replace each \(x^2\) and \(y^2\) according to our given conditions: \[(x + y)^2 = x + xy + yx + y\]
06

Use the given condition (x + y)^2 = x + y

Since \((x + y)^2 = x + y\) for any x and y in R, we have: \[x + y = x + xy + yx + y\]
07

Cancel terms

Notice that we can now cancel out x and y from both sides of the equation: \[xy + yx = 0\]
08

Conclude the proof

Thus, we have shown that for any elements x and y in R, \(xy + yx = 0\), which implies that \(xy = -yx\). However, since \(x^2 = x\), we know that \(yx = -yx\). Hence, we deduce that \(xy = yx\) for any x and y in R. This implies that the ring R is commutative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ring Theory
Ring theory is a fundamental area of study in abstract algebra that deals with the structure and properties of rings. A 'ring' is a mathematical set equipped with two binary operations that resemble addition and multiplication. A ring can be thought of as an abstraction of the integers. In a ring:
  • The addition of two elements produces another element of the ring.
  • There is an additive identity (usually described as 0), such that when added to any element, it returns that element.
  • Every element has an additive inverse, an element that when added to it yields the additive identity.
  • The multiplication of two elements produces another element of the ring.
  • There is a multiplicative identity (usually denoted as 1, not always present), such that when multiplied by any element, it returns that element.

However, not all rings are commutative. For a ring to be classified as commutative, it must satisfy the condition that the product of any two elements is independent of the order in which they are multiplied; that is, for any elements 'a' and 'b', the equation ab = ba must hold. This property is vital for many areas in mathematics and is the focus of the exercise we are looking at.
Associative Law
The associative law is a principle that governs how elements in a set are grouped during addition or multiplication. This law holds in many algebraic structures, including rings, and it states that the way in which elements are grouped does not affect the final result. In terms of addition, for any elements 'a', 'b', and 'c' in a ring, it must always be true that (a + b) + c = a + (b + c). Similarly, for multiplication, (ab)c = a(bc).

This law is critical when we expand expressions and perform operations in ring theory. It assures us that regardless of how we group terms during calculations, the outcome remains consistent, which allows for flexibility and simplification in algebraic manipulations.
Distributive Law
The distributive law is an essential rule in algebra that connects the operations of addition and multiplication. In the context of ring theory, it states that for any elements 'a', 'b', and 'c' in a ring, the following must be true: a(b + c) = ab + ac and (a + b)c = ac + bc.

This property is fundamental when breaking down complex expressions into more manageable parts. In the textbook solution, the distributive law enables us to expand the expression (x + y)^2 correctly, distributing the multiplication over the addition, which leads to the conclusion that the ring is commutative when combined with other ring properties.
Binomial Theorem
The binomial theorem provides a formula for expanding the power of a sum in terms of binomial coefficients. It applies to expressions of the form (a + b)^n, where 'a' and 'b' are any numbers, and 'n' is a nonnegative integer. The theorem states that (a + b)^n can be expanded into the sum of terms in the form C(n, k) * a^(n-k) * b^k, where 'C(n, k)' represents the binomial coefficient, or 'n choose k', and 'k' ranges from 0 to 'n'.

In our exercise, we apply a specific case of the binomial theorem, where 'n' is 2, to expand (x + y)^2. This is a vital step in demonstrating that each term in the expansion follows the laws of ring theory and assists in proving the commutative property of the ring 'R'.

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Most popular questions from this chapter

Let \(R\) be a commutative ring and let \(J_{1}, J_{2}\) be ideals. They are called relatively prime if $$ J_{1}+J_{2}=R $$ Suppose \(J_{1}\) and \(J_{2}\) are relatively prime. Given elements \(a, b \in R\) show that there exists \(x \in R\) such that \(x=a \quad\left(\bmod J_{1}\right) \quad\) and \(\quad x=b \quad\left(\bmod J_{2}\right)\) [This result applies in particular when \(R=\mathbf{Z}, J_{1}=\left(m_{1}\right)\) and \(J_{2}=\left(m_{2}\right)\) with relatively prime integers \(\left.m_{1}, m_{2} .\right]\)

(a) Let \(J_{1}, J_{2}\) be relatively prime ideals in a commutative ring \(R\). Show that the map \(a \bmod J_{1} \cap J_{2} \mapsto\left(a \bmod J_{1}, a \bmod J_{2}\right)\) induces an isomorphism $$ \left.f: R / J_{1} \cap J_{2}\right) \rightarrow R / J_{1} \times R / J_{2} $$ (b) Again, if \(J_{1}, J_{2}\) are relatively prime, show that \(J_{1} \cap J_{2}=J_{1} J_{2}\). Example. If \(m, n\) are relatively prime integers, then \((m) \cap(n)=(m n) .\) (c) If \(J_{1}, J_{2}\) are not relatively prime, give an example to show that one does not necessarily have \(J_{1} \cap J_{2}=J_{1} J_{z}\). (d) In (a), show that \(f\) induces an isomorphism of the unit groups $$ \left(R / J_{1} J_{2}\right)^{*} \stackrel{\circ}{\rightarrow}\left(R / J_{1}\right)^{*} \times\left(R / J_{2}\right)^{*} . $$ (e) Let \(J_{1} \ldots \ldots, I\), be ideals of \(R\) such that \(J_{i}\) is relatively prime to \(J_{1}\) for \(i \neq k\). Show that there is a natural ring isomorphism $$ R / J_{1} \ldots J_{r} \rightarrow \prod R / J_{i} $$

Let \(R\) be the set of numbers of type \(a+b \sqrt{2}\) where \(a, b\) are integers. Show that \(R\) is a ring. but not a ficld.

Let \(R\) be a commutative ring. Let \(L, M\) be ideals. (a) Show that \(L M \subset L \cap M\). (b) Given an example when \(L\). \(M \neq L \cap M\). Note that as a result of (a), if \(J\) is an ideal of \(R\), then we get a sequence of ideals contained in each other by taking powers of \(J\), namely $$ J \Rightarrow J^{2} \supset J^{3} \supset \cdots \supset J^{\prime \prime}=\cdots . $$

(a) L.et \(R\) be a ring. and let \(x, y \in R\) be such that \(x y=y x\). What is \((x+y)^{*}\) ? (Cf. Fxercise 2 of Chapter I, \(\$ 2\).) (b) Recall that an element \(x\) is called nilpotent if there exists a positive integer \(n\) such that \(x^{*}=0 .\) If \(R\) is commutative and \(x, y\) are nilpotent, show that \(x+y\) is nilpotent.
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