Polynomial Ring with coefficients from another ring

I need to show that the set $\mathbb{R}[x]/((x^2+1)\mathbb{R}[x]) $ is a field. From the general properties of polynomial rings I think its easy to say that this set is an abelian group under polynomial addition and that its identity element is the zero polynomial but I'm struggling to see how it could have a multiplicative inverse for all of its elements. I know that $\mathbb{Z}/n\mathbb{Z}$ form a field when $n$ is a prime number as $gcd(a,n)=1 $ for all $a \in \mathbb{Z}/n\mathbb{Z}$ meaning that all elements of the set have a multiplicative inverse but I don't know how I can apply a similar logic to polynomials.

Solution 1:

Observe that $p:=x^2+1$ is irreducible in $\Bbb R[x]$ which is a principal ideal domain, so $p$ behaves exactly like a prime.
(Any proper factor of $x^2+1$ would be of degree one, but it has no real root.)

Euclid's algorithm for the greatest common divisor works the same way in $\Bbb R[x]$ and also provides the inverse for a polynomial $f$ modulo $p$ whenever $\gcd(p,f)=1$.

Alternatively, you can prove that $(x^2+1)\Bbb R[x]$ is not only a prime ideal, but also a maximal ideal.