Consider the ring $A=\mathbb{C}[x,y]/(y^2-x^4+5x-4)$, and consider the ideal $\mathfrak{m}=(y, x+2)$. Is $\mathfrak{m}$ maximal?
My sketched solution: consider an arbitrary element of $A$ and a representative $p(x,y)\in \mathbb{C}[x,y]$ for this element. Then $p(x,y)$ will be of the form $p(x,y)=y(\ldots)+g(x)$. Now by the Euclidean algorithm, there exist $q,r$ so that $g(x)=q(x)(x+2)+r(x)$ with $deg(r)<1$, i.e. $r(x)=c$ for some $c\in\mathbb{C}$. Summing up, $p(x,y)=y(\ldots)+(x+2)(\ldots)+c$ for some complex number $c$. Therefore there exist a surjective homorphism $\mathbb{C}\rightarrow A/\mathfrak{m}$, so $A/\mathfrak{m}\simeq\text{some quotient of $\mathbb{C}$}$. Clearly it must be $\mathbb{C}$ itself (since it's a field), so $A/\mathfrak{m}\simeq \mathbb{C}$ which implies that $\mathfrak{m}$ is maximal.
Is it a valid answer? Thanks in advance.
Edit: $A=\mathbb{C}[x,y]/(y^2-x^4+5x^2-4)$
Best Answer
As Stefan Walter pointed out, you need to show that the quotient is nonzero. To see that, notice that we have $$(y^2-x^4+5x^2-4,y,x+2)=(x^4-5x^2+4,y,x+2)=(y,x+2)$$ So $A/\mathfrak m\cong {\bf C}[x,y]/(x+2,y)\cong {\bf C}[x]/(x+2)\cong {\bf C}$. Since the quotient is a field, $\mathfrak m$ must be maximal.