Question

If \(V=V_{1} \cup \cdots \cup V_{T}\) is the decomposition of an algebraic set into irreducible components, show that \(V_{i} \notin \cup_{j \neq i} V_{j}\).

Short answer

Question: Prove that for any irreducible component \(V_i\) of an algebraic set V, it does not belong to the union of all other irreducible components \(V_j\)'s where \(j \neq i\). Solution: Taking into account the definitions of algebraic sets and irreducible algebraic sets, we can prove this by showing that the assumption \(V_i \subseteq \cup_{j \neq i} V_j\) leads to a contradiction with the irreducibility of \(V_i\) since \(V_i\) can be written as a union of proper algebraic subsets. Thus, our claim is proved.

Step-by-step solution

Definitions

First, let's define an algebraic set and an irreducible algebraic set. An algebraic set \(V \subseteq \mathbb{A}^n\) is a set defined by the vanishing of a collection of polynomials, i.e., there exists a set of polynomials \(S \subseteq k[x_1, \cdots, x_n]\) such that \(V = \{P \in \mathbb{A}^n | f(P) = 0, \forall f \in S\}\). An algebraic set \(V\) is irreducible if it cannot be written as the union of two proper algebraic subsets, i.e., if \(V = V_1 \cup V_2\), then either \(V_1 = V\) or \(V_2 = V\). With these definitions, let's move on to the next step.

Proof by contradiction

We'll prove the claim is true by contradiction. Suppose, for the sake of contradiction, that \(V_i \subseteq \cup_{j \neq i} V_j\) for some index \(i\). It means \(V_i = V_i \cap \cup_{j \neq i} V_j\). Now, note that \(V_i \cap V_j\) is an algebraic set for all \(j \neq i\). This is because the intersection of algebraic sets is an algebraic set. Further, since \(V_i \subseteq \cup_{j \neq i} V_j\), we have: \(V_i = V_i \cap \left( \bigcup_{j \neq i} V_j \right) = \bigcup_{j \neq i} \left( V_i \cap V_j \right)\).

Irreducibility contradiction

Since \(V_i\) is an irreducible algebraic set, its decomposition into irreducible components consists of a single component, namely \(V_i\) itself. Thus, by the definition of irreducibility, \(V_i\) cannot be written as the union of two proper algebraic subsets. However, in Step 2, we showed that \(V_i\) can be written as the union of \(\left( V_i \cap V_j \right)\) where \(j \neq i\). This contradicts the irreducibility of \(V_i\), as it can be written as a union of proper algebraic subsets.

Conclusion

Since assuming \(V_i \subseteq \cup_{j \neq i} V_j\) gives rise to a contradiction, our assumption must be false. Hence, \(V_{i} \notin \cup_{j \neq i} V_{j}\), which proves the claim.

Key concepts

Irreducible Components

In algebra, an irreducible component of an algebraic set is a component that cannot be further decomposed into simpler algebraic subsets. Let's break this down into simpler terms. Imagine an algebraic set as a big puzzle that consists of multiple pieces, each representing a smaller set that fits together to form the whole picture.
Irreducible components are those pieces that are as simple as they can get - they cannot be split into more pieces without breaking their own structure. Even if they are part of a larger puzzle, they remain whole and indivisible. An algebraic set is considered irreducible if it cannot be divided into two smaller non-empty algebraic sets and still remain the same set.

• Irreducible components are fundamental building blocks in algebraic geometry.
• The decomposition of an algebraic set into its irreducible components helps understand its structure.
• An irreducible algebraic set can be thought of as consisting of just one component, itself.

Understanding irreducible components is crucial for analyzing the structure of algebraic sets and how different sets relate to each other.

Vanishing Polynomials

Polynomials are essential tools in algebraic geometry as they help define algebraic sets. A vanishing polynomial is one that evaluates to zero for all points of a given algebraic set. Essentially, this polynomial discerns the set's boundaries and defines the structure of the set by only vanishing on its points.
When you consider vanishing polynomials with respect to an algebraic set, think of them as a kind of rule or equation that the points in the set must follow. If you have a polynomial that vanishes on a particular set, it means whenever you substitute a point from that set into the polynomial, you get zero. This characterizes the points of the set in a concrete, mathematical way.

• Vanishing polynomials serve as a defining factor for algebraic sets.
• They determine whether a given point belongs to the set.
• If several polynomials vanish on different parts of the set, they can help determine the decomposition into irreducible components.

By studying vanishing polynomials, mathematicians can understand the properties and relationships of algebraic sets more deeply.

Proof by Contradiction

Proof by contradiction is a common mathematical technique used to establish the truth of a statement by showing that assuming the opposite leads to a contradiction. It involves taking the negation of the statement you want to prove, and then logically deducing an impossibility from this assumption.
The essence of a proof by contradiction is to demonstrate that the assumption of the opposite ends in a logical paradox, thus confirming the original statement must be true. In the context of irreducible components, this method is used to establish that if an irreducible component could be expressed as part of the union of the others, it would contradict its status as an indivisible whole.

• Proof by contradiction begins with the assumption that the statement is false.
• Logical reasoning is applied to this assumption to find an inconsistency.
• Ending up with a contradiction validates the original statement as true.

Mastering proof by contradiction allows mathematicians to verify complex concepts efficiently, like the unique nature of irreducible components within an algebraic set.