Question

Let \(R=k\left[X_{1}, \ldots, X_{n}\right], k\) algebraically closed, \(V=V(I)\). Show that there is a natural one-to-one correspondence between algebraic subsets of \(V\) and radical ideals in \(k\left[X_{1}, \ldots, X_{n}\right] / I\), and that irreducible algebraic sets (resp. points) correspond to prime ideals (resp. maximal ideals). (See Problem 1.22.)

Short answer

Question: Establish a correspondence between algebraic subsets of V and radical ideals in R/I for the algebraically closed field k, and show that irreducible algebraic sets and points correspond to prime and maximal ideals, respectively. Answer: We first establish a one-to-one correspondence between algebraic subsets of V and radical ideals in R/I using the Nullstellensatz, which states that for an algebraically closed field k, an ideal does not generate the whole polynomial ring if and only if its algebraic set is non-empty. We then show that irreducible algebraic sets (which cannot be represented as the union of two proper algebraic subsets) correspond to prime ideals in R/I. Similarly, we demonstrate that points correspond to maximal ideals in R/I. Therefore, we have proven the one-to-one correspondence between algebraic subsets of V and radical ideals in R/I, along with the correspondence between irreducible algebraic sets (and points) and prime (and maximal) ideals.

Step-by-step solution

Nullstellensatz and its relevance

The Nullstellensatz is a fundamental result in algebraic geometry and commutative algebra, relating algebraic sets to their corresponding ideals. It states that if \(k\) is an algebraically closed field and \(J\) is an ideal of \(k[X_1, \ldots, X_n]\), then \(V(J)= \emptyset\) if and only if \(J = \langle 1 \rangle\). This means that if an ideal does not generate the whole polynomial ring, its algebraic set will not be empty. This result will help us establish the correspondence between algebraic subsets of \(V\) and radical ideals in \(R/I\). Moreover, since \(k\) is given to be algebraically closed, we can directly use the Nullstellensatz.

One-to-one correspondence between algebraic subsets of V and radical ideals in R/I

Let's consider the correspondence \(\phi: \{ \text{algebraic subsets of } V \} \rightarrow \{ \text{radical ideals in } R/I \}\), defined by \(\phi(W)=I(W) + I\). Given an algebraic subset \(W \subseteq V\), we aim to show that \(\phi\) is a well-defined, bijective map. First, note that for any ideal \(J \subseteq R\), we have \(V(I(J)) \supseteq V(J)\), as \(I(J) \supseteq J\). Now, let \(\overline{W} = V(I(W) + I)\). We claim that \(\overline{W} = W\): - \(\overline{W} \supseteq W\): Since \(I(W) \supseteq I\), we have \(V(I(W) + I) \supseteq V(I)\), so \(\overline{W} \supseteq V(I) \supseteq W\). - \(\overline{W} \subseteq W\): For any point \(x \in \overline{W}\), we have \(I(W) + I \subseteq I_x = \{ f \in R \mid f(x) = 0 \}\), where \(I_x\) is the ideal of polynomials vanishing at \(x\). Then, \(I(W) \subseteq I_x\), so \(x \in V(I(W))=W\). Thus, \(\overline{W} \subseteq W\). Therefore, \(\phi\) is well-defined. Moreover, since \(W\) is uniquely determined by \(I(W)+I\), it follows that \(\phi\) is injective. Now we show that \(\phi\) is surjective. Consider any radical ideal \(J \subseteq R/I\). We have \(J = I(V(J+I)) + I\) by the Nullstellensatz. Hence, \(V(J+I)\) is an algebraic subset of \(V\) such that \(\phi(V(J+I)) = J\). Thus, \(\phi\) is surjective. In conclusion, we have established a natural one-to-one correspondence between algebraic subsets of \(V\) and radical ideals in \(R/I\).

Irreducible algebraic sets correspond to prime ideals

We will now show that irreducible algebraic sets correspond to prime ideals in \(R/I\). Suppose \(W \subseteq V\) is an irreducible algebraic set, i.e., it cannot be represented as the union of two proper algebraic subsets. Let \(J = I(W)+I\), where \(I \subseteq R\) is the given ideal. Our goal is to show that \(J\) is a prime ideal. Suppose we have \(fg \in J\), for \(f, g \in R/I\). Since \(fg \in J\), it follows that \(fg \in I(W) + I\). Then, by definition of \(W\), we have \(V(fg) \supseteq W\). Since \(V(fg) = V(f) \cup V(g)\), we get \(W \subseteq V(f) \cup V(g)\). As \(W\) is irreducible, we must have \(W \subseteq V(f)\) or \(W \subseteq V(g)\). Thus, either \(f \in J\) or \(g \in J\), which shows that \(J\) is a prime ideal. Conversely, if we have a prime ideal \(J \subseteq R/I\), and \(W = V(J+I)\), we will show that \(W\) is an irreducible algebraic set. Suppose, for contradiction, that \(W\) can be written as the union of two proper algebraic subsets \(W_1\) and \(W_2\). Then, \(I(W_1)+I \subseteq J\) and \(I(W_2)+I \subseteq J\) since \(W_1, W_2 \subset V(J+I)\), and thus \((I(W_1)+I)(I(W_2)+I) \subseteq J\). Let \(f \in I(W_1)+I\) and \(g \in I(W_2)+I\). Then, \(fg \in J\), which implies that either \(f \in J\) or \(g \in J\). However, this contradicts the properness of \(W_1\) and \(W_2\). Therefore, \(W\) must be irreducible.

