# Is $\mathbb{R}[X]/(P)$ isomorphic to $\mathbb C$ for every irreducible polynomial $P$ of degree $2?$

Trying to solve a problem, I got stuck on the following question.

If $P\in\mathbb R[X]$ is an irreducible polynomial and $\operatorname {deg} P=2$, then is it true that $$\mathbb{R}[X]/(P)\cong\mathbb C?$$

I believe it must be true from the bits I've been hearing throughout my studies, but of which I haven't seen proofs. I've been unable to find anything on the internet other than this single page. It tells me that there is something called Weierstrass' theorem which should be relevant, but can't find even the precise statement of the theorem, let alone the proof.

Here's what I have tried.

First of all, I can assume that $P$ is monic, because for any polynomial of degree $2$ there's a monic polynomial generating the same ideal in $\mathbb R[X].$ So I have

$$P(x)=x^2+bx+c$$

for some real numbers $b,c$ such that $b^2<4c$.

I want to assume there is an isomorphism $\phi:\mathbb R[X]/(P)\to\mathbb C$ and work out what it should look like. I will denote by $[S]$ the coset of $S\in \mathbb R[X]$ in $\mathbb R[X]/(P).$ I'd like to find $S$ such that $\phi([S])=i.$

I have

$$[x^2+bx+c-1]\mapsto -1,$$

So the coset that is mapped to $-1$ has to be a square root of $[x^2+bx+c-1]$ in $\mathbb R[X]/(P).$ Such a coset must be represented by a linear polynomial $kx+l$. Thus $$(kx+l)^2-(x^2+bx+c-1)\in (P),$$

that is $$(kx+l)^2-(x^2+bx+c-1)\equiv a(x^2+bx+c)$$

for some real number $a$, and further

$$(k^2-1-a)x^2+(2kl-b-ab)x+l^2-c-ac+1\equiv 0.$$

This gives

$$\begin{cases} k^2=1+a\\ l^2=(1+a)c+1\\ 2kl=(1+a)b \end{cases}$$

The problem is that solving this gives formulas that are very difficult to handle. And I need to handle them if I want to prove that what I'm constructing is indeed an isomorphism. Is there a better way of doing it?

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Dear ymar: Are you supposed to know the fact that $\mathbb C$ is algebraically closed? – Pierre-Yves Gaillard Mar 3 '12 at 17:48
@Pierre-YvesGaillard I do know this fact from my course in analytic complex functions. It was proven there. – user23211 Mar 3 '12 at 18:07
The non-constructive proof is simple then: $\mathbb{C}$ is an extension of $\mathbb{R}[x]/(P)$, and $\mathbb{R}[x]/(P)$ is an extension of $\mathbb{R}$. Now, look at the degrees of the extensions (i.e. their dimension as $\mathbb{R}$-vector spaces). – Hurkyl Mar 3 '12 at 18:14
Dear ymar: Then you know that any proper algebraic extension of $\mathbb R$ is isomorphic to $\mathbb C$. – Pierre-Yves Gaillard Mar 3 '12 at 18:14
@Hurkyl I think I understand. If $\xi\in \mathbb C$ is a root of $P$, then $\mathbb R[X]/(P)\cong \mathbb R[\xi].$ I have $\mathbb R\subseteq\mathbb R[\xi]\subseteq\mathbb C,$ and since $\operatorname{deg}(P)=2$, it must be that $(\mathbb R[\xi]:\mathbb R)=2$. But also $(\mathbb C:\mathbb R)=2$, so $(\mathbb C:\mathbb R[\xi])=1.$ So I get $\mathbb C=\mathbb R[\xi]\cong \mathbb R[X]/(P).$ But I don't see where I used the fact that $\mathbb C$ is algebraically closed. – user23211 Mar 3 '12 at 18:40

Hint $\$ For an explicit isomorphism proceed as in the quadratic formula. Suppose R is an ordered field which is closed under taking sqrts of positive elements, and suppose that some polynomial $\rm\ a\:X^2 + b\:X + c \in R[X]\:$ of negative discriminant $\rm\:d = b^2 - 4ac < 0\:$ has a root $\rm\:x\:$ in some extension field $\rm\:F\supset R\:.$ Then $\rm\:-1\:$ is a square in $\rm\:F\:$ since $$\rm (b+2ax)^2 =\: b^2+4a(bx+ax^2) =\: b^2+4a(-c) =\: d < 0\ \ \Rightarrow\ \ \left(\frac{b+2ax}{\sqrt{-d}}\right)^2 =\: -1$$

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You're making working with the quotient ring harder than it needs to be. You're familiar with working with the usual presentation of the complex numbers $\mathbb{R}[i] / (i^2 + 1)$, right? Working in $\mathbb{R}[x] / (P)$ isn't much different; it's just $x^2$ reduces to something else.

It might help you to define a new variable $\xi$ to be the image of $x$ in $\mathbb{R}[x] / (P)$, and do arithmetic in terms of $\xi$.

Anyways, it's much, much easier to figure out the image $x$ in $\mathbb{C}$ than it is to find a preimage of $i$ in $\mathbb{R}[x] / (P)$. (the latter shouldn't be all that hard once you get used to arithmetic, though)

Hint: for it to be a homomorphism, you need to find a complex number α such that $\alpha^2 + b \alpha + c = 0$.

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Thank you very much! – user23211 Mar 3 '12 at 18:51

If $\mathbb R[X]/(P)$ injects into $\mathbb C$ (and the injection preserves $\mathbb R$), then $X$ must map to one of the complex roots of $P$ which you can find using the quadratic formula.

So if $p+qi$ is a root of $P$, then the residue class $[\frac{1}{q}X-\frac{p}{q}]$ ought to behave like $i$. Proving that is probably just a matter of inserting the quadratic formula (which splits nicely into real and imaginary terms here) and reducing.

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Thank you. I will try to complete this constructive proof and if I have trouble, I'll ask here again. – user23211 Mar 3 '12 at 18:51