Question

Let \(I=\left(X^{2}-Y^{3}, Y^{2}-Z^{3}\right) \subset k[X, Y, Z]\). Define \(\alpha: k[X, Y, Z] \rightarrow k[T]\) by \(\alpha(X)=\) \(T^{9}, \alpha(Y)=T^{6}, \alpha(Z)=T^{4}\). (a) Show that every element of \(k[X, Y, Z] / I\) is the residue of an element \(A+X B+Y C+X Y D\), for some \(A, B, C, D \in k[Z] .\) (b) If \(F=A+X B+\) \(Y C+X Y D, A, B, C, D \in k[Z]\), and \(\alpha(F)=0\), compare like powers of \(T\) to conclude that \(F=0 .\) (c) Show that \(\operatorname{Ker}(\alpha)=I\), so \(I\) is prime, \(V(I)\) is irreducible, and \(I(V(I))=I\).

Short answer

Answer: The implications are that the ideal I is prime, the variety V(I) is irreducible, and the radical of the ideal I(V(I)) is equal to I (i.e., \(I(V(I)) = I\)).

Step-by-step solution

Form of elements in \(k[X, Y, Z] / I\)

Let's first express elements in \(k[X, Y, Z] / I\) as residue classes of the form \(A + XB + YC + XYD\), where \(A, B, C, D \in k[Z]\). Given the ideal \(I = (X^2 - Y^3, Y^2 - Z^3)\), by reducing modulo \(I\), some polynomial \(F \in k[X, Y, Z]\) can be represented as: $$F \equiv A + XB + YC + XYD \pmod I, \qquad \textrm{with} \, A, B, C, D \in k[Z].$$

Comparing powers of \(T\) if \(\alpha(F) = 0\)

Assume that \(\alpha(F) = 0\), applying \(\alpha\) to \(F = A + XB + YC + XYD\), we get: $$0 = \alpha(F) = \alpha(A) + \alpha(XB) + \alpha(YC) + \alpha(XYD).$$ Now substitute the given values for \(\alpha(X) = T^9\), \(\alpha(Y) = T^6\), and \(\alpha(Z) = T^4\): $$0 = \alpha(A) + T^9\alpha(B) + T^6\alpha(C) + T^{15}\alpha(D).$$ Comparing the powers of \(T\), we obtain the following equations: $$\begin{cases} \alpha(A) = 0 \\ T^9\alpha(B) = 0 \\ T^6\alpha(C) = 0 \\ T^{15}\alpha(D) = 0 \end{cases}.$$ This implies that \(\alpha(A) = \alpha(B) = \alpha(C) = \alpha(D) = 0\) because no terms can cancel each other. The only way this can hold is if \(F = A+XB+YC+XYD = 0\).

Show that \(\operatorname{Ker}(\alpha) = I\)

In order to show that \(\operatorname{Ker}(\alpha) = I\), we'll use the previous result (\(F = 0\) if \(\alpha(F) = 0\)). We already know that if \(F \in I\), then \(\alpha(F) = 0\), so \(I \subset \operatorname{Ker}(\alpha)\). Now let \(F\) be an element in \(\operatorname{Ker}(\alpha)\). From Step 2, we know that \(\operatorname{Ker}(\alpha)\) consists of elements of the form \(A + XB + YC + XYD\) where \(\alpha(A) = \alpha(B) = \alpha(C) = \alpha(D) = 0\). But this is equivalent to saying that \(F \equiv 0 \pmod I\) (i.e., \(F \in I\)). Hence, \(\operatorname{Ker}(\alpha) \subset I\). Combining the two inclusions, we conclude that \(\operatorname{Ker}(\alpha) = I\). This immediately implies that \(I\) is a prime ideal, \(V(I)\) is irreducible, and \(I(V(I)) = I\).

Key concepts

Prime Ideal

In algebraic geometry, the concept of a prime ideal is crucial. A prime ideal in a ring is an ideal such that if the product of two elements is in the ideal, then at least one of those elements must be in the ideal. This is akin to prime numbers in the sense that, just as a prime number can only be evenly divided by 1 and itself, a prime ideal has properties that relate closely to its prime nature.
A prime ideal has a direct connection to the geometric concept of an irreducible variety. In the context of polynomial rings, a prime ideal like the ideal \(I\) in the exercise indicates that the set of solutions or variety associated with it does not split into simpler components.
For example, in our problem, by proving \(\operatorname{Ker}(\alpha) = I\), we establish its primality, illustrating that the variety \(V(I)\) cannot be decomposed into smaller varieties.

Irreducible Variety

An irreducible variety is a geometric object that cannot be broken down into simpler pieces. In more precise terms, this means it is not the union of two smaller non-empty algebraic varieties. This is closely tied to the concept of prime ideals in algebraic geometry.
When the ideal \(I\) is prime in a polynomial ring, it defines an irreducible variety. This means the solutions to the polynomial equations in the ideal are connected in a specific, indivisible way. Thus, when we say that \(V(I)\) is irreducible for our ideal \(I=(X^2-Y^3, Y^2-Z^3)\), it means that the algebraic set described by \(I\) does not have a simpler decomposition.

• The irreducibility of a variety is important because it corresponds to the connectedness and "indivisibility" of the solution set.
• You can think of an irreducible variety like a single "piece" of space that is entirely connected in its solution structure.

Polynomial Rings

Polynomial rings form the backbone of many algebraic structures in algebraic geometry. These rings consist of polynomials with coefficients from a given field, in our exercise, from \(k\).
Polynomial rings, denoted as \(k[X, Y, Z]\) in our exercise, are the basic building blocks. They're generated by polynomials in multiple variables. When we impose an ideal, \(I\), we create a quotient ring \(k[X, Y, Z]/I\), which retains certain linear combinations while modding out others according to the relations defined by \(I\).

• In our problem, the ideal \(I\) generates relations such as \(X^2=Y^3\) and \(Y^2=Z^3\), fundamentally shaping the algebraic structure of \(k[X, Y, Z]/I\).
• This alteration is crucial as it tells us which polynomial expressions are equivalent in this targeted structure.

Homomorphism Kernel

The kernel of a homomorphism, denoted \(\operatorname{Ker}(\alpha)\) in our context, is a core concept in algebra, capturing the idea of the "bug" that maps everything to zero in a ring homomorphism. It collects all elements in the original ring that map to zero in the target ring.
Understanding the kernel is essential because it tells us about the structure we are dealing with. For our given homomorphism \(\alpha: k[X, Y, Z] \rightarrow k[T]\), the kernel includes precisely those elements whose images under \(\alpha\) are zero. This turns out to be the ideal \(I\) itself, showing how \(I\) defines the structure of the map and confirms its primality.

• The significance of the kernel being \(I\) is that it cements \(I\) as a complete and necessary set of equations for describing the transformation \(\alpha\).
• The primality of \(I\) then implies \(V(I)\) is irreducible, interlinking algebraic and geometric perspectives of the solutions.