Show That Dim ⁡ ( A ) ≤ Dim ⁡ ( B ) \dim(A) \leq \dim(B) Dim ( A ) ≤ Dim ( B ) .

Introduction

In the realm of commutative algebra, the Krull dimension is a fundamental concept that measures the complexity of a ring. Given two rings, $A$ and $B$, where $A$ is a subring of $B$, and $B$ is integral over $A$, we aim to establish a relationship between the dimensions of these rings. Specifically, we want to show that the dimension of $A$ is less than or equal to the dimension of $B$. This result is a crucial step in understanding the properties of integral extensions and their impact on the Krull dimension.

Preliminaries

Before diving into the proof, let's recall some essential definitions and properties.

• Krull Dimension: The Krull dimension of a ring $A$, denoted by $\dim(A)$, is the supremum of the lengths of chains of prime ideals in $A$.
• Integral Extension: A ring $B$ is said to be integral over a ring $A$ if every element of $B$ satisfies a monic polynomial equation with coefficients in $A$.
• Chain of Prime Ideals: A chain of prime ideals in a ring $A$ is a sequence of prime ideals $P_0 \varsubsetneq P_1 \varsubsetneq \ldots \varsubsetneq P_n$.

The Proof

Let $A \subseteq B$ be rings, and $B$ be integral over $A$. We want to show that $\dim(A) \leq \dim(B)$. To do this, we'll start by assuming that $\dim(A) = m$ and $\dim(B) = n$. Our goal is to establish a relationship between the dimensions of $A$ and $B$.

Step 1: Establishing a Chain of Prime Ideals in $A$

Let $P_0 \varsubsetneq P_1 \varsubsetneq \ldots \varsubsetneq P_m$ be a chain of prime ideals in $A$. We can extend this chain to $B$ by considering the ideals $Q_i = P_iB$ for each $i$. Since $B$ is integral over $A$, we have that $Q_i$ is a prime ideal in $B$ for each $i$. Moreover, we have that $Q_0 \varsubsetneq Q_1 \varsubsetneq \ldots \varsubsetneq Q_m$ is a chain of prime ideals in $B$.

Step 2: Showing that the Chain of Prime Ideals in $B$ has Length at Most $m$

We want to show that the chain of prime ideals in $B$ has length at most $m$. To do this, we'll use the fact that $B$ is integral over $A$. Let $x \in Q_{m+1}$. Since $B$ is integral over $A$, there exists a monic polynomial $f(x) = x^{n+1} + a_n x^n + \ldots + a_0$ with coefficients in $A$ such that $f(x) = 0$. We can write $f(x) = (x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_{n+1})$, where $\alpha_i \in B$ for each $i$. Since $x \in Q_{m+1}$, we have that $\alpha_i \in Q_{m+1}$ for each $i$. Therefore, we have that $Q_{m+1}$ contains a non-zero divisor of $B$, which implies that $Q_{m+1} = B$. This contradicts the assumption that $Q_0 \varsubsetneq Q_1 \varsubsetneq \ldots \varsubsetneq Q_m$ is a chain of prime ideals in $B$. Therefore, we must have that the chain of prime ideals in $B$ has length at most $m$.

Conclusion

We have shown that $\dim(A) \leq \dim(B)$. This result is a crucial step in understanding the properties of integral extensions and their impact on the Krull dimension. The proof relies on the fact that $B$ is integral over $A$, which allows us to extend chains of prime ideals from $A$ to $B$. The result has important implications for the study of commutative algebra and the properties of rings.

References

• [1] Atiyah, M. F., & Macdonald, I. G. (1969). Introduction to commutative algebra. Addison-Wesley Publishing Company.
• [2] Bourbaki, N. (1959). Commutative algebra. Hermann.
• [3] Eisenbud, D. (1995). Commutative algebra with a view toward algebraic geometry. Springer-Verlag.

Further Reading

• [1] Krull, W. (1935). Dimensionstheorie in Stellenringen. Mathematische Annalen, 111(1), 1-30.
• [2] Zariski, O., & Samuel, P. (1958). Commutative algebra. Vol. 1. D. Van Nostrand Company.
• [3] Matsumura, H. (1970). Commutative algebra. W.A. Benjamin, Inc.
  Krull Dimension and Integral Extensions: Q&A =====================================================

Introduction

In our previous article, we explored the relationship between the Krull dimension of a ring and its integral extensions. We showed that if $A \subseteq B$ are rings, and $B$ is integral over $A$, then $\dim(A) \leq \dim(B)$. In this article, we'll delve deeper into the topic and answer some frequently asked questions.

Q: What is the Krull dimension of a ring?

A: The Krull dimension of a ring $A$, denoted by $\dim(A)$, is the supremum of the lengths of chains of prime ideals in $A$. In other words, it's the maximum length of a chain of prime ideals in $A$.

Q: What is an integral extension?

A: A ring $B$ is said to be integral over a ring $A$ if every element of $B$ satisfies a monic polynomial equation with coefficients in $A$. In other words, every element of $B$ is a root of a monic polynomial with coefficients in $A$.

Q: How does the Krull dimension of a ring relate to its integral extensions?

A: If $A \subseteq B$ are rings, and $B$ is integral over $A$, then $\dim(A) \leq \dim(B)$. This means that the Krull dimension of $A$ is less than or equal to the Krull dimension of $B$.

Q: What are some examples of integral extensions?

A: Some examples of integral extensions include:

• $\mathbb{Z} \subseteq \mathbb{Q}$: The ring of integers is integral over the rational numbers.
• $\mathbb{Z} \subseteq \mathbb{Z}[i]$: The ring of integers is integral over the Gaussian integers.
• $\mathbb{Z} \subseteq \mathbb{Z}[x]$: The ring of integers is integral over the polynomial ring in one variable.

Q: What are some examples of rings that are not integral extensions?

A: Some examples of rings that are not integral extensions include:

• $\mathbb{Q} \subseteq \mathbb{Q}(x)$: The rational numbers are not integral over the polynomial ring in one variable.
• $\mathbb{Z} \subseteq \mathbb{Z}[x,y]$: The ring of integers is not integral over the polynomial ring in two variables.

Q: How does the Krull dimension of a ring relate to its prime ideals?

A: The Krull dimension of a ring is related to its prime ideals in the following way: the Krull dimension of a ring is the maximum length of a chain of prime ideals in the ring.

Q: What are some applications of the Krull dimension?

A: The Krull dimension has many applications in algebra and geometry, including:

• Algebraic geometry: The Krull dimension is used to study the properties of algebraic varieties.
• Commutative algebra: The Krull dimension is used to study the properties of commutative rings.
• Number theory: The Krull dimension is used to study the properties of number fields.

Conclusion

In this article, we've explored the relationship between the Krull dimension of a ring and its integral extensions. We've answered some frequently asked questions and provided examples of integral extensions and rings that are not integral extensions. We've also discussed the applications of the Krull dimension in algebra and geometry.

References

• [1] Atiyah, M. F., & Macdonald, I. G. (1969). Introduction to commutative algebra. Addison-Wesley Publishing Company.
• [2] Bourbaki, N. (1959). Commutative algebra. Hermann.
• [3] Eisenbud, D. (1995). Commutative algebra with a view toward algebraic geometry. Springer-Verlag.

Further Reading

• [1] Krull, W. (1935). Dimensionstheorie in Stellenringen. Mathematische Annalen, 111(1), 1-30.
• [2] Zariski, O., & Samuel, P. (1958). Commutative algebra. Vol. 1. D. Van Nostrand Company.
• [3] Matsumura, H. (1970). Commutative algebra. W.A. Benjamin, Inc.