Show that $R^* = R\setminus \mathfrak{m}.$

Question

Here is the question I want to answer:

A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$

$(a)$ Let $(R, \mathfrak{m})$ be a local ring. Show that $R^* = R\setminus \mathfrak{m}.$

Hint: The quotient $R/ \mathfrak{m}$ is a field.

Here is a trial which I feel that it is not correct specifically because I did not use the hint:

Let $R$ be a commutative ring and let $(R, \mathfrak{m})$ be a local ring, we want to show that $R^* = R\setminus \mathfrak{m}.$

Let $x \in R\setminus \mathfrak{m}$ s.t. $X \notin R^*.$ Then $\langle x \rangle$ is a non-zero proper ideal which is contained in a maximal ideal different from $M,$ which is a contradiction. Therefore $R^* = R\setminus \mathfrak{m}.$

So could anyone criticize this solution to me and tell me the correct and inaccurate/incorrect statements in it please?

Another trial which I feel like it is not written logically correct(specifically the last paragraph and how all the proof before it proves the required):

Suppose that $R$ is a local ring with maximal ideal $\mathfrak{m}$. Then $R/\mathfrak{m} = \{r + \mathfrak{m} \mid r \in R \}$ and the zero element of $R/\mathfrak{m}$ is $0 + \mathfrak{m} = \mathfrak{m}.$\

We will show that every non-zero element in $R / \mathfrak{m}$ has a multiplicative inverse in $R / \mathfrak{m}$ i.e. $R / \mathfrak{m}$ is a field.

Let $ r + \mathfrak{m} \neq \mathfrak{m}$ be a non-zero element in $R / \mathfrak{m}.$Then $r \notin \mathfrak{m}.$\

Clearly, $rR$ is an ideal of $R$ and $ \mathfrak{m} + rR$ is an ideal of $R$ containing $\mathfrak{m}$ properly.

Then $\mathfrak{m} \subset \mathfrak{m} + rR \subset R$ and since $\mathfrak{m}$ is maximal and $r \notin \mathfrak{m}$ then $\mathfrak{m} \neq \mathfrak{m} + rR $ and $\mathfrak{m} + rR = R.$\

Now, since $1 \in R,$ then $1 = m + rs$ for some $m \in \mathfrak{m}$ and $s \in R.$ So, $1 - rs = m \in \mathfrak{m} $ which means $1 - rs + \mathfrak{m} = \mathfrak{m}.$ Therefore, $$1 + \mathfrak{m} = rs + \mathfrak{m} = (r + \mathfrak{m})(s + \mathfrak{m})$$

Hence, for every $\mathfrak{m} \neq r + \mathfrak{m}$ in $R / \mathfrak{m},$ there exists $ s + \mathfrak{m}$ in $R / \mathfrak{m},$ such that $(r + \mathfrak{m})(s + \mathfrak{m}) = 1 + \mathfrak{m}$ where $1 + \mathfrak{m}$ is the multiplicative identity element of $R / \mathfrak{m}.$ And since $R$ is commutative then $$(r + \mathfrak{m})(s + \mathfrak{m}) = (s + \mathfrak{m})(r + \mathfrak{m}) = 1 + \mathfrak{m}$$

Hence, every non-zero element of $R / \mathfrak{m}$ is a unit and since the zero element of $R / \mathfrak{m}$ is $\mathfrak{m}$ then $R^* = R\setminus \mathfrak{m}.$

In this second proof I do not see if I really proved the required or not, so could anyone help me in proving the required if I did not do that or correct my mistakes if I was doing that?

A trivial question on the second proof, why $rR$ is clearly an ideal? could anyone explains that to me please?

In general, I would appreciate if I saw a smart, elegant, simple and logically correct proof of the required.

Answer 1

Given any commutative ring $A$, the relation $A \setminus \mathrm{U}(A)=\displaystyle\bigcup \mathscr{Max}(A)$ holds, where $\mathrm{U}(A)$ indicates the units of the ring and $\mathscr{Max}(A)$ the maximal spectrum, i.e. the collection of all maximal ideals.

On the one hand, it is elementary that any proper ideal must be disjoint from the subset of units, which applies in particular to maximal ideals. This justifies the inclusion $\displaystyle\bigcup\mathscr{Max}(A) \subseteq A \setminus \mathrm{U}(A)$.

Conversely, consider a nonunit $x \in A \setminus \mathrm{U}(A)$. The principal ideal $Ax$ will then be proper and thus included in a certain maximal ideal $M \in \mathscr{Max}(A)$ by virtue of Krull's theorem. This proves the converse inclusion $A \setminus \mathrm{U}(A) \subseteq \displaystyle\bigcup\mathscr{Max}(A)$.

When $A$ is local $\mathscr{Max}(A)$ is a singleton, and the above general relation reduces to precisely the claim you seek to prove.