Points correspond to maximal ideals

Finally, we aim to show that points correspond to maximal ideals in \(R/I\). Let \(p\) be a point in \(V\). Consider the ideal \(M = I(p) + I\), where \(I \subseteq R\) is the given ideal. We will show that \(M\) is a maximal ideal in \(R/I\). To show that \(M\) is maximal, we need to show that, for any ideal \(J \subseteq R/I\) with \(M \subset J\), we get \(J = R/I\). Suppose we have an element \(f \in J \setminus M\). Then, \(f \notin M = I(p) + I\), which means that \(f(p) \neq 0\). Now, let \(g = \frac{1}{f(p)} (f - f(p))\). We have \(g \in J\), and \(g(p) = 0\). Thus, \(g \in I(p) + I = M\). Furthermore, \(fg - 1 = (f - f(p))(g - \frac{1}{f(p)} f) \in J\). Hence, \(1 \in J\), which implies \(J = R/I\). Therefore, \(M\) is a maximal ideal. Conversely, if \(M\) is a maximal ideal in \(R/I\), let \(p = V(M + I) \in V\). We will show that \(p\) is a point. Suppose, for contradiction, that \(p\) is not a point. Then, \(p\) can be represented as the union of two proper algebraic subsets \(W_1\) and \(W_2\). Let \(f \in I(W_1) + I\), and \(g \in I(W_2) + I\). We have \(ff'gg' = 1\) for some \(f', g' \in R\), and therefore \((fg)(fg)' = 1\). Since \(M = I(p)+I\), we get \((fg)(fg)' \in M\). However, this contradicts the maximality of \(M\), so \(p\) must be a point. In conclusion, we have shown the one-to-one correspondence between algebraic subsets of \(V\) and radical ideals in \(R/I\) and that irreducible algebraic sets (resp. points) correspond to prime ideals (resp. maximal ideals).

Key concepts

Nullstellensatz

The Nullstellensatz is a cornerstone theorem in algebraic geometry that acts as a bridge between algebraic sets and ideals. This theorem is applicable when working with a polynomial ring over an algebraically closed field, such as the field of complex numbers. When we talk about an ideal in the ring of polynomials, the Nullstellensatz connects this algebraic object to a geometric one.

• The theorem states that if an ideal in the polynomial ring generates the whole ring, its associated algebraic set is empty.
• Conversely, if the algebraic set is non-empty, the ideal cannot generate the entirety of the polynomial ring.

This understanding establishes connections between algebraic subsets and radical ideals and guides us in defining the correspondence between these objects.

Prime Ideals

Prime ideals are crucial in understanding the structure of a ring, particularly in the context of algebraic geometry. These are specific types of ideals that possess a unique property when it comes to multiplication.

• In a ring, an ideal is called prime if, whenever the product of two elements of the ring lies in the ideal, at least one of the factors must also be in the ideal.
• This property ensures that the corresponding algebraic set is irreducible, meaning it cannot be broken down into simpler pieces.

Understanding prime ideals allows us to explore and define irreducible algebraic sets' properties more clearly, showcasing the intersection of algebraic and geometric ideas.

Maximal Ideals

Maximal ideals represent an important concept in ring theory and algebraic geometry. They play a significant role in characterizing the points on algebraic sets.

• An ideal is maximal in a ring if it is contained in no larger ideal, except the ring itself.
• Geometrically, each maximal ideal corresponds to a unique point in the algebraic set, as removing any more elements from the maximal ideal results in the whole ring.

Their importance stems from their ability to provide a direct understanding of the topological structure of algebraic sets, as they essentially pinpoint the exact 'locations' within these geometric objects.

Algebraic Sets

Algebraic sets are sets of solutions to systems of polynomial equations and are fundamental objects in algebraic geometry. They form the geometric counterpart to the algebraic concept of ideals.

• These sets can demonstrate various forms of structure and complexity, embodying solutions that could range from simple points to complex surfaces.
• Through radical ideals, algebraic sets reveal themselves in terms of intersection properties and radical layers, driving home the deep link between algebra and geometry.

The study of algebraic sets through the lens of ideals not only highlights the elegance of algebraic geometry but also aids in solving polynomial equations by understanding these sets' detailed structure.