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Let \(\rho: R \rightarrow R^{\prime}\) be a ring homomorphism, and let \(a \in R\). Show that \(\rho(a \boldsymbol{R})=\rho(a) \rho(\boldsymbol{R})\)

Short Answer

Expert verified
Question: Prove that the image of the principal ideal generated by a in R under the given ring homomorphism ρ is the principal ideal generated by ρ(a) in R', i.e. ρ(aR) = ρ(a)ρ(R). Answer: To prove ρ(aR) = ρ(a)ρ(R), we showed that ρ(aR) ⊆ ρ(a)ρ(R) and ρ(a)ρ(R) ⊆ ρ(aR). By verifying the containment in both directions, it's concluded that ρ(aR) = ρ(a)ρ(R).

Step by step solution

01

Show that \(\rho(aR) \subseteq \rho(a)\rho(R)\)

Let's take an arbitrary element \(x\) from \(\rho(aR)\). This means that \(x = \rho(ar)\) for some \(r \in R\). Now, the ring homomorphism property states that \(\rho(gh) = \rho(g)\rho(h)\) for any elements \(g, h\) from the ring \(R\). Using this property, we can rewrite \(x\) as \(x = \rho(a)\rho(r)\), where \(\rho(r) \in \rho(R)\). Since \(\rho(a)\rho(r)\) is an element of the ideal generated by \(\rho(a)\) in \(\rho(R)\), we have \(x \in \rho(a)\rho(R)\). Thus, we have proved that \(\rho(aR) \subseteq \rho(a)\rho(R)\).
02

Show that \(\rho(a)\rho(R) \subseteq \rho(aR)\)

Now, let's take an arbitrary element \(y\) from \(\rho(a)\rho(R)\). This means that \(y = \rho(a)s\) for some \(s \in \rho(R)\). Notice that for every \(r \in R\), we must have a corresponding element \(\rho(r) \in \rho(R)\). So, given \(s \in \rho(R)\), there must be an element \(r \in R\) such that \(s = \rho(r)\). Rewriting \(y\), we have \(y = \rho(a)\rho(r)\). Using the ring homomorphism property, we can write \(y = \rho(ar)\). By definition, \(ar \in aR\) and therefore \(y \in \rho(aR)\). Thus, we have proved that \(\rho(a)\rho(R) \subseteq \rho(aR)\).
03

Concluding the proof

Since we have shown that \(\rho(aR) \subseteq \rho(a)\rho(R)\) and \(\rho(a)\rho(R) \subseteq \rho(aR)\), we can conclude that \(\rho(aR) = \rho(a)\rho(R)\). So, the image of the principal ideal generated by \(a\) in \(R\) under the ring homomorphism \(\rho\) is equal to the principal ideal generated by \(\rho(a)\) in \(R'\).

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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 foundational concept in abstract algebra. A 'ring' is a set equipped with two operations, usually called addition and multiplication. These operations must satisfy specific properties:
  • Addition is commutative and associative. There is an additive identity element, usually denoted by 0.
  • Multiplication is associative. There is no requirement for a multiplicative identity, but many rings do have one.
  • Distributive property: Multiplication is distributive over addition.
Rings generalize fields and include structures like integers, polynomials, and matrices. They are essential in many areas of mathematics, providing the framework to work with algebraic structures that aren't groups.
One key aspect of rings is the existence of subrings which inherit the ring properties. Ring homomorphisms play a vital role in ring theory. These are functions between rings that respect both the addition and multiplication operations. They provide a way to map one ring onto another while preserving its structure. Understanding how elements inside rings transform helps in studying the nature of the rings themselves.
Ideal Theory
Ideal theory extends the idea of normal subgroups in group theory to rings. An ideal is a special subset of a ring.
  • It is closed under addition and subtraction.
  • Multiplying any element in the ring with an element of an ideal results in an element of the ideal.
This means if we take any ring element and multiply it with any ideal element, the product stays within the ideal.
Ideals are crucial in analyzing the structure of rings. They help in constructing 'quotient rings', similar in concept to quotient groups. Through ideals, you can simplify complex rings for easier analysis by modding out certain parts.
When we talk about the ideal generated by an element, we visualize the set of all possible products formed by ring elements and this generator element. Understanding transformations of these ideals under functions like ring homomorphisms allows for deeper insights into ring structure and behavior.
Principal Ideal
In the world of ring theory, a principal ideal is one of the simplest kinds of ideals. It originates from a single element in a ring.
  • This element 'a' generates the entire ideal.
  • Mathematically, this is written as \(aR = \{ar : r \in R\}\).
This means a principal ideal consists of all elements obtainable by multiplying 'a' by each member of the ring.
Principal ideals are easy to handle and provide a clear path to understanding more complicated ideal structures. Notably, in the study of ring homomorphisms, principal ideals behave predictably under mappings. As evidenced by the exercise, when you apply a ring homomorphism to an element generating a principal ideal, it maps to an ideal generated by the image of the element.
Understanding principal ideals deepens the comprehension of how rings can be manipulated and studied, offering insights into their underlying structure.

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Most popular questions from this chapter

Write down the multiplication table for \(\mathbb{Z}_{2}[X] /\left(X^{2}+X\right)\). Is this a field?

Let \(\rho: R \rightarrow R^{\prime}\) be a ring homomorphism. Let \(S\) be a subring of \(R,\) and let \(\tau: S \rightarrow R^{\prime}\) be the restriction of \(\rho\) to \(S\). Show that \(\tau\) is a ring homomorphism and that \(\operatorname{Ker} \tau=\operatorname{Ker} \rho \cap S\)

Let \(n\) be a positive integer, and consider the natural map that sends \(a \in \mathbb{Z}\) to \(\bar{a}:=[a]_{n} \in \mathbb{Z}_{n},\) which we may extend coefficient-wise to a ring homomorphism from \(\mathbb{Z}[X]\) to \(\mathbb{Z}_{n}[X],\) as in Example \(7.47 .\) Show that for every \(f \in \mathbb{Z}[X],\) we have a ring isomorphism \(\mathbb{Z}[X] /(f, n) \cong \mathbb{Z}_{n}[X] /(\bar{f})\).

Consider the ring \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) of functions \(f: \mathbb{R} \rightarrow \mathbb{R},\) with addition and multiplication defined point-wise. (a) Show that \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) is not an integral domain, and that \(\operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\) consists of those functions that never vanish. (b) Let \(a, b \in \operatorname{Map}(\mathbb{R}, \mathbb{R}) .\) Show that if \(a \mid b\) and \(b \mid a,\) then \(a r=b\) for some \(r \in \operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\) (c) Let \(\mathcal{C}\) be the subset of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) of continuous functions. Show that \(\mathcal{C}\) is a subring of \(\operatorname{Map}(\mathbb{R}, \mathbb{R}),\) and that all functions in \(\mathcal{C}^{*}\) are either everywhere positive or everywhere negative. (d) Find elements \(a, b \in \mathcal{C},\) such that in the ring \(\mathcal{C},\) we have \(a \mid b\) and \(b \mid a,\) yet there is no \(r \in \mathcal{C}^{*}\) such that \(a r=b\).

Let \(R\) be a ring of characteristic \(m>0,\) and let \(n\) be an integer. Show that: (a) if \(\operatorname{gcd}(n, m)=1,\) then \(n \cdot 1_{R}\) is a unit; (b) if \(1<\operatorname{gcd}(n, m)

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