Show that if K sube F is an extension of degree k, every element alpha in F has a minimal polynomial over K of degree <=k. Find theminimal polymials of complex numers over RR explicitly to verify this fact

Josalynn 2020-11-09 Answered

Show that if \(\displaystyle{K}\subseteq{F}\) is an extension of degree k, every element \(\displaystyle\alpha\in{F}\) has a minimal polynomial over K of degree <=k. Find theminimal polymials of complex numers over \(\displaystyle\mathbb{R}\) explicitly to verify this fact

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nitruraviX
Answered 2020-11-10 Author has 26531 answers
To show that the degree of the minimal polynomial of any complex number over R is of degree at most 2.
This is the problem statement.
Now, \(\displaystyle\mathbb{C}=\mathbb{R}{\left({i}\right)}={\left\lbrace{a}+{i}{b}:{a},{b}\in\mathbb{R}\right\rbrace}.\)
So, \(\displaystyle{d}{e}{g{{\left(\frac{\mathbb{C}}{\mathbb{R}}\right)}}}={2}\), (with I, i as a basis).
To show that if \(\displaystyle{z}\in\mathbb{C}\), then z satisfies a polynomial f(z) with coefficients in \(\displaystyle\mathbb{R}\) with deg(f)
Proof of the statement. To start with any z satisfies a polynomial equation of degree less than or equal to 2.
Let \(\displaystyle{z}={a}+{i}{b},{a},{b}\in\mathbb{R}\)
\(\displaystyle\Rightarrow{\left({z}-{a}\right)}-{i}{b}\Rightarrow{\left({z}-{a}\right)}^{{2}}=-{b}^{{2}}\Rightarrow{z}^{{2}}-{2}{a}{z}+{\left({a}^{{2}}+{b}^{{2}}\right)}={0}\)
Completing the proof
If z=a (b=0) is real, the minimal polynomial is z-a, deg=1
If z=a+ib, with \(\displaystyle{b}\ne{0}\), then \(\displaystyle{z}\notin\mathbb{R}\), and the minimal polynomial is \(\displaystyle{f{{\left({z}\right)}}}={z}^{{2}}-{2}{a}{z}+{\left({a}^{{2}}+{b}^{{2}}\right)}\), (deg=2).
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