-1

I want to prove that $\mathbb{R}[X]/(ax^2 + bx +c) \cong \mathbb{C}$ if $b^2 - 4ac < 0$. I belief I have all of the components needed for this exercise:

  • Knowing the first isomorphic theorem

  • $(2ax + b)^2 = b^2 - 4ac$

  • $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

But I just don't see what the bigger picture is

José Carlos Santos
  • 440,053
Omatikaya
  • 1
  • What did you try? Do you know that $\mathbb{R}[X]/(X^2+1)\simeq \mathbb{C}$? – Zanzi Oct 06 '17 at 08:38
  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made some effort. – José Carlos Santos Oct 06 '17 at 08:42
  • Yeah I know that $\mathbb{R}[X] /(X^2 + 1) \cong \mathbb{C}$. – Omatikaya Oct 06 '17 at 08:44
  • Notice that $b^2-4ac<0$ implies $a\neq 0$. Thus any element of $\mathbb{R}[X]/(ax^2+bx+c)$ has a unique representation as a polynomial of degree at most $1$ (by the division algorithm). It's easy to see that this space is a two-dimensional real vector space. You want to show this is isomorphic to $\mathbb{C}$ as rings. To which linear polynomial should you map $i$? – Mathematician 42 Oct 06 '17 at 08:48
  • Just a heads up, the equality $R[X]/(aX^2 + bX + C) = \big{ f \in \mathbb{R}[X] \mid aX^2 + bX+ C = 0\big}$ is not true! – Mathematician 42 Oct 06 '17 at 08:55
  • so by the division algorithm you'll get a polynomial of the form $r_0 + r_1X$ with $r_0, r_1 \in \mathbb{R}$ with the vector $(r_0, r_1)$. It might have something to do with the evaluation homomophism. With $i$ evaluated in this polynomial $r_0 + r_1X$. – Omatikaya Oct 06 '17 at 08:57
  • Look at the isomorphism given in José Carlos Santos' answer. If we go the other way, it's clear that $i$ gets mapped to the class of the polynomial $X$. – Mathematician 42 Oct 06 '17 at 09:09

2 Answers2

3

Define$$\begin{array}{rccc}f\colon&\mathbb{R}[X]&\longrightarrow&\mathbb C\\&P(X)&\longrightarrow&R\left(-\frac b{2a}+\frac{\sqrt{4ac-b^2}}{2a}i\right),\end{array}$$where $R(X)$ is the remainder of the division of $P(X)$ by $aX^2+bX+c$. Then $f$ is a ring homorphism which is surjective and whose kernel is $\langle aX^2+bX+c\rangle$.

José Carlos Santos
  • 440,053
  • Okay yeah that makes sence – Omatikaya Oct 06 '17 at 09:02
  • Could you actually explain the reasoning of the evaluated complex number in $R(X)$? – Omatikaya Oct 07 '17 at 18:10
  • @Omatikaya What do you mean? Do you want to know where does $-\frac b{2a}+\frac{\sqrt{ac-b^2}}{2a}i$ come from? – José Carlos Santos Oct 07 '17 at 18:13
  • yes, i understand that it has something to do with the zero of the function. – Omatikaya Oct 07 '17 at 18:17
  • @Omatikaya Oops! Now I see I made a typo: I should have written $4ac$ instead of $ac$. I've edited my answer. Was that your problem? – José Carlos Santos Oct 07 '17 at 18:20
  • Oh yeah. But what is the reasoning behind choosing the zero of the function $aX^2 + bX +c$ and evaluating it in R(X). – Omatikaya Oct 07 '17 at 18:24
  • @Omatikaya Note that $R\left(-\frac b{2a}+\frac{\sqrt{4ac-b^2}}{2a}i\right)=P\left(-\frac b{2a}+\frac{\sqrt{4ac-b^2}}{2a}i\right)$, precisely because $-\frac b{2a}+\frac{\sqrt{4ac-b^2}}{2a}i$ is a root of $aX^2+bX+c$. So,\begin{align}\ker f&=\left{Q(x)\in\mathbb{Q}[x],\middle|,Q\left(-\frac b{2a}+\frac{\sqrt{4ac-b^2}}{2a}i\right)=0\right}\&=\left{Q(x)\in\mathbb{Q}[x],\middle|,P(x)\mid Q(x)\right}\&=\bigl\langle P(x)\bigr\rangle.\end{align} – José Carlos Santos Oct 07 '17 at 18:35
  • Thank you very much for this answer! I completely understand it now. Thank you for your time! – Omatikaya Oct 07 '17 at 18:52
1

Recall that if $f$ is irreducible over a field $F$, then $F[x]/\langle f \rangle \cong F(\alpha)$, where $\alpha$ is any root of the polynomial. Since $b^2 - 4ac < 0$ for this polynomial $ax^2 + bx + c$ in $\mathbb{R}[x]$, its roots are imaginary, and the polynomial is irreducible over $\mathbb{R}$. From this, we have $\mathbb{R}[x]/\langle ax^2 + bx + c \rangle \cong \mathbb{R}(\alpha)$, where $\alpha$ is one of its roots.

Thus, the problem boils down to showing that $\mathbb{R}(\alpha) \cong \mathbb{C}$, where $\alpha = a + bi$ with $a, b \in \mathbb{R}$ and $b \neq 0$. By definition, $\mathbb{R}(\alpha) = \{f(\alpha) \ | \ f \in \mathbb{R}[x]\}$, from which it is clear that, at least, $\mathbb{R}(\alpha) \subseteq \mathbb{C}$. To show that these fields are actually equal, we want to demonstrate that we can produce any given complex number by finding a polynomial in $\mathbb{R}[x]$ and evaluating it at $\alpha$. I'll leave it to you to explicitly write down a (linear) polynomial that outputs the generic complex number $c + di$ when evaluated at $\alpha$.

Kaj Hansen
  • 33,511