Finite Dimensional Division Algebras - Bresar Lemma 1.1

[Math Amateur]

I am reading Matej Bresar's book, "Introduction to Noncommutative Algebra" and am currently focussed on Chapter 1: Finite Dimensional Division Algebras ... ...

I need help with the an aspect of the proof of Lemma 1.1 ... ...

Lemma 1.1 reads as follows:
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In the above text, at the start of the proof of Lemma 1.1, Bresar writes the following:" ... ... Since the dimension of $$D$$ is $$n$$, the elements $$1, x, \ ... \ ... \ , x^n$$ are linearly dependent. This means that there exists a non-zero polynomial $$f( \omega ) \in \mathbb{R} [ \omega ]$$ of degree at most $$n$$ such that $$f(x) = 0$$ ... ... "My question is as follows:

How exactly (rigorously and formally) does the elements $$1, x, \ ... \ ... \ , x^n$$ being linearly dependent allow us to conclude that there exists a non-zero polynomial $$f( \omega ) \in \mathbb{R} [ \omega ] $$ of degree at most $$n$$ such that $$f(x) = 0$$ ... ?Help will be much appreciated ...

Peter

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In order for readers of the above post to appreciate the context of the post I am providing pages 1-2 of Bresar ... as follows ...https://www.physicsforums.com/attachments/6193
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[Opalg]

My question is as follows:

How exactly (rigorously and formally) does the elements $$1, x, \ ... \ ... \ , x^n$$ being linearly dependent allow us to conclude that there exists a non-zero polynomial $$f( \omega ) \in \mathbb{R} [ \omega ] $$ of degree at most $$n$$ such that $$f(x) = 0$$ ... ?

The definition of dimension tells you that the maximum number of elements in a linearly independent set is $n$. The set $\{1,x,\ldots,x^n\}$ contains $n+1$ elements and must therefore be linearly dependent.

Now go back to the definition of linear dependence for the set $\{1,x,\ldots,x^n\}$. This says that there exist real numbers $\alpha_0,\alpha_1,\ldots,\alpha_n$, not all zero, such that $\alpha_01 + \alpha_1x + \alpha_2x^2 + \ldots + \alpha_nx^n = 0.$

Define the polynomial $f(\omega)$ by $f(\omega) = \alpha_0 + \alpha_1\omega + \alpha_2\omega^2 + \ldots + \alpha_n\omega^n.$ This polynomial is nonzero (because not all its coefficients are zero), and clearly $f(x) = 0.$

 

[Math Amateur]

The definition of dimension tells you that the maximum number of elements in a linearly independent set is $n$. The set $\{1,x,\ldots,x^n\}$ contains $n+1$ elements and must therefore be linearly dependent.

Now go back to the definition of linear dependence for the set $\{1,x,\ldots,x^n\}$. This says that there exist real numbers $\alpha_0,\alpha_1,\ldots,\alpha_n$, not all zero, such that $\alpha_01 + \alpha_1x + \alpha_2x^2 + \ldots + \alpha_nx^n = 0.$

Define the polynomial $f(\omega)$ by $f(\omega) = \alpha_0 + \alpha_1\omega + \alpha_2\omega^2 + \ldots + \alpha_n\omega^n.$ This polynomial is nonzero (because not all its coefficients are zero), and clearly $f(x) = 0.$

Thanks Opalg ... very clear and very helpful ...

Appreciate your help ...

Peter