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Zeros of a Polynomial Function 📂Abstract Algebra

Zeros of a Polynomial Function

Definition 1

f(x):=k=0nakxk=a0+a1x++anxn f(x) : = \sum_{k=0}^{n} a_{k} x^{k} = a_{0} + a_{1} x + \cdots + a_{n} x^{n} Let us define the evaluation function ϕα:F[x]E\phi_{\alpha} : F [ x ] \to E in αE\alpha \in E for a polynomial function fF[x]f \in F [x] and a field FEF \le E as follows. ϕα(f(x)):=a0+a1α++anαn=f(α) \phi_{\alpha} ( f(x) ) : = a_{0} + a_{1} \alpha + \cdots + a_{n} \alpha^n = f (\alpha) f(α)=0f( \alpha ) = 0 satisfying in αE\alpha \in E is called the zero f(x)f(x) of f(x)f(x).

Explanation

Evaluation Function

As a fact, ϕα\phi_{\alpha} becomes a homomorphism.

If the definition feels too vague, consider the simple example of ϕi:R[x]C\phi_{i} : \mathbb{R} [ x ] \to \mathbb{C}. For instance, if we have ϕi(x23x+2)\phi_{i} ( x^2 - 3 x + 2), it simply becomes f(x)=x23x+2 f(x) = x^2 - 3 x + 2 by substituting ii into i23i+2=1i2C i^2 - 3 i + 2= 1 -i2 \in \mathbb{C} Of course, it does not matter if RC\mathbb{R} \le \mathbb{C}.

Motive of Zero

Thinking of kernel, ker(ϕα)\ker ( \phi_{\alpha} ) would be a set of functions satisfying f(α)=0f(\alpha) = 0. Continuing with the above example, elements of ker(ϕi)R[x]\ker ( \phi_{i} ) \subset \mathbb{R} [ x ] would be polynomial functions that have (xi)(x-i) as a factor.

Thus, it is undoubtedly natural to call α\alpha the zero of f(x)f(x) when satisfying f(α)=0f( \alpha ) = 0. Equivalently, if ϕα(f(x))=0\phi_{\alpha} ( f(x) ) = 0 then α\alpha is said to be the zero of f(x)f(x).

The reason for explaining the concepts of substitution and roots by involving functions is precisely to define the concept of ‘solution of equation’ rigorously. f(x)=g(x) f(x) = g(x) Consider the equation above, for instance. There’s no reason not to think of a set of such equations, but it’s much more straightforward to consider f(x)f(x) and g(x)g(x) separately rather than collecting relations. If a set of equations XX is a collection of such equations, it should be able to be expressed as (f(x)=g(x))X \left( f(x) = g(x) \right) \in X But after all, since reducing it comes down to f(x)=g(x)    f(x)g(x)=0 f(x) = g(x) \iff f(x) - g(x) = 0 there’s no need to keep the right side messy and free. Having the set XX follow a structure that holds functions rather than equations, and having interest in when they hold, is no different than collecting equations with 00 on the right side.

This expansion of thought will lead to facts such as ‘polynomial functions with real coefficients can have imaginary roots’ being abstracted and generalized.


  1. Fraleigh. (2003). A first course in abstract algebra(7th Edition): p201, 204. ↩︎