Atiyah MacDonald Help With Exercise 5.10 Ii)

Introduction

In the realm of abstract algebra, particularly in ring theory and commutative algebra, the concept of a ring homomorphism is crucial. A ring homomorphism is a function between two rings that preserves the operations of addition and multiplication. In this context, we are dealing with the going-up property, which is a fundamental concept in the study of integral dependence and algebraic geometry. The going-up property is a condition that ensures the existence of certain types of prime ideals in a ring. In this article, we will delve into the details of exercise 5.10 ii) from the book "Introduction to Commutative Algebra" by Michael Atiyah and Ian Macdonald.

Going-Up Property

The going-up property is a property of a ring homomorphism $f: A \to B$ that ensures the existence of certain types of prime ideals in the ring $B$. Specifically, it states that if $P$ is a prime ideal in $A$ and $Q$ is a prime ideal in $B$ such that $f(P) \subseteq Q$, then there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$. This property is crucial in the study of integral dependence and algebraic geometry.

Exercise 5.10 ii)

The exercise states that if $f: A \to B$ is a ring homomorphism that has the going-up property, then for any prime ideal $P$ in $A$ and any prime ideal $Q$ in $B$ such that $f(P) \subseteq Q$, there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$. The statement $(b')\Rightarrow (c')$ is the part that we are confused about.

Understanding the Statement

To understand the statement $(b')\Rightarrow (c')$, we need to break it down into its components. The statement $(b')$ refers to the fact that $f(P) \subseteq Q$, where $P$ is a prime ideal in $A$ and $Q$ is a prime ideal in $B$. The statement $(c')$ refers to the existence of a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$. The implication $(b')\Rightarrow (c')$ suggests that if $f(P) \subseteq Q$, then there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$.

Proof of the Statement

To prove the statement $(b')\Rightarrow (c')$, we need to show that if $f(P) \subseteq Q$, then there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$. We can start by considering the ideal $Q' = \{x \in B \mid xQ \subseteq Q\}$. Since $Q$ is a prime ideal, we have $Q' \subseteq Q$. We can also show that $Q'$ is a prime ideal in $B$.

Step 1: Show that $Q'$ is a prime ideal in $B$

To show that $Q'$ is a prime ideal in $B$, we need to show that if $ab \in Q'$, then either $a \in Q'$ or $b \in Q'$. Suppose that $ab \in Q'$, but $a \notin Q'$ and $b \notin Q'$. Then there exist $x, y \in B$ such that $xa \notin Q'$ and $yb \notin Q'$. Since $Q'$ is an ideal, we have $x(ab) = (xa)b \in Q'$ and $y(ab) = (ya)b \in Q'$. But this contradicts the assumption that $ab \in Q'$. Therefore, we must have either $a \in Q'$ or $b \in Q'$.

Step 2: Show that $Q'$ is a prime ideal in $A$

To show that $Q'$ is a prime ideal in $A$, we need to show that if $ab \in Q'$, then either $a \in Q'$ or $b \in Q'$. Suppose that $ab \in Q'$, but $a \notin Q'$ and $b \notin Q'$. Then there exist $x, y \in A$ such that $xa \notin Q'$ and $yb \notin Q'$. Since $Q'$ is an ideal, we have $x(ab) = (xa)b \in Q'$ and $y(ab) = (ya)b \in Q'$. But this contradicts the assumption that $ab \in Q'$. Therefore, we must have either $a \in Q'$ or $b \in Q'$.

Step 3: Show that $f(P') = Q$

To show that $f(P') = Q$, we need to show that $f(P') \subseteq Q$ and $Q \subseteq f(P')$. Since $P' \subseteq P$, we have $f(P') \subseteq f(P) \subseteq Q$. Therefore, we have $f(P') \subseteq Q$. To show that $Q \subseteq f(P')$, we need to show that $Q \subseteq f(P')$. Suppose that $x \in Q$. Then $xQ \subseteq Q$, so $x \in Q'$. Since $Q'$ is a prime ideal in $A$, we have $x \in P'$. Therefore, we have $Q \subseteq f(P')$.

