Prove that $\mathbb{R}[x]/ I$ is a field Let $f(x) = x^2 + 1 ∈ \Bbb R[x]$ and let $ I = \{ \,s(x)f(x) \mid s(x) ∈ \Bbb R[x] \,\} $.
(i) Prove that $ \mathbb {R}[x]/ I $ is a field, i.e., every non-zero element of $ \Bbb R[x]/I $ has an inverse.
(ii) Find the dimension of $ \mathbb{R}[x]/ I $.
I'm struggling to prove this where $f(x)$ is an element of the reals. When $f(x)$ is an element of the $\mathbb {Z}_2[x]$, we can prove by contradiction that one element does not have an inverse, namely $x + 1 + I$.  
The second part I believe I have an answer, but I'm not positive it is correct. I believe that the dimension is two and the elements are $x+1$ and $x+I$. 
Any help would be greatly appreciated!
 A: I understand you are starting an algebra course and studying concepts from the scratch, so what you are doing is practising with ideal quotients:
1) Pick any polynomial $f\in\mathbb{R}[x]$, and divide it by $x^2+1$ to get $$f(x)=q(x)(x^2+1)+r(x).$$
Observe that $$f(x)-r(x)=q(x)(x^2+1)$$ is in $I$, so the equivalence class of $f$ is the same as the equivalence class of $r$, and $r$ has degree at most 1, i.e., $r(x)=ax+b$ for some $a,b\in\mathbb{R}$. Note that, since we have total freedom when choosing $f$, we can get all linear polynomials (of the form $ax+b$) inside $I$. Moreover, not two of them belong to the same equivalence class, since if $ax+b-(cx+d)=g(x)(x^2+1)$ then it must be $g(x)=0$ and $cx+d=ax+b$ by a degree argument. What we have proved is that the equivalence classes of $I$ are in one-to-one correspondence with linear polynomials. Addition in $I$ is just addition of linear polynomials. What happens with multiplication? Well, it is the usual multiplication, but having into account that $x^2+1=0$, i.e., that $x^2=-1$, so that $$(ax+b)(cx+d) = acx^2+(ad+bc)x+bd =$$ $$= ac(-1)+(ad+bc)x+bd = (ad+bc)x+(bd-ac)$$
(we obtain a linear polynomial, as expected).
Note: If you stare long enough the multiplication rule above, you may discover you already know the ring $\mathbb{R}[x]/I$.
Now let us show that every nonzero element of $\mathbb{R}[x]/I$ has an inverse: given $ax+b$, its inverse must be of the form $cx+d$. By the multiplication rule above,  we need $ad+bc=0$ in order for the degree 1 term to dissapear, and $bd-ac=1$ in order to get the inverse. Remember that $a,b$ are fixed and $c,d$ are unknowns. This may seem a bit difficult to solve (although you can do it), so I will take a smarter road: if we remember the useful trick $(a+b)(a-b)=a^2-b^2$, then we find that $$(ax+b)(ax-b)=(ax)^2-b^2 = a^2x^2-b^2 = a^2(-1)-b^2 = -(a^2+b^2).$$
The number $-(a^2+b^2)\in\mathbb{R}$ is not zero unless $a=0=b$. So if $ax+b\neq 0$ then we can divide by $-(a^2+b^2)$ and get
$$(ax+b)\frac{ax-b}{-(a^2+b^2)} = 1,$$
i.e., we have found an inverse for every nonzero element of $\mathbb{R}[x]/I$.
2) I understand you want the dimension of $\mathbb{R}[x]/I$ as an $\mathbb{R}$-vector space. Well, by the exposition above, we already know that $\{1,x\}$ is a generator system for $\mathbb{R}[x]/I$, so it is only natural that we try to show that it is also linearly independent and so it forms a basis: Suppose on the contrary that there exist $0\neq a,b\in\mathbb{R}$ such that $ax+b=0$. Then we have two different linear polynomials with the same equivalence class (namely, $ax+b$ and $0x+0$), something we proved impossible in point 1). Therefore $\{1,x\}$ is a basis and $\dim_{\mathbb{R}}\mathbb{R}[x]/I = 2$.
Question: Have you already guessed which ring $\mathbb{R}[x]/I$ actually is?
A: There are a few ways to do this:


*

*Note that the ideal generated is maximal, since $x^2+1$ is irreducible over $\mathbb R$ (if it factors nontrivially, then it factors into linear parts, what are the possibilities for this?) and hence the quotient is a field

*Another way (still using irreducibility) is to use the fact that the division algorithm tells you that any polynomial $p(x)$ of degree $\geq 2$ can be rewritten as $p(x)+(x^2+1)q(x)+r(x)$ where $r(x)$ is linear, and so all elements in the quotient ring are of the form $\{a+bx\}$. Furrthermore, let $\alpha \in \mathbb C$ be a root of $x^2+1$.
Consider the field extension $\mathbb R[\alpha]=\{a+b\alpha \mid a,b \in \mathbb R\}$. We also know that the evaluation homomorphism 
$$\mathbb R[X]/I \to E $$
given by $X \mapsto \alpha$
has trivial kernel, so it is an isomorphism.
Now, we need to verify that $\mathbb R[\alpha]$ is a field, which amounts to finding an inverse.
If you know Bezout's identity for polynomials (it follows from the division algorithm: Since $x^2+1$ is irreducible, $gcd(x^2+1,p(x))=1$, so there exist polynomials $m(x),n(x)$ so that
$$1=m(x)(x^2+1)+n(x)p(x).$$
Letting $x \mapsto \alpha$ we have solutions for $n(\alpha)p(\alpha)=1$, and in particular we see that $a+b\alpha=p(\alpha)$ has an inverse.
But, we can also do this explicitly. Under the assumption that $a+b\alpha$ is nonzero, we have that
$(a+b \alpha)(c+d \alpha)=1 \iff ac-bd=1, bc+ad=0 $
using these equations, we can deduce that
$$c=a/(a^2+b^2),\,\, d=-b/(a^2+b^2).$$
If the previous calculation looked familiar, then this should lead naturally to perhaps the clearest way to see that it is a field 


*The field is in fact well known, and has appeared (disguised) somewhere in this answer.

