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## Tag: commutative algebra

### Nakayama’s lemma

The Nakayama lemma as a concept is present throughout Commutative Algebra. And truth be told, learning it is not easy. The proof contains a small trick that is deceptively simple, but throws off many people. Also, it is easy to dismiss this lemma as unimportant. But as one would surely find out later, this would be an error in judgement. I am going to discuss this theorem and its proof in detail.

The statement of the theorem, as stated in Matsumura, is:

Let $I$ be an ideal in $R$, and $M$ be a finitely generated module over $R$. If $IM=M$, then there exists $r\in R$ such that $r\equiv 1\mod I$, and $rM=0$.

What does this statement even mean? Why is it so important? Why are the conditions given this way? Are these conditions necessary conditions? These are some questions that we can ask. We will try and discuss as many of them as we can.

$M$ is probably finitely generated so that we can generate a matrix, which by definiton has to be finite dimensional. Where the matrix comes in will become clear when we discuss the proof. What does $IM=M$ imply? This is a highly unusual situation. For instance, if $M=\Bbb{Z}$ and $I=(2)$, then $(2)\Bbb{Z}\neq\Bbb{Z}$. I can’t think of examples in which $I\neq (1)$, and $IM=M$. However, that does not mean that there do not exist any. What does it mean for $r\equiv 1\mod I$? It just means that $r=1+i$ for some $i\in I$. That was fairly simple! Now let’s get on with the proof.

Let $M$ be generated by the elements $\{a_1,a_2,\dots,a_n\}$. If $IM=M$, then for each generator $a_i$, we have $a_i=b_{i1}a_1+b_{i2}a_2+\dots+b_{in}a_n$, where all the $b_{ij}\in I$. We then have $b_{i1}a_1+b_{i2}a_2+\dots+(b_{ii}-1)a_i+\dots+b_{in}a_n=0$. Let us now create a matrix of these $n$ equations in the natural way, in which the rows are indexed by the $i$‘s. The determinant of this matrix will be $0$, as for any column vector that we multiply this matrix with, we will get $0$. On expanding this determinant, we will get an expression of the form $(-1)^n+ i$, where $i\in I$. If $n$ is odd, then just multiply the expression by $-1$. In either case, you get $1+i'$, where $i\in I$ ($i'=i$ or $i'=-i$).

Now as $1+i'$ is $0$, we have $(1+i')M=0$. Hence, $r=1+i'$ such that $r\equiv 1\mod I$ and $rM=0$

The reason why the proof is generally slightly confusing is that it is done more generally. It is first assume that there exists a morphism $\phi:M\to M$ such that $\phi(M)\subset IM$. Cayley-Hamilton is then used to give a determinant in terms of $\phi$, and then it is assumed that $\phi=1$. Here I have directly assumed that $\phi=1$, which made matters much simpler.

### Atiyah-Macdonald Part II

Let me start by first talking about the proof of the fact that the intersection of all prime ideals in a ring is the nilradical. The proof is constructive, and hence perhaps non-trivial. I will attempt to generalize the methodology of the proof.

This proof deals with creating a sort of border around the number objects that satisfy a certain property. Then it takes two external objects (that clearly don’t satisfy the property), and gets a third object. Now this third object may or may not lie inside the order. It is easily proven that it does not. Hence, _theorem_

Also, the mechanism of the proof could be used to prove that the maximal ideal containing all odd numbers is prime. The maximal ideal containing all even numbers is prime. The maximal ideal containing all powers of 2 is prime, etc.

12. Let $a\in R$ be an idempotent element. As $(a)$ is a proper ideal of $R$, it is contained within $M$, which is the unique maximal ideal. Hence, as $a$ is part of the Jacobson ring, $1-a$ is a unit. Now $(1-a)=(1-a^2)$. This implies that $a=0$, which is obviously a contradiction.

15. i) Easy

ii) Easy

iii) It is easy to prove. Note that this is not true only for the set of **prime** ideals containing $\{\bigcup E_i\}$. Let $W(A)$ be the set of ideals (any ideal) containing the set $A$. Then $W(\bigcup E_i)=\bigcap W(E_i)$. This is an important generalization.

iv) I have proved this in an earlier post. For what kinds of ideals do we have $V(a\cap b)=V(ab)$? Thinking about these things is what will give your thinking depth. The set of ideals which satisfy the following property: if $m^2\in I\implies m\in I$. Prime ideals and even radical ideals are vastly specialized examples of this very general condition.

For what kinds of ideals do we have $V(a)\cup V(b)=V(ab)$? Ideals which exhibit the following property: if $a\cap b\subseteq I$, then $a\subseteq I$ or $b\subseteq I$. Prime ideals are just one of the many types of ideals which may potentially fulfil this property. In fact, one may name a new class of ideals which are defined in such a way so as to fulfil this property.

Also, what kinds of sets mimic the behaviour of closed sets? Infinite intersections should be closed, finite unions should be closed, $\emptyset$ and the whole space $X$ (arbitrary name) should be closed, etc.

We see that $V(E)$ exhibits the properties of a closed set. How do we ensure that the infinite union of such closed sets is not closed? Refer to this question I asked on stackexchange- http://math.stackexchange.com/questions/846020/a-question-about-the-zariski-topology/846034#846034. The argument does not work for an infinite number of ideals because….well let me first give the argument for a finite number of ideals. Assume we have the ideals $a_1,a_2,\dots,a_n$, and $V(a_1a_2\dots a_n)$ does not contain $a_1,a_2,\dots,a_{n-1}$. Then it must contain $a_n$. Now if we have an infinite number of ideals, can we claim that $V$ does not contain all but one ideal? Yes. But can we prove that the ideal left aside is indeed contained within $V$? For that we need to deal with infinite products of elements. Such products are not defined. Hence, this argument does not translate to the infinite case.

