Every element of $\mathbb{R}[x,y]/\langle x,x^2+y^2+1 \rangle$ is unit Show that every (no zero) element in the ring $R=\mathbb{R}[x,y]/\langle x,x^2+y^2+1 \rangle$ is a unit.
Answer: In my mind came the 1st isomorphism theorem: to find a field $\mathbb{K}$ such that $\mathbb{R}[x,y]/\langle x,x^2+y^2+1 \rangle \cong \mathbb{K}$. Possibly $\mathbb{K=C}$. But it is difficult to find an epimorphism $φ$ with kernel the ideal $I=\langle x,x^2+y^2+1 \rangle$.
Another idea is to use the definition; $$ \forall f(x,y)+I \in R,\exists g(x,y)+I  \in R-\{0_R \}: (f(x,y)+I)(g(x,y)+I)=1+I$$
so $f(x,y)g(x,y)+I=1+I \iff f(x,y)g(x,y)-1 \in I$.
But both ideas are difficult to use. 
 A: $$R=\frac{\mathbb{R}[x,y]}{\big\langle x,x^2+y^2+1\big\rangle}=\frac{\mathbb{R}[x,y]}{\big\langle x,y^2+1\big\rangle}\cong\frac{\mathbb{R}[x,y]\big/\langle x\rangle}{\big\langle x,y^2+1\big\rangle\big/\langle x\rangle}\cong\frac{\mathbb{R}[y]}{\big\langle y^2+1\big\rangle}\cong\mathbb{C}\,.$$
A: Since $\mathbb{R}[x,y]/\langle x,x^2+y^2+1 \rangle$ is a finitely generated $\mathbb{R}$-algebra, it being a field means that it is isomorphic to either $\mathbb{R}$ or $\mathbb{C}$. The former possibility is unlikely, so we can try with $\mathbb{C}$. Consider the unique ring homomorphism $\varphi\colon\mathbb{R}[x,y]\to\mathbb{C}$ such that $\varphi(r)=r$ for $r\in\mathbb{R}$, $\varphi(x)=0$ and $\varphi(y)=i$.
It is clear that $\varphi$ is surjective, $x\in\ker\varphi$ and $y^2+1\in\ker\varphi$. Therefore also $x^2+y^2+1\in\ker\varphi$.
Suppose $f(x,y)\in\ker\varphi$; write it as
$$
f(x,y)=f_0(y)+f_1(y)x+\dots+f_n(y)x^n
$$
Since $x\in\ker\varphi$, we deduce that $f_0(y)\in\ker\varphi$, in particular $f_0(i)=0$. Thus $f_0(y)$ has $i$ as root and so also $-i$. Hence $y^2+1$ divides $f_0(y)$.
As a consequence, $f(x,y)\in\langle x,y^2+1\rangle$.
It is obvious that $\langle x,y^2+1\rangle=\langle x,x^2+y^2+1\rangle$.
By the homomorphism theorem
$$
\mathbb{R}[x,y]/\langle x,x^2+y^2+1 \rangle=
\mathbb{R}[x,y]/\ker\varphi\cong\mathbb{C}
$$
A: Let me say something about how you can approach this by finding an epimorphism $\varphi:\mathbb{R}[x,y]\to\mathbb{C}$ with kernel $I$, since there is a nice general technique for these sorts of problems that is not obvious.  As suggested in the comments, you can define $\varphi$ by $\varphi(f)=f(0,i)$.  Then it is easy to see that $\ker(\varphi)$ contains $I$.  The tricky part is showing that $\ker(\varphi)$ is no larger than $I$.  That is, we want to prove that the induced homomorphism $\bar\varphi:\mathbb{R}[x,y]/I\to\mathbb{C}$ is injective.
The trick to prove a statement like this is to take any element of $\mathbb{R}[x,y]$ and "reduce" it to an element of $\mathbb{C}$ using only elements of $I$.  Here's how it works rigorously.  Suppose $f\in \mathbb{R}[x,y]$, and let $\bar{f}$ denote the image of $f$ in the quotient $\mathbb{R}[x,y]/I$.  We can separate $f$ into its monomials which contain $x$, and its monomials which do not contain $x$.  All the monomials which contain $x$ are multiples of $x$ and hence in $I$.  This means that $\bar{f}=\bar{f_0}$, where $f_0$ is the sum of the monomials in $f$ that don't contain $x$.
Now $f_0$ is a polynomial only in $y$.  Since $y^2+1\in I$ so $\overline{y^2}=-1$ in $\mathbb{R}[x,y]/I$, we can reduce each term of $f_0$ to contain no power of $y$ greater than $1$, by repeatedly replacing $y^2$ with $-1$.  In this way, we obtain a polynomial $f_1$ of the form $a+by$ with $a,b\in\mathbb{R}$ such that $\bar{f_0}=\bar{f_1}$.
Thus we have shown that $\bar{f}=\bar{f_1}$ for some polynomial $f_1$ of the form $a+by$.  Now suppose $\bar{f}$ were in the kernel of $\bar{\varphi}$.  This means $$0=\bar{\varphi}(\bar{f})=\bar{\varphi}(\bar{f_1})=\varphi(f_1)=a+bi.$$  This can only happen if $a=b=0$, so $f_1=0$ and hence $\bar{f}=0$.  Thus the kernel of $\bar{\varphi}$ is trivial.
