Question

Let \(F\) be a field, let \(f \in F[X, Y],\) and let \(E:=F[X, Y] /(f)\) Define \(V(f):=\\{(x, y) \in F \times F: f(x, y)=0\\}\) (a) Every element \(\alpha\) of \(E\) naturally defines a function from \(V(f)\) to \(F,\) as follows: if \(\alpha=[g]_{f},\) with \(g \in F[X, Y],\) then for \(P=(x, y) \in V(f),\) we define \(\alpha(P):=g(x, y) .\) Show that this definition is unambiguous, that is, \(g \equiv h(\bmod f)\) implies \(g(x, y)=h(x, y)\) (b) For \(P=(x, y) \in V(f),\) define \(M_{P}:=\\{\alpha \in E: \alpha(P)=0\\} .\) Show that \(M_{P}\) is a maximal ideal of \(E,\) and that \(M_{P}=\mu E+v E,\) where \(\mu:=[X-x]_{f}\) and \(v:=[Y-y]_{f}\)

Short answer

In this exercise, we had to prove properties of a quotient ring E for a given field F and polynomial f. We successfully showed that (a) the definition of an element α of E is unambiguous for all points in V(f); and (b) the set M_P is a maximal ideal of E for any point P. By breaking down the problem into smaller subproblems and utilizing basic definitions and properties of polynomial rings, ideals, and quotient rings, we provided a step-by-step proof for each claim. The final result established that M_P is equal to µE + νE, which gives further insight into the structure of the quotient ring E.

Step-by-step solution

Basic definitions

Before diving into the exercise, let's briefly define the key concepts used in this problem. - \(F[X,Y]\): The polynomial ring in two variables X and Y over the field F. - Ideal: A subset \(I\) of a ring \(R\) is an ideal if it is closed under addition and multiplication by any element of \(R\). - Quotient ring: Given a ring \(R\) and an ideal \(I\) of \(R\), the quotient ring \(R/I\) is the set of equivalence classes of elements in \(R\) under the relation defined by \(r \sim s\) if and only if \(r-s \in I\). - Maximal ideal: An ideal \(M\) of a ring \(R\) is maximal if there exists no proper ideal of \(R\) containing \(M\) and \(R\). Now, let's proceed with solving the given problem.

Prove that the definition of α(P) is unambiguous

Let \(\alpha = [g]_f\) and \(\alpha = [h]_f\). By the definition of equivalence classes, \(g \equiv h \pmod{f}\) implies that there exists a polynomial \(k \in F[X,Y]\) such that \(f \cdot k = h - g\). Now, let \(P = (x,y) \in V(f)\), which means \(f(x,y) = 0\). Then, \(0 = f(x,y) \cdot k(x,y) = (h(x,y) - g(x,y))\). This implies that \(g(x, y) = h(x, y)\) for all \((x,y) \in V(f)\), and thus, the definition of \(\alpha(P)\) is indeed unambiguous.

Prove that M_P is a maximal ideal of E

Let \(P=(x, y) \in V(f)\) and \(M_P = \{ \alpha \in E : \alpha(P) = 0 \}\). First, we prove that \(M_P\) is an ideal of \(E\). Let \(\alpha, \beta \in M_P\), and let \(g, h\) be the polynomials corresponding to the equivalence classes of \(\alpha\) and \(\beta\). Then, \(\alpha(P) = g(x,y) = 0\) and \(\beta(P) = h(x,y) = 0\). If we now add the two elements, we have \((\alpha + \beta)(P) = (g + h)(x,y) = g(x,y) + h(x,y) = 0\), which means \(\alpha + \beta \in M_P\). Similarly, for any \(\gamma \in E\), let \(k\) be the polynomial corresponding to the equivalence class of \(\gamma\). Then, \((\alpha \cdot \gamma)(P) = (g \cdot k)(x,y) = g(x,y) \cdot k(x,y) = 0\), which implies that \(\alpha \cdot \gamma \in M_P\). Thus, \(M_P\) is indeed an ideal of \(E\). Now, let's prove that \(M_P\) is maximal in \(E\). Suppose there is an ideal \(N\) in \(E\) such that \(M_P \subset N\) and \(N \neq E\). Let \(\beta \in N \setminus M_P\), with corresponding polynomial \(h\). Since \(\beta \notin M_P\), we have \(h(x,y) \neq 0\). Consider the ideals \(I = (f, X-x)\) and \(J = (f, Y-y)\) in the ring \(F[X,Y]\). Since \((x,y) \in V(f)\), we know that \((x,y)\) is a common root of both \(f(x,y) = 0\) and \(X-x=0\), and similarly, \(f(x,y) = 0\) and \(Y-y = 0\). Now, since \(h(x,y) \neq 0\), there must exist a polynomial \(g \in F[X,Y]\) such that either \(g \equiv X - x\ (\bmod\ f)\) or \(g \equiv Y - y\ (\bmod\ f)\). Without loss of generality, let's assume \(g \equiv X - x\ (\bmod\ f)\). Then, \([g]_f = [X-x]_f = \mu \in E\). Since \(\beta = [h]_f \in N\) and \(N\) is an ideal, we also have \(\beta \cdot \mu = [h \cdot g]_f \in N\). This is a contradiction because the product of two nonzero elements is zero. Consequently, \(M_P\) must be a maximal ideal of \(E\).

