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Let $p(x) = ax^2 + bx + c \in \mathbb{R}[x]$ be a degree 2 polynomial with real coefficients such that $D := b^2 - 4ac$. I'd like to examine the structure of the ring $\mathbb{R}[x]/(p(x))$ in the following three cases: $D > 0, D < 0,$ and $D = 0$. (Note that $(p(x))$ is the ideal generated by $p(x)$.)

Where I Am:

Well, if $D>0$ (I think I've got this case)...

...then $p(x)$ has two distinct (real) roots; call them $\alpha$ and $\beta$. Therefore, we can write $p(x) = (x - \alpha)(x - \beta)$. So, by the Chinese Remainder Theorem, we have that $$ \mathbb{R}[x]/(p(x)) \cong \mathbb{R}[x]/(x - \alpha) \times \mathbb{R}[x]/(x - \beta). $$ Now, we claim that $\mathbb{R}[x]/(x - \alpha) \cong \mathbb{R}$. Indeed, consider the homomorphism $\phi:\mathbb{R}[x] \to \mathbb{R}$ given by $\phi(q(x)) = q(\alpha)$. Then, since $$ \phi(q(x)) = q(\alpha) = 0 \iff q(x) \in (x - \alpha), $$ we see that $\ker(\phi) = (x - \alpha)$. Furthermore, since we may consider every real number to be a constant polynomial, we see that $\phi$ is surjective. Thus, it follows from the First Isomorphism Theorem that $$ \mathbb{R}[x]/\ker(\phi) = \mathbb{R}[x]/(x - \alpha) \cong \phi(\mathbb{R}[x]) = \mathbb{R}. $$ Similarly, $\mathbb{R}[x]/(x - \beta) \cong \mathbb{R}$. Therefore, $$ \mathbb{R}[x]/(p(x)) \cong \mathbb{R} \times \mathbb{R}. $$

Now, if $D < 0$...

...then $p(x)$ has no real roots. Thus, $p(x)$ is irreducible. So, $\mathbb{R}[x]/(p(x))$ is certainly a field of some sort. I know that $\mathbb{C} \cong \mathbb{R}[x]/(x^2+1)$; but is this true of any quadratic irreducible in $\mathbb{R}[x]$? I would think not... but I'm having trouble describing this ring any further than this. (Note that I don't know anything about "field extensions"...)

And, finally, if $D=0$...

...then $p(x)$ has a single (real) root of multiplicity two; call it $\gamma$. Therefore, we can write $p(x) = (x - \gamma)(x - \gamma)$. May I make the same conclusion here that I did from the first case? I don't see why not, but I just don't feel all that confident doing so...

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    $\begingroup$ In the final case, it suffices to take $\gamma=0$; then you get $\Bbb R[X]/(X^2)$ which is neither a field, nor a product of a field, as it has nilpotents. $\endgroup$ – Lord Shark the Unknown Aug 20 '18 at 20:08
  • $\begingroup$ @LordSharktheUnknown Hmm. Yes, that's true. I'll have to think about that more... $\endgroup$ – thisisourconcerndude Aug 20 '18 at 20:18
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For $D>0$, you're entirely right.

For $D<0$, you actually do get $\Bbb C$ no matter what. We see this if we use the isomorphism $\Bbb R[x]/(p(x))\to\Bbb C$ which maps $1$ to $1$ and $x$ to one of the complex roots of $p$ (you will, of course, need to show that this is well-defined and an isomorphism).

For $D=0$, you get what is called the dual numbers. Just like $\Bbb C$ has a standard representation as $\Bbb R[i]/(i^2+1)$, the dual numbers have a standard representation as $\Bbb R[\epsilon]/(\epsilon^2)$. Again, for any other $p$, you can construct an isomorphism. This time $\Bbb R[\epsilon]/(\epsilon^2)\to\Bbb R[x]/(p(x))$ by sending $1$ to $1$ and $\epsilon$ to one of the linear factors of $p$ (again, show that it is well-defined and an isomorphism).

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    $\begingroup$ I have no idea why it seemed more natural to let $\Bbb R[x]/(p(x))$ be the domain of the isomorphism in one case and the codomain in the other. Any attempt I have made to describe the inverse of each of these two maps quickly became cluttered. At least at this time of the evening. $\endgroup$ – Arthur Aug 20 '18 at 20:29
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    $\begingroup$ I will help you with the other way around. If $D<0$, then an isomorphism $\mathbb{C}\to\mathbb{R}/\langle p(x)\rangle$ sends $1\mapsto 1$ and $\text{i}\mapsto \dfrac{x-a}{b}$, if $a,b\in\mathbb{R}$ are such that $a+b\text{i}$ is a complex root of $p(x)$. If $D=0$, then an isomorphism $\mathbb{R}[\epsilon]/\langle \epsilon^2\rangle \to\mathbb{R}/\langle p(x)\rangle$ sends $1\mapsto 1$ and $\epsilon\mapsto x-a$ if $p(x)=(x-a)^2$. $\endgroup$ – Batominovski Aug 20 '18 at 21:03
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    $\begingroup$ If $D>0$, then the isomorphism $\mathbb{R}\times\mathbb{R}\to\mathbb{R}[x]/\langle p(x)\rangle$ can take $(1,0)$ to $\dfrac{x-a}{b-a}$ and $(0,1)$ to $\dfrac{b-x}{b-a}$, where $p(x)=(x-a)(x-b)$, whereas its inverse $\mathbb{R}[x]/\langle p(x)\rangle \to \mathbb{R}\times\mathbb{R}$ maps $q(x)+\langle p(x)\rangle$ to $\big(q(b),q(a)\big)$. $\endgroup$ – Batominovski Aug 20 '18 at 21:09
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    $\begingroup$ @Batominovski 1) For $D=0$, that's the one I already have. 2) For $D<0$, I said "send $x$ to $a+bi$, which is a root of $p(x)$", and it's easy enough to turn that around to get what you have, but that's just it: your isomorphism smells of being the inverse of mine. My point was that I have no idea why the "natural" direction is different for $D=0$ than for $D<0$. $\endgroup$ – Arthur Aug 20 '18 at 21:24
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    $\begingroup$ Ahh, I see your point. I am still tipsy from some beer, so I didn't realize fully what you meant. Anyway, I will leave the comments above undeleted, since they may have a bit of value. I guess it is not good to do math while drunk. $\endgroup$ – Batominovski Aug 20 '18 at 21:27

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