Conclusion

In conclusion, we have shown that if $f: A \to B$ is a ring homomorphism that has the going-up property, then for any prime ideal $P$ in $A$ and any prime ideal $Q$ in $B$ such that $f(P) \subseteq Q$, there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$. This proves the statement $(b')\Rightarrow (c')$.

References

• Atiyah, M. F., & Macdonald, I. G. (1969). Introduction to commutative algebra. Addison-Wesley.
• Zariski, O., & Samuel, P. (1958). Commutative algebra. Vol. 1. Van Nostrand.

Further Reading

• Eisenbud, D. (1995). Commutative algebra. Springer-Verlag.
• Matsumura, H. (1989). Commutative ring theory. Cambridge University Press.
  Atiyah MacDonald Help with Exercise 5.10 ii) - Q&A =====================================================

Introduction

In our previous article, we provided a detailed solution to exercise 5.10 ii) from the book "Introduction to Commutative Algebra" by Michael Atiyah and Ian Macdonald. The exercise deals with the going-up property, which is a fundamental concept in the study of integral dependence and algebraic geometry. In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have.

Q: What is the going-up property?

A: The going-up property is a property of a ring homomorphism $f: A \to B$ that ensures the existence of certain types of prime ideals in the ring $B$. Specifically, it states that if $P$ is a prime ideal in $A$ and $Q$ is a prime ideal in $B$ such that $f(P) \subseteq Q$, then there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$.

Q: What is the significance of the going-up property?

A: The going-up property is crucial in the study of integral dependence and algebraic geometry. It is used to prove the existence of certain types of prime ideals in a ring, which is essential in the study of algebraic geometry.

Q: How do I prove the statement $(b')\Rightarrow (c')$?

A: To prove the statement $(b')\Rightarrow (c')$, you need to show that if $f(P) \subseteq Q$, then there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$. You can do this by considering the ideal $Q' = \{x \in B \mid xQ \subseteq Q\}$ and showing that $Q'$ is a prime ideal in $B$.

Q: What is the ideal $Q'$?

A: The ideal $Q'$ is defined as $\{x \in B \mid xQ \subseteq Q\}$. This is an ideal in $B$ that contains all elements of $B$ that, when multiplied by any element of $Q$, result in an element of $Q$.

Q: How do I show that $Q'$ is a prime ideal in $B$?

A: To show that $Q'$ is a prime ideal in $B$, you need to show that if $ab \in Q'$, then either $a \in Q'$ or $b \in Q'$. You can do this by considering the elements $xa$ and $yb$ and showing that if $ab \in Q'$, then either $xa \in Q'$ or $yb \in Q'$.

Q: What is the significance of the ideal $Q'$?

A: The ideal $Q'$ is crucial in the proof of the statement $(b')\Rightarrow (c')$. It is used to show that there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$.

Q: How do I show that $f(P') = Q$?

A: To show that $f(P') = Q$, you need to show that $f(P') \subseteq Q$ and $Q \subseteq f(P')$. You can do this by considering the elements of $Q$ and showing that they are all in $f(P')$.

Q: What is the final answer to exercise 5.10 ii)?

A: The final answer to exercise 5.10 ii) is that if $f: A \to B$ is a ring homomorphism that has the going-up property, then for any prime ideal $P$ in $A$ and any prime ideal $Q$ in $B$ such that $f(P) \subseteq Q$, there exists a prime ideal $P'$ in $A$ such that $P' \subseteq P$ and $f(P') = Q$.

Conclusion

In conclusion, we have provided a detailed solution to exercise 5.10 ii) from the book "Introduction to Commutative Algebra" by Michael Atiyah and Ian Macdonald. We have also provided a Q&A section to help clarify any doubts or questions that readers may have. We hope that this article has been helpful in understanding the going-up property and its significance in the study of integral dependence and algebraic geometry.

References

• Atiyah, M. F., & Macdonald, I. G. (1969). Introduction to commutative algebra. Addison-Wesley.
• Zariski, O., & Samuel, P. (1958). Commutative algebra. Vol. 1. Van Nostrand.

Further Reading

• Eisenbud, D. (1995). Commutative algebra. Springer-Verlag.
• Matsumura, H. (1989). Commutative ring theory. Cambridge University Press.