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Chapter 2: Problem 5

Universelle Eigenschaft der Bruchringe: Sei \(R\) ein Ring und \(S \subset R\) ein multiplikatives System. Man zeige: Zu jodem Ringhomomorphismus \(\varphi: R \longrightarrow R^{\prime}\) mit \(\varphi(S) \subset R^{\prime *}\) gibt es genau einen Ringhomomorphismus \(\bar{p}: S^{-1} R \longrightarrow R^{\prime}\) mit \(\varphi=\bar{q} \circ r ;\) dabei bezeichne \(r: R \longrightarrow S^{-1} R\) den kanonischen Homomotphismus, gereben durch \(a+\frac{d}{1}\).

Short Answer

Expert verified
We defined the map 饾憺虆 : S鈦宦 R 鈫 R' by 饾憺虆(a/s) = 蠁(a)蠁(s)鈦宦 for all a/s 鈭 S鈦宦 R and showed that it is well-defined. Next, we proved that 饾憺虆 is a ring homomorphism by verifying the necessary conditions. Finally, we showed that 饾憺虆 is unique by comparing it with another homomorphism 饾憺虆鈧 that satisfies the same property (i.e., 蠁 = 饾憺虆鈧 鈭 r). Therefore, there exists a unique ring homomorphism 饾憺虆 : S鈦宦 R 鈫 R' such that 蠁 = 饾憺虆 鈭 r, proving the universal property of localization for a ring R and a multiplicative system S in R.

Step by step solution

01

Define 饾憺虆 and show that it is well-defined

: Consider the map 饾憺虆 : S鈦宦 R 鈫 R' defined by 饾憺虆(a/s) = 蠁(a)蠁(s)鈦宦 for all a/s 鈭 S鈦宦 R. Now, we need to show that 饾憺虆 is well-defined. Consider two equivalent fractions a/s = b/t, where a, b 鈭 R and s, t 鈭 S. By definition, there exists u 鈭 S such that u(at - bs) = 0. Since 蠁 is a ring homomorphism and 蠁(S) 鈯 R'*, we have: \( \phi(u) (\phi(a) \phi(t) - \phi(b) \phi(s) ) = \phi(uat) - \phi(ubs) = \phi(u(at-bs)) = \phi(0) = 0 \) This implies that 蠁(a)蠁(t)鈦宦 = 蠁(b)蠁(s)鈦宦, so 饾憺虆(a/s) = 饾憺虆(b/t). Hence, 饾憺虆 is well-defined.
02

Prove that 饾憺虆 is a ring homomorphism

: Now, we need to show that 饾憺虆 is a ring homomorphism, i.e., it satisfies the following conditions: 1. 饾憺虆(1) = 1 2. 饾憺虆(a/s + b/t) = 饾憺虆(a/s) + 饾憺虆(b/t) for all a/s, b/t 鈭 S鈦宦 R 3. 饾憺虆((a/s)(b/t)) = 饾憺虆(a/s)饾憺虆(b/t) for all a/s, b/t 鈭 S鈦宦 R First, we have 饾憺虆(1) = 饾憺虆(1/1) = 蠁(1)蠁(1)鈦宦 = 1. This proves condition 1. For condition 2, let a/s, b/t 鈭 S鈦宦 R. We have: \( \begin{aligned} 饾憺虆(a/s + b/t) &= 饾憺虆\left(\frac{at + bs}{st}\right) \\ &= \phi(at+bs) \phi(st)^{-1} \\ &= \frac{\phi(a)\phi(t) + \phi(b)\phi(s)}{\phi(s)\phi(t)} \\ &= 饾憺虆(a/s) + 饾憺虆(b/t) \end{aligned} \) This proves condition 2. Finally, for condition 3, let a/s, b/t 鈭 S鈦宦 R. We have: \( \begin{aligned} 饾憺虆\left(\frac{a}{s}\cdot \frac{b}{t}\right) &= 饾憺虆\left(\frac{ab}{st}\right) \\ &= \phi(ab) \phi(st)^{-1} \\ &= \left(\phi(a)\phi(s)^{-1}\right)\left(\phi(b)\phi(t)^{-1}\right) \\ &= 饾憺虆\left(\frac{a}{s}\right)饾憺虆\left(\frac{b}{t}\right) \end{aligned} \) This proves condition 3. Since 饾憺虆 satisfies all the conditions for a ring homomorphism, 饾憺虆 is a ring homomorphism.
03

Show that 饾憺虆 is unique

: To prove that 饾憺虆 is unique, let 饾憺虆鈧 : S鈦宦 R 鈫 R' be another ring homomorphism such that 蠁 = 饾憺虆鈧 鈭 r. We will show that 饾憺虆 = 饾憺虆鈧. For all a 鈭 R, let a/1 be its canonical image in S鈦宦 R. Then, we have: \( \begin{aligned} 饾憺虆(a/1) &= 饾憺虆(r(a)) \\ &= \phi(a) \\ &= 饾憺虆鈧(r(a)) \\ &= 饾憺虆鈧(a/1) \end{aligned} \) Thus, 饾憺虆 = 饾憺虆鈧 on the canonical image of R in S鈦宦 R. By the universal property of localization, 饾憺虆 and 饾憺虆鈧 coincide on the entire S鈦宦 R, so 饾憺虆 = 饾憺虆鈧. So, we have shown that there is a unique ring homomorphism 饾憺虆 : S鈦宦 R 鈫 R' such that 蠁 = 饾憺虆 鈭 r. This completes the proof of the universal property of localization.

