Can't understand a proof of maximal ideal

Question

The question is

Let $R = \{a+b\sqrt{2} \mid a,b \text{ integers}\}$. Let $M = \{a+b\sqrt{2} ~|~ 5|a \text{ and } 5|b\}$. Prove $M$ is a maximal ideal.

We already had the knowledge $M$ is an ideal. Here is the proof.

Pf: Let $N$ ideal of $R$, $M \subset N$, $N \not= M$.

Then $\forall r+s\sqrt{2} \in N, r,s\in \Bbb Z, 5\nmid r \lor 5 \nmid s$, let $t = (r+s\sqrt{2})(r-s\sqrt{2}) = r^2-2s^2 \in N$

$$\dots$$

So $5 \nmid t$.

$\color{red}{ut(\in N) + 5v(\in N) = 1 \text{ for some } u,v\in \Bbb Z, \text{ so } 1\in N \Rightarrow N=R.}$ Q.E.D.

I don't understand the last sentence. It seems not so surprising, but I can't think of why. Is it a theorem that an ideal containing $1$ is a maximal ideal? I did not find such thing in the book.

Answer 1

Generally, we can prove that ideals $M$ are maximal by showing that any strictly greater ideal $N \supsetneq M$. must contain 1.

Any ideal containing 1 must be the whole ring $R$, since ideals $I$ satisfy $ri\in I$ for all $r\in R, i \in I$. If we can set $i = 1$, then we see that $r\cdot 1 = r \in I$ for all $r\in R$.