## 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

## 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).