Identifying $ \mathbb{R}[x]/(x^2+k) $ for $k<0$ and $k>0$ Suppose we have a quotient ring:
$$ \mathbb{R}[x]/(x^2+k) $$
How would I:


*

*Show the quotient ring is isomorphic to $\mathbb{C}$ for $k>0$

*Identify the ring by an isomorphism to some familiar ring (not defined in terms of quotients or cosets) for $k<0$


I'm not really sure how to go about either. Should I be explicitly constructing an isomorphism?
 A: For the first part you already know that you're trying to show that your ring $\Bbb R[x]/(x^2+k)$ is isomorphic to $\Bbb C$, so a natural way to do this is to construct an isomorphism by hand.  Two facts that you should consider:


*

*Most nice ring maps $f: \Bbb R[x] \to \Bbb C$ take $\Bbb R$ to the copy of $\Bbb R$ already existing in $\Bbb C$ (ie, $f(r) = r$ for all $r\in R$).  If that is the case, then the map is uniquely determined by choosing a value for $f(x)$.

*Given any ideal $I$ of $\Bbb R(x)$, such a ring map $f: \Bbb R[x] \to \Bbb C$ naturally gives rise to a well-defined map $\overline{f}: \Bbb R[x]/I \to \Bbb C$ with domain the quotient $\Bbb R(x)/I$ if and only if $f(y) = 0$ for all $y\in I$.  Here the relevant ideal $(x^2+k)$ is given by polynomial  multiples of $x^2+k$.
The second part is more difficult, since you need to determine identity of the desired 'familiar ring'.  Some hints:


*

*If $k<0$ is written as $k = -c^2$, then $x^2+k= x^2-c^2$ factors as $(x-c)(x+c)$.  This polynomial factorization also lets you factor the ideal $(x^2-c^2)$ as the product of the two ideals $(x-c)$ and $(x+c)$.

*If two ideals $I_1$ and $I_2$ are coprime, then the general form of the Chinese Remainder Theorem shows that $\Bbb R[x]/I_1I_2$ is isomorphic to the product $\Bbb R[x]/I_1 \times \Bbb R[x]/I_2$.

A: If $k>0$, define $f:\mathbb{R}[x]\rightarrow \mathbb{C}$ by $f([x])=\sqrt k i$, $f(1)=1$ where $[x]$ is the class of the polynomial $x$ in $\mathbb{R}[x]/(x^2+k)$.
If $k=-c^2$, then $x^2-k=x-c^2=(x+c)(x-c)$, $\mathbb{R}/(x-c)(x+c)$ is isomorphic to $\mathbb{R}/(x-c)\times \mathbb{R}/(x+c)=\mathbb{R}\times \mathbb{R}$.
A: If $k>0$ then $x^2 + k$ has no roots in $\mathbb{R}$. 
Let $\beta \in \mathbb{C}$ be one root of the (irreducible over $\mathbb{R}$) polynomial $x^2 + k$, therefore the map
$\varphi : \mathbb{R}[X] \longrightarrow \mathbb{C}$ defined by $\varphi(f(x)) = f(\beta)$ is a surjective homomorphism (check it) with kernel $Ker(\varphi) = (x^2+k)$ (to see this just use the fact that $x^2 + k$ generates a maximal ideal), therefore by the First Isomorphism Theorem we get that
$\dfrac{\mathbb{R}[X]}{(x^2+k)} \simeq Im(\varphi) = \mathbb{C}$
Another way: $\dfrac{\mathbb{R}[X]}{(x^2+k)} $ ia quadratic (field) extension of $\mathbb{R}$, therefore it must be isomorphic to $\mathbb{C}$.
