Question

Continuing with the previous exercise, now assume that the characteristic of \(F\) is \(\operatorname{not} 2,\) and that \(f=Y^{2}-\phi,\) where \(\phi \in F[X]\) is a non-zero polynomial with no multiple roots in \(F\) (see definitions after Exercise 7.18 ). (a) Show that if \(P=(x, y) \in V(f),\) then so is \(\bar{P}:=(x,-y),\) and that \(P=\bar{P} \Longleftrightarrow y=0 \Longleftrightarrow \phi(x)=0\) (b) Let \(P=(x, y) \in V(f)\) and \(\mu:=[X-x]_{f} \in E .\) Show that \(\mu E=M_{P} M_{\bar{P}}\) (the ring-theoretic product). Hint: use Exercise \(7.43,\) and treat the cases \(P=\bar{P}\) and \(P \neq \bar{P}\) separately.

Short answer

In summary, we have shown that for any points \(P = (x, y) \in V(f)\) and \(\bar{P} = (x, -y) \in V(f)\), it holds that \(P = \bar{P}\) if and only if \(y = 0\) if and only if \(\phi(x) = 0\). Furthermore, we have established that the relationship \(\mu E = M_P M_{\bar{P}}\) always holds, regardless of whether \(P \neq \bar{P}\) or \(P = \bar{P}\).

Step-by-step solution

The given polynomial equation is \(f = Y^2 - \phi\). Since \(P = (x, y) \in V(f)\), we know that \(y^2 - \phi(x) = 0\). Now we want to check if \(\bar{P} = (x, -y)\) satisfies the equation: substituting \(Y = -y\) in the equation, we obtain: $$ (-y)^2 - \phi(x) = y^2 - \phi(x) = 0 $$ This equality holds, which means that \(\bar{P} = (x, -y) \in V(f)\). **Step 2: Show \(P = \bar{P} \Longleftrightarrow y = 0 \Longleftrightarrow \phi(x) = 0\)**

If \(P = \bar{P}\), then \((x, y) = (x, -y)\), implying that \(y = -y\). Since the characteristic of \(F\) is not 2, this means \(y = 0\). Now, we know that \(y^2 - \phi(x) = 0\), so we get \(0 = \phi(x)\). Conversely, if \(y = 0\), then \(P = (x, 0)\) and \(\bar{P} = (x, -0) = (x, 0)\). Since \(y = 0\), then the equation \(y^2 - \phi(x) = 0\) implies \(\phi(x) = 0\). So, \(P = \bar{P} \Longleftrightarrow y = 0 \Longleftrightarrow \phi(x) = 0\). **Step 3: Prove \(\mu E = M_P M_{\bar{P}}\) for \(P \neq \bar{P}\)**

Suppose \(P \neq \bar{P}\), which means \(y \neq 0\). Using Exercise 7.43, we know that \(\mu E = M_P + M_{\bar{P}}\). Since \(E[x] = \{p(x)\}_{p(X) \in E}\), we have: $$ \mu E = \{[(X - x)p(x)]_f \mid p(X) \in E\} $$ Now, let's compute the product of \(M_P\) and \(M_{\bar{P}}\), which can be written as: $$ M_P M_{\bar{P}} = \{[(X - x)p(x)][(X - x)q(x)]_f \mid p(X),q(X) \in E\} $$ Let \([(X - x)p(x)]_f = [(X - x)(yp(x) + y\phi(x))]_f \in \mu E\). We have: $$ [(X - x)p(x)]_f = [(X - x)(yp(x) + y\phi(x))]_f = M_P M_{\bar{P}} $$ Thus, \(\mu E = M_P M_{\bar{P}}\) when \(P \neq \bar{P}\). **Step 4: Prove \(\mu E = M_P M_{\bar{P}}\) for \(P = \bar{P}\)**

Suppose \(P = \bar{P}\), which means \(y = 0\). Since \(P = \bar{P}\), then \(M_P = M_{\bar{P}}\). Using Exercise 7.43, we know that \(\mu E = 2M_P\). Now, let's compute the product of \(M_P\) and \(M_{\bar{P}}\), which can be written as: $$ M_P M_{\bar{P}} = \{[(X - x)p(x)][(X - x)q(x)]_f \mid p(X),q(X) \in E\} $$ Let \([(X - x)p(x)]_f = [(X - x)(0p(x) + 0\phi(x))]_f \in \mu E\). We have: $$ [(X - x)p(x)]_f = M_P M_{\bar{P}} $$ Since \(M_P = M_{\bar{P}}\), we can rewrite the equation as \(\mu E = 2M_P = M_P M_{\bar{P}}\). Thus, \(\mu E = M_P M_{\bar{P}}\) when \(P = \bar{P}\). Now, we have shown that for both cases \(P \neq \bar{P}\) and \(P = \bar{P}\), the relationship \(\mu E = M_P M_{\bar{P}}\) holds.

Key concepts

Characteristic of a Field

Understanding the characteristic of a field is essential for delving into higher-level algebra. The characteristic of a field is the smallest positive number of times you need to add the multiplicative identity, 1, to itself to get 0. If adding 1 to itself repeatedly never results in 0, then the field is said to have characteristic 0. However, if there is such a number, the field will have a characteristic equal to that number.

For instance, the field of real numbers has a characteristic of 0 because adding 1 to itself will never give you 0. In contrast, the field with elements {0, 1}, which adheres to arithmetic modulo 2, has characteristic 2, as 1 + 1 = 0 in this system. In the context of algebraic geometry, knowing the characteristic is useful to understand symmetries, solutions to equations, and the structure of algebraic varieties.

Polynomial Equations

Polynomial equations form the backbone of algebraic geometry as they define algebraic sets and varieties. A polynomial equation in a field F can be written in the form of P(X) = 0, where P is a polynomial composed of variables and coefficients from F. These equations can have a varying number of solutions depending on the degree of the polynomial and the properties of the field F.

For example, a polynomial equation like \(Y^2 - \(phi\)(X) = 0\) can describe a curve in a two-dimensional space. Solving such equations often involves finding the set of all possible solutions, known as the variety or zeros of the polynomial. These solutions are vital in understanding the geometric and algebraic properties of shapes defined by polynomials.

Exercise Solutions

When providing exercise solutions, especially for complex subjects like algebraic geometry, clarity and step-by-step explanations are paramount. A well-crafted solution will not only provide the answer but also the rationale behind each step taken towards that answer. It reinforces understanding by breaking down the problem into manageable parts. For instance, in the provided exercise, demonstrating that a point \(P=(x, y)\) and its counterpart \(\bar{P}=(x, -y)\) lie on the variety \(V(f)\) requires substituting the coordinates into the polynomial equation and proving they satisfy the equation. Such meticulous proof-steps, when detailed correctly, help students grasp underlying concepts, fostering their problem-solving abilities.

Ring Theory

Ring theory, a fundamental area of abstract algebra, studies structures called rings that generalize the arithmetic concepts of addition and multiplication. A ring is a set equipped with two binary operations, typically called addition and multiplication, where the set is an abelian group under addition and a monoid under multiplication. Rings have applications across mathematics, including number theory, topology, and algebraic geometry.

In our exercise, ring theory principles are applied to show the relationship between the element \(\mu\) in a ring extension E and the ideals \(M_P\) and \(M_{\bar{P}}\). The ring-theoretic product, denoted by \(M_P M_{\bar{P}}\), illustrates the multiplication of two ideals, which in turn helps in understanding the structure and properties of the solutions to the polynomial equation and their interactions within the ring.