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I am starting to study rings. One of the first examples in my book about ring factors is

$$\mathbb{R}[x] / \langle x^2+1 \rangle = \{ ax + b + \langle x^2 +1 \rangle \mid a,b \in \mathbb{R} \}$$

I am wondering if

$$\mathbb{R}[x] / \langle x^2+1 \rangle = \mathbb{R}[x] / \langle x^2+2 \rangle\text{ ?}$$

and more generally

$$\mathbb{R}[x] / \langle x^2+1 \rangle = \mathbb{R}[x] / \langle a x^2+b x + c \rangle \, \, \forall (a,b,c) \in \mathbb{R}^3 \setminus \{(0,0,0) \}?$$

How can i show that they are equal? (or that they are not?)

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  • $\begingroup$ Both rings are isomorphic to $\mathbb C$. $\endgroup$ – Georges Elencwajg May 8 '14 at 18:24
  • $\begingroup$ Certainly not for arbitrary $a,b,c$, take $a=b=c=0$. $\endgroup$ – Dietrich Burde May 8 '14 at 18:25
  • $\begingroup$ Are all instance of $R[x]$ meant to be $\mathbb R[x]$? $\endgroup$ – Hagen von Eitzen May 8 '14 at 18:25
  • $\begingroup$ @HagenvonEitzen Yes, thanks. Corrected. $\endgroup$ – José D. May 8 '14 at 18:32
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    $\begingroup$ Note that if you replaced $\mathbb{R}$ by $\mathbb{Z}$ or even $\mathbb{Q}$, then the rings are not isomorphic since one ring has a square root of $-2$ and the other does not. $\endgroup$ – alex.jordan May 8 '14 at 19:32
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For any $u,v\in\mathbb R$ (i.e. for any $z=u+iv\in\mathbb C$), there is a ring homomorphism $f\colon \mathbb R[X]\to \mathbb C$ that is the identity on $\mathbb R$ and that maps $X\mapsto u+iv$. (This is the universal peroperty of the polynomial ring). If $v=0$, obviously the image is just $\mathbb R$, not all of $\mathbb C$. If $v\ne 0$, then $f$ is onto; in fact $f(\frac bvX-\frac {ub}v+a)=a+bi$. One verifies that $f(X^2)=u^2-v^2+2uvi=2uf(X)-(u^2+v^2)$ and hence $X^2-2uX+u^2+v^2$ is in the kernel of $f$. As $\mathbb R[X]$ is a principal ideal domain and no linear polynomial can be in the kernel, we conclude that the kernel is $\langle X^2-2uX+u^2+v^2\rangle$ and by the isomorphism theorems $$ \mathbb R[X]/\langle X^2-2uX+u^2+v^2\rangle\cong \mathbb C\qquad\text{if }v\ne0.$$ The discriminant of $X^2-2uX+u^2+v^2$ is simply $(-2u)^2-4(u^2+v^2)=-4v^2<0$. We note that for any quadratic polynomial $X^2+pX+q$ with discriminant $D=p^2-4q<0$ we can simply let $u=-\frac p2$ and $v=\frac12\sqrt{-D}$ and thus find that $$ \mathbb R[X]/\langle X^2+pX+q\rangle\cong \mathbb C\qquad\text{if }D=p^2-4q<0.$$


What is $\mathbb R[X]/\langle X^2+pX+q\rangle$ if the discriminant is nonnegative? If $D=0$, then $\epsilon:=X+\frac p2+\langle X^2+pX+q\rangle$ is a strange element: it is nonzero, but its square is $\epsilon^2=X^2+pX+\frac{p^2}4+\langle X^2+pX+q\rangle$, and that is just zero (because $q=\frac{p^2}4$). This ring is often just written $\mathbb R[\epsilon]$ with the mnemonic that $\epsilon$ is very small, but $\epsilon^2$ is so negligibly small that it really equals $0$. $$ \mathbb R[X]/\langle X^2+pX+q\rangle\cong \mathbb R[\epsilon]\qquad\text{if }D=p^2-4q=0.$$


If $D>0$, $X^2+pX+q$ has two distinct real roots $a,b$. Can you see what the ring looks like now? As the previous example, it has zero divisors, so at least it is definitely not isomorphic to $\mathbb C$.

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You want your polynomial to have two conjugate complex roots. Does this give you an idea on the isomorphism?

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  • $\begingroup$ No really, I have a lot to study, yet. But that means that the last equation of my question holds, right? (any 2nd order polynomial with real coefficients has two conjungate complex roots, right?) $\endgroup$ – José D. May 8 '14 at 18:37
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    $\begingroup$ @Trollkemada Not if the roots are real. In other words, you need to restrict your $a$, $b$, $c$ so that the discriminant is strictly negative. $\endgroup$ – Nicholas Stull May 8 '14 at 18:43
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They are not equally set theoretically, but are isomorphic as fields. As long as $ax^2+bx+c$ is an irreducible polynomial over $\mathbb{R}[x]$, then the ideal $\langle ax^2+bx+c\rangle$ is maximal, thus the quotient $\mathbb{R}[x]/\langle ax^2+bx+c\rangle$ is a field. In fact the quotient is isomorphic to $\mathbb{C}$ whenever the irreducible polynomial is degree 2, and is isomorphic to $\mathbb{R}$ whenever the irreducible polynomial is degree 1.

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Once you get to the isomorphism theorem (which should be very soon), this is fairly easy to prove. Unfortunately, this might not make a whole lot of sense right now, but it should be rather illuminating in a week or two. The isomorphism theorem says that, given a homomorphism $\phi:R \rightarrow S$, then:

$$R/\ker(\phi) \cong Im(\phi)$$

Typically for polynomial rings, the evaluation homomorphism $ev_a$ does the trick, which basically evaluates all polynomials in the ring at $a$. That is, $f(x) \mapsto f(a)$. You should prove that this is indeed a homomorphism.

So for the first, we are going to look at the evaluation homomorphism $ev_i:\mathbb{R}[x] \rightarrow \mathbb{C}$. Note that its kernel is $\langle x^2 + 1 \rangle$, since that is the minimal polynomial with $i$ as a root. Further, this is surjective onto $\mathbb{C}$ since $(ax + b) \mapsto (ai + b)$ for all $a, b \in \mathbb{R}$. So by the isomorphism theorem, we get $\mathbb{R}[x]/\langle x^2 + 1 \rangle \cong \mathbb{C}$.

Now let's look at the second. This time, we will consider the evaluation homomorphism $ev_{i\sqrt{2}}: \mathbb{R}[x] \rightarrow \mathbb{C}$. Note that the kernel is $\langle x^2 + 2 \rangle$, since that is the minimal polynomial with $i\sqrt{2}$ as a root. Again, this is surjective onto $\mathbb{C}$ since $(a\sqrt{2}x + b) \mapsto (ai + b)$ for all $a, b \in \mathbb{R}$. So by the isomorphism theorem, we get $\mathbb{R}[x]/\langle x^2 + 2 \rangle \cong \mathbb{C}$.

Thus, both quotient rings are isomorphic to $\mathbb{C}$, and so we conclude $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{R}[x]/\langle x^2+2 \rangle$.

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