$$ \usepackage{amssymb} \newcommand{\N}{\mathbb{N}} \newcommand{\C}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\ZZ}{\ooalign{Z\cr\hidewidth\kern0.1em\raisebox{-0.5ex}{Z}\hidewidth\cr}} \newcommand{\colim}{\text{colim}} \newcommand{\weaktopo}{\tau_\text{weak}} \newcommand{\strongtopo}{\tau_\text{strong}} \newcommand{\normtopo}{\tau_\text{norm}} \newcommand{\green}[1]{\textcolor{ForestGreen}{#1}} \newcommand{\red}[1]{\textcolor{red}{#1}} \newcommand{\blue}[1]{\textcolor{blue}{#1}} \newcommand{\orange}[1]{\textcolor{orange}{#1}} \newcommand{\tr}{\text{tr}} \newcommand{\id}{\text{id}} \newcommand{\im}{\text{im}\>} \newcommand{\res}{\text{res}} \newcommand{\TopTwo}{\underline{\text{Top}^{(2)}}} \newcommand{\CW}[1]{\underline{#1\text{-CW}}} \newcommand{\ZZ}{% \ooalign{Z\cr\hidewidth\raisebox{-0.5ex}{Z}\hidewidth\cr}% } % specific for this document \newcommand{\cellOne}{\textcolor{green}{1}} \newcommand{\cellTwo}{\textcolor{red}{2}} \newcommand{\cellThree}{\textcolor{brown}{3}} \newcommand{\cellFour}{\textcolor{YellowOrange}{4}} $$

Prove that \(\mathbb{R}[x]/(x^2+bx+c)\) is isomorphic to \(\mathbb{R}^2\) or \(\mathbb{C}\) - A proof

math
algebra
abstract algebra
ring theory
meme
Author

Luca Leon Happel

Published

October 21, 2021

The exercise

Around one year ago I tried to solve the following riddle involving abstract algebra, but I failed and did not finish the question at that

Prove that Rx/(x^2+bx+c) is isomorphic to R^2 or C

The proof

Let \(f=x^2+bx+c\in\mathbb{R}[x]\). \(f\) can either have one double root, iff the discriminant \(\Delta f=b^2-4c\) is zero, or it can have two real roots iff \(\Delta f>0\), or it can have two complex roots iff \(\Delta f<0\).

Negative discriminant

If \(\Delta f>0\) we have \(f(x)\in\mathbb{R}\) with \(x\not\in\mathbb{R}\). We can rewrite \(f(x)=x^2+bx+c=0\) to be \(x^2+bx=-c\), and therefor \(x^2+bx\in\mathbb{R}\). Also, because \(\Delta f>0\), we know that that we can split \(f(x)\) into its linear factors \(f(x)=(x-w)(x-v)\) with \(w,v\in\mathbb{R}\). This means that our number \(x\) will be equal to \(0\) if we subtract a real number from it, meaning it must be real, thereofr concluding that \(\mathbb{R}/(x^2+bx+c)\cong \mathbb{R}\times \mathbb{R}\)

Positive discriminant

In case our discriminant \(\Delta f\) is positive we know that \(f(x)\) cannot be split into real linear monomial factors, which means that \(x\) cannot be a real number. Also, because \(f\) only has purely complex roots, \(x\) must be a purely complex number as well. Because \(\mathbb{R}[x]/(f)\) is a vectorspace over \(\mathbb{R}\) we know that \(\alpha x+\beta=i\) for \(\alpha, \beta\in \mathbb{R}\), concluding that \(\mathbb{R}[x]/(x^2+bx+c) \cong \mathbb{R}[i]=\mathbb{C}\)

Conclusion

Reading Bosch Algebra really helped me grasp some of the concepts like field extensions some more. I assume that this knowledge also took some time to ripen inside my brain, but after some time algebra finally makes more sense to me. Also I finished my exam about abstract algebra a few months ago! (While also studying for commutative algebra and already having finished my introductory course in algebraic geometry).