### Some facts, better explained, from Atiyah-Macdonald

Today we shall discuss some interesting properties of elements of a ring.

1. If $a\in R$ is not a unit, then it is present in some maximal ideal of the ring $R$. Self-explanatory.

2. If $a$ is present in every maximal ideal, then $1+xa$ is a unit for all $x\in R$. Proof: Let $1+xa$ not be a unit. Then it is present in some maximal ideal (from 1). Let $1+xa=m$, where $m$ is an element from the maximal ideal $1+xa$ is a part of. Then $1=m-xa$. Hence, $1$ is also a member of the maximal ideal, which is absurd.

Let’s break down this theorem into elementary steps, and present a better proof (than given on pg.6 of “Commutative Algebra” by Atiyah-Macdonald). If $x\in M_1$ for some maximal ideal $M_1$, then $1\pm xy\notin M_1$ for all $y\in R$. Similarly, If $x\in M_2$ for some maximal ideal $M_2$, then $1\pm xy\notin M_2$ for all $y\in R$. This argument can then be extended to the fact that if $x\in$ all maximal ideals, then $1\pm xy\notin$ any maximal ideal for all $y\in R$. An element not there in any maximal ideal is a unit. Hence, $1\pm xy$ is a unit.

3. If $1-xy$ is a unit for all $y\in R$, then $x$ is part of every maximal ideal in $R$. Proof: Let is assume $x$ is not part of some maximal ideal. Then there exists some $m\in$ that maximal ideal such that $m+xy=1$. This implies that $m=1-xy$, which is impossible as $1-xy$ is a unit. The same argument can be used for $1+xy$.

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On pg.6 of Atiyah-Macdonald, it is mentioned that if $a,b,c$ are ideals, then $a\cap (b+c)=a\cap b+a\cap c$ if $b\subseteq a$ and $c\subseteq a$. It is not elaborated in the book, and the flippant reader may be confused. I’d like to elaborate on this concept. $a\cap (b+c)$ consists of those elements in $b+c$ which are also there in $a$. Now elements of $a\cap (b+c)=[(b+c)$ wrt those elements of both $b$ and $c$ that are in $a$] $\bigcup$ [ $(b+c)$ wrt those elements of $b$ that are in $a$ and those elements of $c$ that are not] $\bigcup$ [ $(b+c)$ wrt those elements of $b$ that are not in $a$ and those elements of $c$ that are] $\bigcup$ [ $(b+c)$ wrt those elements of both $b$ and $c$ that are not in $a$]

The second and the third terms are null sets, as can be easily seen.

The fourth term is NOT necessarily empty. However, it becomes an empty set if $b,c\subseteq a$. It may also become an empty set under other conditions which ensure that if both $b$ and $c$ are not in $a$, then $b+c\notin a$.

In summation, $a\cap (b+c)=a\cap b+a\cap c$ is definitely true when $b,c\subseteq a$. However, it is also true under other conditions which ensure that the fourth term is an empty set.

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I have extended a small paragraph in Atiyah-Macdonald to a full-fledged exposition: ( $a,b,c$ and $d$ are ideals in commutative ring $R$)

1. $a\cup (b+c)\subseteq(a\cup b)+(a\cup c)$– Both sides contain all elements of $a$ and $b+c$. Remember that $b\cup c\subseteq b+c$. However, the right hand side also contains elements of the form $a+b$ and $a+c$, which the left hand side does not contain.

2. $a\cap (b+c)$– This has already been explained above.

3. $a+(b\cup c)=(a+b)\cup (a+c)$– Both are exactly the same.

4. $a+ (b\cap c)\subseteq (a+b)\cap (a+c)$– There might be $b_1,c_1\notin b\cap c$ such that $a'+b_1=a''+c_1$. However, any element in $a+(b\cap c)$ will definitely be present in $(a+ b)\cap (a+c)$.

5. $a(b\cup c)\supseteq ab\cup ac$– LHS contains elements of the form $a'b_1+a''c_1$, which the RHS doesn’t. In fact, LHS is an ideal while the RHS isn’t. You might wonder how LHS is an ideal. I have just extended the algorithm used to make $AB$ an ideal when $A$ and $B$ are both ideals to situations in which $A$ is an ideal and $B$ is any subset of $R$.

6. $a(b\cap c)\subseteq ab\cap ac$– The RHS may contain elements of the form $a'b_1=a''c_1$ for $b_1,c_1\notin b\cap c$.

7. $a(b+c)=ab+ac$– Easy enough to see.

8. $(a+b)(c\cap d)=(c\cap d)a+(c\cap d)b\subseteq(ac\cap ad)+(bc\cap bd)$

From this formula, we have $(a+b)(a\cap b)\subseteq (a\cap ab)+(b\cap ab)\subseteq ab$.
This fact is mentioned on pg.7 of Atiyah-Macdonald.

9. $(a+b)(c\cup d)= (c\cup d)a+(c\cup d)b= (ca\cup da)+(cb\cup db)$.

From this formula, we have $(a+b)(a\cup b)=(a^2\cup ab)+(b^2\cup ab)\supseteq ab$.