$\mathbb R[x]/\langle x^2+1\rangle$ is a field How to show that $\mathbb R[x]/\langle x^2+1\rangle$ is a field.  
I wrote the representation of $\mathbb R[x]/\langle x^2+1\rangle$ =$\{a+bx+\langle x^2+1\rangle\big|a,b\in \mathbb R\}$. 
Now how should I prove this...  
kindly help.
 A: Any polynomial in the ring can be reduced to a first degree polynomial. So a general element is of the form $a + bx + \langle x^2 + 1\rangle$, or $a + bx$ for short ($a, b \in \Bbb R$).
I want to show that any non-zero element has an inverse. So, assume that $a + bx$ and $c + dx$ are inverses of one another. I want to solve for $c$ and $d$, showing that no matter what $a$ and $b$ are, as long as they are not both zero, there is an inverse to $a + bx$.
We begin by expanding the relation between $a + bx$ and $c +dx$:
$$
1 = (a + bx)(c + dx) = ac + (ad + bc)x + bdx^2\\
= ac - bd + (ad + bc)x
$$
Since this is an equality of polynomials, we must have
$$
\cases{
1 = ac - bd\\0 = ad + bc
}
$$
If $a = 0$, this has the solution $d = -1/b$ and $ c = 0$. If $b = 0$, we have the solution $d = 0$ and $c = 1/a$.
Now, assume that both $a$ and $b$ are different from zero. The lower equation gives $d = -\frac{bc}{a}$. This makes the top equation give
$$
1 = ac - bd\\
1 = ac + \frac{b^2c}{a}\\
\frac{1}{c} = a + \frac{b^2}{a}\\
c = \frac{1}{a + \frac{b^2}{a}} = \frac{a}{a^2 + b^2}
$$
which exists because neither $a$ nor $b$ is equal to $0$, and $\Bbb R$ is a field. This also gives
$$
d = -\frac{b}{a^2 + b^2}
$$
and that's why $\Bbb R[x]/\langle x^2 + 1\rangle$ is a field.
A: Depending on how much theory you know, it can be shown in a few lines:
If $R$ is a ring, $R/I$ is a field if and only if the ideal $I\subseteq R$ is maximal. For polynomials, the ideal $\langle f\rangle$ is maximal if and only if $f$ is an irreducible polynomial in $R$, and since $\deg(f)=2$, it is irreducible if and only if it does not have any roots.
The polynomial $x^2+1$ has no real roots, so $\mathbb{R}/\langle x^2+1\rangle$ must be a field.
A: $\mathbb R[x]  $ is an euclidean domain and $x^2+1\ is\ irriducible$ then the ideal $(x^2+1)$ is prime then the ideal is maximal then the quotient is a field
A: What's important here is that polynomial $x^2+1$ is irreducible in $R[x]$. Irreducibility is all you need.
Indeed, take any non-zero element $g(x) + \langle x^2+1\rangle \in R[x]/\langle x^2+1\rangle$. Since this element is non-zero in $R[x]/\langle x^2+1\rangle$, $g(x)$ isn't divisible by $x^2+1$. Since $x^2+1$ is irreducible, it follows that $gcd(g(x), x^2+1) = 1$. Then there exist polynomials $a(x), b(x) \in R[x]$ such that $1 = a(x)g(x) + b(x)(x^2+1)$. Observe that the following holds in the quotient ring:
$$
\begin{align*}
(a(x) + \langle x^2+1\rangle)\cdot(g(x) + \langle x^2+1\rangle) &= a(x)g(x) + \langle x^2 + 1\rangle \\
 &= 1 + \langle x^2+1\rangle.
\end{align*}
$$
So, our element $g(x) + \langle x^2 + 1\rangle$ has an inverse, therefore $R[x]/\langle x^2 + 1 \rangle$ is a field, QED.
PS: indeed, it's not even important that we're dealing with polynomials. We could have used a general fact like this: if $K$ is a PID, and $g \in K$ is an irreducible element, then $K/\langle g \rangle$ is a field.
