Proposition 1.6 Atiyah

Question

Let $A$ be a ring and $\mathcal{m}$ a maximal ideal of $A$, such that every element of $1+\mathcal{m}$ is a unit in $A$. Then $A$ is a local ring.

Let $x\in A\setminus\mathcal{m}$ (It is not the quotient ring, but $A$ setminus $\mathcal{m}$). Since $m$ is maximal, the ideal generated by $x$ and $\mathcal{m}$ is $A=(1)$, that is $$(1)=(m,\{x\})$$

Edit $(m,\{x\})$ denotes the ideal generated by $\mathcal{m}$ and $x$.

Question. Why there exist $y\in A$ and $t\in\mathcal{m}$ such that $xy+t=1$?

@Nico I clarified what was asked in the question. Sorry if I was unclear. — , Jul 04 '21 at 09:33
Exact dupe of: Let $M$ be a maximal ideal in $R$ such that for all $x\in M$, $x+1$ is a unit. Show that $R$ is a local ring with maximal ideal $M$. Please search before posting questions. — Bill Dubuque, Jul 04 '21 at 09:42
By here the ideal in $,A,$ generated by $,M,x,$ is the ideal sum $,M+xA,,$ therefore $,1\in M+xA,\Rightarrow, 1=m+x:!a,$ for some $,m\in M,,a\in A.,$ See esp. the universal property of the ideal sum explained there, and its analogy with the gcd universal property. That thread should also be added to the dupe list. — Bill Dubuque, Jul 04 '21 at 10:39

score 3 · Answer 1 · answered Jul 04 '21 at 09:23

3

If $a\in A$, I denote $\bar a$ the class of $a$ in $A/\mathfrak m$. Let $x\in A\setminus \mathfrak m$. Since $\mathfrak m$ is maximal, $A/\mathfrak m$ is a field. Therefore there is $\bar y\in A/\mathfrak m$ s.t. $\bar x\bar y=1$, i.e. $1=(x+m_1)(y+m_2)=xy+t$ where $t=m_1y+m_2x+m_1m_2\in \mathfrak m$.

answered Jul 04 '21 at 09:23

joshua

1,001

Thanks for the answer, but $A\setminus\mathcal{m}$ is not the quotient ring, but $A$ setminus $\mathcal{m}$. – Jul 04 '21 at 09:32
@Nat.: I know... – joshua Jul 04 '21 at 09:37

score 1 · Accepted Answer · answered Jul 04 '21 at 09:34

1

In general, given a ring $A$, an element $x \in A$, and an ideal $I \subset A$, we have $$(\{x\}, I) = \{ax + t : a \in A,\, t \in I\}.$$ The above is easy to show. First, check that the set on the right is indeed an ideal. Second, show that any ideal containing $x$ and $I$ must contain the right set.

Apply it to your case by noting that $(\{x\}, \mathfrak{m}) = A \ni 1$.

answered Jul 04 '21 at 09:34

Aryaman Maithani

15,771

Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Jul 04 '21 at 09:42
2

OP's question was specifically about one particular step. The linked question is not necessary, @BillDubuque. – Aryaman Maithani Jul 04 '21 at 09:45
But the answer(s) include this step (and more, including conceptual motivation). – Bill Dubuque Jul 04 '21 at 09:50
Could you link the particular answer which does that? @BillDubuque – Aryaman Maithani Jul 04 '21 at 09:51

score 1 · Answer 3 · answered Jul 08 '21 at 10:49

Another solution I thought of: if you're familiar with $J=\operatorname{JacRad}(R)$, this is the radical who's equal to the intersection of all maximal ideals of $R$. One can easily prove that $x\in J\iff$ for every $y\in R$, $1+xy$ is a unit.

We prove $J=m$ in this particular case. The inclusion $J\subseteq m$ is straight from the definition of $J$. Now let $x\in m, y\in R$. From the definition of an ideal, $xy\in m$, and therefore $1+xy$ is a unit. This is true for all $y\in R$, so $J=m$.

Why does this imply that $R$ is a local ring? well assume for a contradiction that it's not. Then we have two maximal ideals, $m, m'$ s.t $m\not\subset m'$ and $m'\not\subset m$. By definition, $J\subseteq m\cap m'\subsetneq m$ - contradiction.

Proposition 1.6 Atiyah

3 Answers3