hI PJPu17. The ring \(\displaystyle{R}\) is commutative. We observe that
\(\displaystyle{x^2+y^21=x\cdot x+(y+1)\cdot (y1)\in\langle{x,y1\rangle}}\), so,
\(\displaystyle{\langle{x^2+y^21\rangle}\leq \langle{x,y1\rangle}\leq \mathbb{K}[x,y]}\)
and according to the 3rd Ring Isomorphism Theorem, we get
\(\displaystyle{\left(\mathbb{K}[x,y]/\langle{x^2+y^21\rangle}\right)/\langle{x,y1\rangle}/\langle{x^2+y^21\rangle}\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}}\)
or equivalently,
\(\displaystyle{R/\mu\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}}\) as rings.
Now, the map \(\displaystyle{\Phi:\mathbb{K}[x,y]\to \mathbb{K}\,,f(x,y)\mapsto f(0,1)}\)
is a ring homomorphism, which is onto \(\displaystyle{\mathbb{K}}\) and satisfies \(\displaystyle{\rm{Ker}(\Phi)=\langle{x,y1\rangle}}\)
Then,
\(\displaystyle{R/\mu\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}=\mathbb{K}[x,y]/\rm{Ker}(\Phi)\cong Im(\Phi)=\mathbb{K}}\)
that is, the ideal \(\displaystyle{\mu}\) is maximal. If \(\displaystyle{\mu}\) is a free \(\displaystyle{R}\) module
of rank \(\displaystyle{1}\), then \(\displaystyle{R\cong \mu}\) as \(\displaystyle{R}\) modules
and more specifically, \(\displaystyle{R=\mu\,\,\land\,\,\mu\subset R\implies \mu=R}\),
a contradiction, since \(\displaystyle{\mu}\) is a maximal ideal of \(\displaystyle{R}\).
