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

If 𝜌 is injective, are 𝜑 and 𝜑' equal? Answer: Yes, the minimal polynomial 𝜑' divides the minimal polynomial 𝜑. In addition, if 𝜌 is injective, then 𝜑' and 𝜑 are equal.

Step-by-step solution

Recall the definitions of homomorphism, minimal polynomial, and injective function

We need to recall these concepts as they will be used throughout the solution: 1. A homomorphism \(\rho: E \rightarrow E^{\prime}\) is a function that preserves the algebraic structure, meaning that: - \(\rho(x+y) = \rho(x) + \rho(y)\) and - \(\rho(xy) = \rho(x)\rho(y)\) for all \(x, y \in E\). 2. The minimal polynomial \(\phi\) of \(\alpha\) over a field \(F\) is the monic polynomial of smallest degree that has coefficients in \(F\) and \(\phi(\alpha) = 0.\) 3. An injective function (also called one-to-one) is a function that maps distinct elements of its domain to distinct elements of its codomain.

Apply the homomorphism to the minimal polynomial

Applying the homomorphism \(\rho\) to the equation \(\phi(\alpha) = 0\), we get: $$\rho(\phi(\alpha)) = \rho(0)$$ Using the properties of homomorphism, we can write: $$\phi(\rho(\alpha)) = \rho(0) = 0$$ This shows that \(\rho(\alpha)\) is a root of \(\phi\).

Show that \(\phi^{\prime}\) divides \(\phi\)

From Step 2, we've shown that \(\phi(\rho(\alpha)) = 0\). Thus, \(\rho(\alpha)\) is a root of the polynomial \(\phi\). We know that \(\phi^{\prime}\) is the minimal polynomial of \(\rho(\alpha)\) over \(F\), meaning it is the polynomial of the smallest degree with coefficients in \(F\) such that \(\phi^{\prime}(\rho(\alpha)) = 0\). Since \(\rho(\alpha)\) is a root of \(\phi\), the polynomial \(\phi\) can be expressed as the product of \(\phi^{\prime}\) and a polynomial \(g\): $$\phi = \phi^{\prime}g$$ Therefore, \(\phi^{\prime}\) divides \(\phi\).

Show that \(\phi^{\prime} = \phi\) if \(\rho\) is injective

Suppose that \(\rho\) is injective. If \(\phi^{\prime}\) strictly divides \(\phi\), meaning that \(g\) is not a constant polynomial, then \(g(\alpha) = 0\) since \(\phi(\alpha) = 0\). Now, let's consider the equation \(\phi^{\prime}(\rho(\alpha)) = 0\). Since \(\rho\) is injective, \(\rho(\alpha_1) = \rho(\alpha_2)\) implies \(\alpha_1 = \alpha_2\). Thus, the minimal polynomial \(\phi^{\prime}\) of \(\rho(\alpha)\) must have the same degree as the minimal polynomial \(\phi\) of \(\alpha\). In other words, the degrees of \(\phi\) and \(\phi^{\prime}\) must be equal, which implies that \(g\) must be a constant polynomial. Consequently, \(\phi^{\prime} = \phi\) when \(\rho\) is injective. We have now shown that \(\phi^{\prime}\) divides \(\phi\), and \(\phi^{\prime} = \phi\) if \(\rho\) is injective, as desired.