I have a problem set in which a question in its premise assumes that $<7,\sqrt{10}>$ the principal ideal generated by 7 and $\sqrt{10}$ is a maximal ideal of $\mathbb{Z}[\sqrt{10}]=\{a+b\sqrt{10}:a,b$ belong to $\mathbb{Z}\} $.
But I showed that $<7,\sqrt{10}>$ is equal to $\mathbb{Z}[\sqrt{10}]$. So it can't be maximal, since in the definition a maximal ideal must be a proper ideal.
My work: I showed that $<7,\sqrt{10}> = \{7a+10b+c\sqrt{10} : a,b, c\in \mathbb{Z} \} = \{s+t\sqrt{10} : t, s \in \mathbb{Z}\}=\mathbb{Z}[\sqrt{10}].$
I used the fact that 7 and 10 are relatively prime, so we can write any integer $n$ as a combination of them. Is my work false? And if it is in fact a maximal ideal, how I can show that?
Edit: in the question it asks me to show that $<7,\sqrt{10}>$ is maximal by considering the quotient ring.
Presumably a typo in the question: the ideal $\langle 7,\sqrt{10}\rangle$ is actually principal, but for the simple reason that it equals $\mathbb{Z}[\sqrt{10}]$, so it's definitely not maximal.
The question is probably about the principal ideal $\langle 7+\sqrt{10}\rangle$, which is a different beast.
Consider the homomorphism $\mathbb{Z}[X]\to\mathbb{Z}[\sqrt{10}]$ that maps $X$ to $\sqrt{10}$; its kernel is generated by the polynomial $X^2-14X-39$ and is so contained in the ideal generated by $X$ and $3$. So the idea might be to show that $3+\langle7+\sqrt{10}\rangle$ is not invertible in the quotient ring.
From $3(x+y\sqrt{10})=1+(u+v\sqrt{10})(7+\sqrt{10})$ we obtain $$ 3x=1+7u+10v,\qquad 3y=u+7v $$ Work modulo $3$ and find a contradiction.