Show that M_P = µE + νE

To prove this, we need to show that each element in \(M_P\) can be written as a linear combination of µ and ν, and that each element of the form \(a\cdot\mu + b\cdot\nu\) belongs to M_P. Let \(\alpha \in M_P\), with the corresponding polynomial \(g(x,y)\). Since \(g(x,y) \equiv 0\ (\bmod\ f)\), we can write \(g = a \cdot (X - x) + b \cdot (Y - y) + f \cdot k\) for some polynomials \(a, b, k \in F[X,Y]\). Then, \(\alpha(P) = g(x,y) = a(x,y) \cdot (X - x) + b(x,y) \cdot (Y - y) = 0\). Thus, \(\alpha \in M_P\) can be written as \(\alpha = \mu \cdot [a]_f + \nu \cdot [b]_f\), which implies that \(M_P \subseteq \mu E + \nu E\). On the other hand, let \(\alpha = a\cdot\mu + b\cdot\nu\) for some \([a]_f, [b]_f \in E\). Let \(g(x,y)\) be the polynomial representative of \([a]_f\) and \(h(x,y)\) be the polynomial representative of \([b]_f\). Then, we have \(\alpha(P) = (a\cdot\mu + b\cdot\nu)(P) = a\cdot\mu(P)+ b\cdot\nu(P) = a(x,y) \cdot (X-x) + b(x,y) \cdot (Y-y)\). Since \((x,y) \in V(f)\), we have \(X-x=0\) and \(Y-y=0\), thus \(\alpha(P) = 0\). Therefore, any \(\alpha = a\cdot\mu + b\cdot\nu\) belongs to \(M_P\), which means that \(\mu E + \nu E \subseteq M_P\). Combining both observations, we obtain that \(M_P = \mu E + \nu E\), which completes the exercise.

Key concepts

Quotient Ring

When we consider a polynomial ring like \( F[X, Y] \), it consists of polynomials with variables \( X \) and \( Y \) over a field \( F \). A quotient ring, denoted \( R/I \), is formed by pairing this polynomial ring with an ideal \( I \). To understand quotient rings, you should know that:

• An ideal \( I \) is a subset of \( R \) where any element multiplied by an element of \( R \) still belongs to \( I \).
• In forming the quotient ring, elements of \( R \) are considered equivalent if their difference is an element of \( I \).

A common way to form a quotient ring in field theory is by considering the ring of all polynomials modulo an ideal generated by a given polynomial. Here, if \( f \) is a polynomial, the quotient ring \( F[X, Y]/(f) \) consists of equivalence classes or sets of polynomials that have the same remainder when divided by \( f \). This differs fundamentally from the simpler arithmetic modulo, as it takes into account the ring structure of polynomials.
Understanding and building upon the idea of quotient rings allows mathematicians to effectively study polynomial equations and their roots.

Maximal Ideal

Maximal ideals are pivotal in ring theory because they help establish conditions under which a ring can be transformed into a simple structure known as a field. An ideal \( M \) in a ring \( R \) is called maximal if:

• There are no other ideals in \( R \) that contain \( M \), except for \( R \) itself.
• It is as large as possible without being the entire ring.

In our exercise, the set \( M_P = \{ \alpha \in E : \alpha(P) = 0 \} \) is an example of a maximal ideal within the quotient ring \( E \). It captures all the elements \( \alpha \) in \( E \) that evaluate to zero at a particular point \( P \).
This notion of maximality means that if you add any element from outside \( M \) to it, you get the entire ring, highlighting its critical role in dividing ring structures into simpler pieces. In algebraic geometry, this concept helps in translating geometric problems into algebraic ones by associating points (like \( P \)) on a variety with maximal ideals.

Field Theory

Field theory is a central area of mathematics, exploring structures known as fields. Fields are algebraic structures where every non-zero element has a multiplicative inverse, allowing division by non-zero numbers. Here are some key aspects:

• The basic examples include real numbers \( \mathbb{R} \), complex numbers \( \mathbb{C} \), rational numbers \( \mathbb{Q} \), and finite fields.
• Fields possess both addition and multiplication defined such that these operations adhere to the familiar laws of arithmetic (associativity, commutativity, and distributivity).
• Every field is a ring, but not every ring is necessarily a field.

The exercise you're tackling embeds its groundwork in field theory by using the field \( F \) over which the polynomials are defined. Especially in the context of quotient rings, understanding how fields function helps to understand how these constructed algebraic objects behave, particularly how they might relate back to more familiar concepts like functions and equations.
When a quotient ring forms a field, it provides a neat and fully reversible framework for dealing with polynomial equations, linking ideals and geometry in a fascinating and functional way.