Problem 2
Question
Let \(\mathrm{p}\) be a prime ideal, and \(\mathrm{a}, \mathrm{b}\) ideals of \(A\). If \(\mathrm{ab} \subset \mathrm{p}\), show that \(a \subset \mathrm{p}\) or \(\mathrm{b} \subset \mathrm{p}\).
Step-by-Step Solution
Verified Answer
Assuming for contradiction that neither \(a\) nor \(b\) is a subset of the prime ideal \(\textit{p}\), we find elements \(x \in a\) and \(y \in b\) such that \(x \notin p\) and \(y \notin p\). Since the product of ideals \(ab \subseteq p\), we know \(xy \in p\). This contradicts the definition of a prime ideal, which states that either \(x \in p\) or \(y \in p\). Therefore, our assumption must be false, and if \(ab \subseteq p\), then either \(a \subseteq p\) or \(b \subseteq p\).
1Step 1: Definition of a prime ideal
Recall the definition of a prime ideal: An ideal p in a commutative ring A is called a prime ideal if for any elements x and y in A, if their product xy is an element of p, then either x is in p or y is in p.
2Step 2: Assume the contrary
Let's assume for the sake of contradiction that neither a nor b is a subset of the prime ideal p. In other words, there exist elements x ∈ a and y ∈ b such that x ∉ p and y ∉ p.
3Step 3: Consider the product of x and y
Since x and y belong to a and b respectively, and a and b are ideals, their product xy belongs to the product of the ideals ab.
Since ab ⊆ p, we have xy ∈ p.
4Step 4: Contradiction with the definition of a prime ideal
However, this contradicts the definition of a prime ideal. We know that xy ∈ p, and since p is a prime ideal, this implies that either x ∈ p or y ∈ p. But we have assumed that x ∉ p and y ∉ p in Step 2, leading to a contradiction.
5Step 5: Conclusion
Therefore, our initial assumption that neither a nor b is a subset of the prime ideal p must be false. Hence, if ab ⊆ p, then either a ⊆ p or b ⊆ p.
Key Concepts
Commutative RingIdealsContradiction Method
Commutative Ring
A commutative ring is a fundamental concept in algebra that helps us understand many structures in mathematics. In a commutative ring, the result of multiplying two elements doesn’t change regardless of their order. This property is similar to ordinary multiplication of numbers, where for any two numbers, say 2 and 3, you have 2 × 3 = 3 × 2.
Commutative rings contain two binary operations: addition and multiplication, and these operations must satisfy several key properties:
Commutative rings contain two binary operations: addition and multiplication, and these operations must satisfy several key properties:
- Associativity: For all elements, say a, b, and c, in the ring, we have (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c).
- Commutativity: As mentioned, a + b = b + a and a × b = b × a for all elements a and b.
- Identity: There must be an additive identity (usually 0) and a multiplicative identity (usually 1) such that a + 0 = a and a × 1 = a for all elements a.
- Distributivity: Multiplication distributes over addition, meaning a × (b + c) = a × b + a × c for all elements a, b, and c.
Ideals
Ideals are special subsets of a commutative ring that have properties making them crucial in ring theory. An ideal behaves somewhat like a number zero in modular arithmetic, allowing us to "reduce" elements.
For a subset I of a ring R to be an ideal, it must satisfy a couple of important properties:
For a subset I of a ring R to be an ideal, it must satisfy a couple of important properties:
- I must be a subgroup with respect to addition, meaning that for any two elements a and b in I, a + (-b) is also in I.
- If you take any element r from the ring R and an element a from the ideal I, the product r × a must also be in the ideal I.
Contradiction Method
The contradiction method is a powerful tool in mathematical proofs, where one assumes the opposite of what's to be proven and shows that this assumption leads to an impossibility or a contradiction.
To apply this method, follow these steps:
To apply this method, follow these steps:
- Begin by assuming that the statement you want to prove is false. This is called your 'contrary assumption'.
- Use logical reasoning and the properties of the system you’re working within (like properties of rings or ideals) to explore the consequences of your assumption.
- Demonstrate that these consequences clash with some known fact or axiom, thereby leading to a contradiction.
Other exercises in this chapter
Problem 1
Let \(A\) be a commutative ring. Let \(M\) be a module, and \(N\) a submodule. Let \(N=Q_{1} \cap \cdots \cap Q\), be a primary decomposition of \(N\). Let \(\b
View solution Problem 3
Let \(q\) be a primary ideal. Let \(a, b\) be ideals, and assume \(a b \subset\). Assume that \(b\) is finitely generated. Show that \(a \subset q\) or there ex
View solution Problem 5
Let \(A\) be an arbitrary commutative ring and let \(S\) be a multiplicative subset. Let \(p\) be a prime ideal and let \(q\) be a p-primary ideal. Then p inter
View solution Problem 6
If \(a\) is an ideal of \(A\), let \(a_{s}=S^{-1} a\). If \(\varphi_{s}: A \rightarrow S^{-1} A\) is the canonical map, abbreviate \(\varphi_{s}^{-1}\left(a_{s}
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