Is the following algebraic structure a field? Is this a field?
\begin{array}{l}{(M,+, \cdot) \text { with } M=\{m+n \sqrt{5}: m, n \in \mathbb{Z}\} \text { and }+\text { and } \cdot \text { is the usual addition and}} \\ {\text { multiplication of numbers in } \mathbb{R}. }\end{array}
I found out, that the inverse element of $(M,\cdot)$ for $a$ is $a^{-1}=\frac{1}{m+n\sqrt{5}}$ but that is not in $M$, because there are no numbers $n,m$ to get $\frac{1}{m+n\sqrt{5}}$, right? (It wouldn't be a field then.)
 A: Here is one way to do the problem.
$M$ is contained in the field $\mathbb R$. The inverse to $n + m \sqrt 5$ in $\mathbb R$ has the form $a + b \sqrt 5$ for some rational numbers $a,b \in \mathbb R$. Compute these numbers and see $a$ and $b$ are not integers. 
Now suppose $n + m \sqrt 5$ has an inverse $n' + m' \sqrt 5$ in $M$. Then it is also  an inverse in $\mathbb R$. Since inverses in a field are unique we have $n' + m' \sqrt 5 = a + b \sqrt 5$. 
Exercise: Suppose $a,b,c,d$ are rational and $a + b \sqrt 5 = c + d \sqrt 5$. Prove $a=c$ and $b=d$.
The above implies $n'=a$ and $m'=b$. This contradicts how $n',m'$ are integers and $a,b$ are not. We conclude there is no inverse in $M$.
Edit: Of course Robert's answer proves $M$ is not a field with much less work. The above shows something much stronger. Namely that $1,-1$ are the ONLY invertible elements of $M$.
A: Existence of inverse for non-zero elements is required for $M$ to be a field, so you are correct.
For a counter-example it's usually more clear to pick one counter-example (e.g. $2\in M$ but $\frac{1}{2}\notin M$), but your reasoning is correct.