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Key Concepts

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

Universal Property
Localization in ring theory refers to the process of adapting a ring structure by focusing on certain elements, specifically through a multiplicative system. The universal property in this context means that given any ring homomorphism \( \varphi \colon R \longrightarrow R^\prime \) with \( \varphi(S) \subset R^{\prime *} \), there exists a unique homomorphism \( \bar{\varphi} \colon S^{-1}R \rightarrow R^\prime \) such that \( \varphi = \bar{\varphi} \circ r \), where \( r \) is the canonical homomorphism.

  • The universal property guarantees that \( \bar{\varphi} \) extends \( \varphi \) uniquely through the localized ring \( S^{-1}R \).
  • This property allows \( S^{-1}R \) to serve as a bridge, ensuring that mappings through \( R \) remain consistent even after localization.
Understanding this property helps to ensure that transformations respecting the multiplicative set produce no conflicts, fortifying the structure created by localization.
Ring Homomorphism
A ring homomorphism is a algebraic structure-preserving map between two rings. It fulfills key properties which align with ring operations such as addition and multiplication. Ensuring these properties remain intact after a transformation is key to understanding the behavior of homomorphisms.

When mapping fractions through a homomorphism, such as \( \bar{\varphi} \colon S^{-1}R \rightarrow R^\prime \), it must preserve:
  • Identity: \( \bar{\varphi}(1) = 1 \)
  • Addition: \( \bar{\varphi}(a/s + b/t) = \bar{\varphi}(a/s) + \bar{\varphi}(b/t) \)
  • Multiplication: \( \bar{\varphi}((a/s) \cdot (b/t)) = \bar{\varphi}(a/s) \cdot \bar{\varphi}(b/t) \)
These properties ensure that the algebraic operations between ring elements are everywhere consistent, vital for engaging with localized constructions.
Multiplicative System
In localization, a multiplicative system \( S \) is a key player, constituting a set of elements in a ring \( R \) closed under multiplication and containing the identity. The multiplicative system facilitates the creation of new ring elements in \( S^{-1}R \) by inverting the elements of \( S \).

It enables a kind of algebraic flexibility, allowing some elements to become invertible which otherwise wouldn鈥檛 be. Important properties include:
  • Every element in \( S \) can multiply any ring element and still reside in the ring.
  • This supports constructing fractions \( a/s \) for \( a \in R, s \in S \), leading to a wider algebraic manipulation.
Employing a multiplicative system is central, giving rise to a more robust algebraic framework through localization.
Equivalence of Fractions
Equivalence in the fraction form used in localization ensures consistency in defining elements. Two elements \( a/s \) and \( b/t \) in \( S^{-1}R \) are equivalent if there exists some \( u \in S \) such that \( u(at - bs) = 0 \). This guarantees that under any homomorphism, such as \( \bar{\varphi} \), both representations yield the same result.

  • This equivalence skews into preserving the structure across inevitable manipulations.
  • The requirement ensures algebraic operations like addition and multiplication remain meaningful.
Understanding equivalence of fractions in localization is key as it maintains integrity, giving assurance that operations on these abstract fractions exhibit the expected behaviors.

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Most popular questions from this chapter

Sel \(K\) ein K枚rper und sei \(f=X^{3}+a X+b \in K[X]\) ein Polynom. welches in \(K[X]\) vollstandig in Linearfaktoren zerfallt. Man zeige: Die Nullstellen von \(f\) sind genau dann paarweise verschieden, wenn die "Diskriminante" \(\Delta=-4 a^{3}-27 b^{2}\) nicht verschwindet.

Man fithre die in Satz 4 beschiriebene Division nit Rest im Polynomiring \(Z[X]\) in folgenden F盲llen explizit durch: (i) \(f=3 X^{5}+2 X^{4}-X^{3}+3 X^{2}-4 X+7, \quad g=X^{2}-2 X+1\) (ii) \(f=X^{5}+X^{4}-5 X^{3}+2 X^{2}+2 X-1, \quad g=X^{2}-1 .\)

Sei \(K\) ein K么rper und \(\varphi: K\left[X_{1} \ldots . X_{m}\right] \longrightarrow K\left[X_{1}, \ldots, X_{n}\right]\) ein Ringisomorphismus mit \(\left.\varphi\right|_{K}=\operatorname{id}_{K}\). Man zeige, es gilt \(m=n\).

Es sei \(R\) ein faktorieller Ring. Man zeige: (i) Ist \(S \subset R\) ein maltiplikatives System, so ist auch der Bruchring \(S^{-1} R\). faktoriell. Wie verhalten sich die Primelemente von \(R z u\) denen von \(S^{-1} R ?\) (ii) Fur Primelemente \(p \in R\) setze man \(R_{p}:=S_{p}^{-1} R\) mit \(S_{p}=R-(p)\). Ein Polynom \(f \in R[X]\) ist genau dann primitiv, wenn f眉r jedes Primelement \(p \in R\) das induzierte Polynom \(f_{p} \in R_{p}[X]\) primitiv ist.

Sei \(K\) ein Korper und \(f \in K\left[X_{1}, \ldots . X_{n}\right]\) ein homogenes Polynom vom Totalgrad \(d>0\). Man zeige, dass fur jede Primfaktorzerlegung \(f=p 1 \ldots p_{r}\) die Faktoren \(p_{i}\) homogen sind.
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