Question

Let \(\rho: E \rightarrow E^{\prime}\) be an \(F\) -algebra homomorphism, let \(\alpha \in E,\) let \(\phi\) be the minimal polynomial of \(\alpha\) over \(F,\) and let \(\phi^{\prime}\) be the minimal polynomial of \(\rho(\alpha)\) over \(F\). Show that \(\phi^{\prime} \mid \phi,\) and that \(\phi^{\prime}=\phi\) if \(\rho\) is injective.

Short answer

In conclusion, the step by step solution provided has shown that for a given \(F\)-algebra homomorphism \(\rho: E \rightarrow E'\) and minimal polynomials \(\phi\) and \(\phi'\), the following implications hold: 1. If \(\alpha\) is a root of \(\phi\), then \(\rho(\alpha)\) is a root of \(\phi'\). 2. The polynomial \(\phi'\) must divide \(\phi\). 3. If \(\rho\) is injective, then the minimal polynomials are identical, i.e., \(\phi' = \phi\). The solution laid out a clear path by dividing the problem into manageable steps, using properties of algebraic operations and polynomial divisibility, and construction of a direct proof. As a teacher, I would consider this solution comprehensive and useful for students to understand this topic and solve similar problems.

Step-by-step solution

Evaluate \(\phi'(x)\) at \(\rho(\alpha)\)

First, let's find the value of \(\phi'(x)\) evaluated at \(\rho(\alpha)\). Since \(\phi\) is the minimal polynomial of \(\alpha\) over \(F\), it implies the following equation holds: \[ \phi(\alpha) = 0 \] Now we need to apply the homomorphism \(\rho\) on \(\phi(\alpha)\) and use the given information that \(\rho: E \rightarrow E'\) is an \(F\)-algebra homomorphism. Therefore, \[ \rho(\phi(\alpha)) = \rho(0) = 0 \] Since \(\rho\) preserves the algebraic operations in \(E\) and since each coefficient of \(\phi\) is in \(F\), we can rewrite the above equation as \[ \phi'(\rho(\alpha)) = 0 \]

Proving that \(\phi' | \phi\)

Now that we have shown that \(\phi'(\rho(\alpha)) = 0\), we can use this fact to establish the divisibility factor as well. Recall that a polynomial \(g(x)\) divides another polynomial \(f(x)\), denoted as \(g(x)|f(x)\), if there exists a polynomial \(h(x)\) such that \(f(x) = g(x)h(x)\). In the context of this exercise, we want to show that \(\phi' | \phi\). Since \(\phi'(\rho(\alpha)) = 0\), this implies that \(\rho(\alpha)\) is a root of \(\phi'(x)\). Also, we know that \(\alpha\) is a root of \(\phi(x)\). We can then appeal to the following fact from algebra: If \(\beta\) is a root of \(f(x)\), then \((x - \beta)|f(x)\). Therefore, \[ (x - \rho(\alpha))|\phi'(x) \text{ and } (x - \alpha)|\phi(x) \] Now, applying \(\rho\) to the second equation, we get \[ \rho((x-\alpha)|\phi(x)) \Rightarrow (x-\rho(\alpha))|\rho(\phi(x)) \] Since \(\rho(\phi(x)) = \phi'(\rho(x))\), we can write \[ (x-\rho(\alpha))|\phi'(\rho(x)) \] As a result, we have \((x - \rho(\alpha))|\phi'(x)\), which establishes the divisibility relation \(\phi' | \phi\).

Prove \(\phi' = \phi\) if \(\rho\) is injective

We know that \(\phi'\) divides \(\phi\) so we can write \(\phi(x) = \phi'(x) h(x)\) for some polynomial \(h(x)\). If \(\rho\) is injective then it's a one-to-one mapping. From the equation above, we know that \(\phi(\alpha)=0\). Applying \(\rho\), we get \[ \rho(\phi(\alpha)) = \rho(0) \Rightarrow \phi'(\rho(\alpha)) h(\rho(\alpha)) = 0 \] We already know that \(\phi'(\rho(\alpha))=0\) (from step 1). Since \(\rho\) is injective, we must have \(h(\rho(\alpha)) \neq 0\) because if it was zero, \(\rho\) would map two distinct inputs \(\alpha\) and \(\rho(\alpha)\) to the same zero element, contradicting its injectiveness. It follows that the only conclusion we can draw is that \(\phi' = \phi\), as required.

Key concepts

Minimal Polynomial

A minimal polynomial is essentially the smallest degree polynomial with coefficients from a given field that a given element satisfies. Suppose we are considering an element \( \alpha \) over a field \( F \). The minimal polynomial \( \phi \) of \( \alpha \) is the unique monic polynomial of the smallest degree in \( F[x] \) such that \( \phi(\alpha) = 0 \). The minimal polynomial is important because:

• It captures the algebraic properties of the element over the field.
• It helps determine the field extension generated by the element.
• It allows us to understand structures like splitting fields.

Every polynomial has a minimal polynomial if the element is algebraic over the field. This concept is particularly useful when dealing with field extensions and algebraic structures, because it simplifies the evaluation of properties related to the element in the algebraic context. For example, we can utilize the minimal polynomial to test polynomial divisibility or roots.

Injective Homomorphism

An injective homomorphism is a special kind of function between algebraic structures, like groups or fields, that is both a homomorphism and injective. A homomorphism means it preserves operations; for example, it will respect addition and multiplication if the context is fields or rings. Being injective means each element in the image is mapped from a unique element in the domain.

• When \( \rho: E \rightarrow E' \) is injective, \( \rho(a) = \rho(b) \) implies \( a = b \).
• Injective homomorphisms preserve distinctness, which is crucial for the integrity of algebraic structures.
• They also assure that the image under the homomorphism retains the structure's properties without collapsing it into fewer elements.

This property is vital when showing that two polynomials such as \( \phi' \) and \( \phi \) are equal if \( \rho \) is injective. Because no other distinct polynomial is squashed into a result that resembles \( \phi \), maintaining the degree and form is guaranteed. Thus, homomorphic maps keep these polynomial relationships intact without simplifying or losing important algebraic data.

Polynomial Divisibility

Polynomial divisibility involves determining if one polynomial can be divided by another, leaving no remainder. For polynomials \( f(x) \) and \( g(x) \), we say \( g(x) \mid f(x) \) if there exists another polynomial \( h(x) \) such that \( f(x) = g(x)h(x) \). This concept is important because it allows us to simplify and understand complex polynomial equations.

• It helps identify common roots and factors between polynomials.
• Factorization provides a clearer insight into the structure and roots of a polynomial.
• It plays a key role in solving polynomial equations and inequalities.

In the context of field theory and algebraic extensions, polynomial divisibility can illustrate the relationship between minimal polynomials. It shows how one root can complement or extend another, particularly under homomorphisms. In the considered exercise, demonstrating that the minimal polynomial \( \phi' \) divides \( \phi \) uses these principles to verify that homomorphic images are structurally consistent with their origin fields.