Also note that the infinite union of closed sets may indeed be closed! Take the union of $[a_k,a_{k+1}]$, where $a_i$ is the $i^{th}$ natural number. However, the infinite union need not be closed. Take closed sets of the form $[\frac{1}{2^k},\frac{1}{2^{k-1}}]$ for example.

17. i) $X_f\cup X_g=(V(f)\cup V(g))'=(V(fg))'=X_{fg}$

A little word about **why** $V(f)\cup V(g)=V(fg)$ is true. If a prime ideal contains $fg$, and does not contain $f$, it HAS to contain $g$! This is also true if $f$ and $g$ are ideals, as mentioned in an earlier problem.

ii) If every prime ideal contains $f$, then $f$ has to be nilpotent. If $f$ is not nilpotent, there is a prime ideal which does not contain it, as the intersection of all prime ideals is the nilradical.

iii) If no prime ideal contains $f$, then $f$ is a unit. This is because if $f$ is not a unit, then $(f)$ is a proper ideal, and hence contained inside some maximal ideal. Every such maximal ideal is prime.

A little detour here. Why is every maximal ideal prime? Let $ab\in M$, where $M$ is maximal, and $a\notin M$. Then $as+m=1$, where $m\in M$ and $s\in R$ ($R$ is the commutative ring under consideration). Now multiply by $b$ to get $abs+mb=b$. Note that $abs$ and $mb$, both are in $M$, and $M$ is closed under addition (like every other ideal). Hence, $b\in M$.

iv) If $X_f=X_g$, then $V(f)=V(g)$. Now every ideal (not just prime ideals) containing $f$ will also contain $(f)$. Also, every prime ideal containing $(f)$ will also contain $r((f))$. Is that all? Will the intersection not contain any other element? Yes, we’re sure that the intersection will not contain any other element. Which brings me to my proof of this fact (not mentioned in Atiyah-Macdonald).

Let us consider an ideal containing $(f)$, which does not contain any power of $a$. Is this set nonempty? No, $(f)$ is one such ideal. Let us assume that the intersection of all prime ideals containing $f$ contains an element $a\notin r(f)$. It is easy to check that Zorn’s lemma can be used to create a maximum ideal not containing any power of $a$. It is also easy to prove that this maximum ideal is prime. Hence there is a prime ideal which does not contain $a$.

Now if $r((f))=r((g))$, then every prime ideal containing $f$ (and hence containing $r((f))$) will contain $r((g))$, and hence $g$. Hence $V(f)\subseteq V(g)$. We may similarly prove $V(g)\subseteq V(f)$.

v) The first thing to note here is that $V(f,g)=V(f)\cap V(g)$. Also: let $\{X_f,X_g,X_h,\dots\}$ be an open covering of $X$. Then $V(f)\cap V(g)\cap V(h)\cap\dots$ is empty. Note that the infinite intersection of closed sets is again a closed set. Also, that closed set is $V(f,g,h,\dots)$. If there is no prime ideal which contains ALL these elements, then there must be a finite number of them which add up to give $1$. Let those elements be $a_1,a_2,a_3,\dots a_n$. Then $X_{a_1},X_{a_2},\dots X_{a_n}$ also covers $X$.

vi) Let $X_f$ have a covering $\{X_m,X_n,\dots\}$. Then $V(m)\cap V(n)\cap\dots$ contains no ideal from $X_f$. Hence every ideal it contains contains $f$. In other words, $V(m,n,\dots)$ contains $f$. If that is true, then a finite number of them must add up to form $f$. Hence, the intersection of a finite number of $V(a)$ will lie completely inside $V(f)$. Taking their complements will give a finite cover of $X_f$.

What really is the gist of this argument? When we’re talking about an open set $X_f$, we’re really just talking about its complement, or the prime ideals which contain $f$. When we’re talking about an open cover, we’re really just talking about the intersection of the prime ideals in the complement. Hence, when one is a subset of another, we have a useful result. All prime ideals which contain these many elements also contain $f$…something like that. At the end of the day, all we’re doing is talking about various prime ideals containing certain elements, and what we can extract from that.

What if we had an open set of the form $X_f\cup X_g$? Then the complement is $V(f,g)$. If $X_f\cup X_g$ has an open cover, then there are finite expressions in $V(f,g)$ that are equal to $f$ and $g$. Let $a_1+a_2+\dots a_m=f$ and $b_1+b_2+\dots b_m=g$. Then taking taking complement of $V(a_1,a_2,\dots a_m,b_1,b_2,\dots b_n)$ gives an open cover of $X_f\cup X_g$.

vii)

18. i) The set is closed if it contains all ideals containing a certain element, say $f$. If $x$ was not maximal, then we could have made it bigger, and that bigger ideal would still contain $f$.

ii) This is by definition. It can be rather misguiding that there is no topological “proof” for this. For example, $\overline\{a,b,c\}=V(a,b,c)$ for any $a,b,c$ in commutative ring $R$.

iii) $\overline{\{x\}}$ is the set of all prime ideals containing $p_x$. Hence if $y\in\overline{\{x\}}$, then $p_y$ must contain $p_x$.

iv) Let us take two prime ideals $m$ and $n$. Also, suppose that $X_m$ contains $X_n$. This implies that $n$

### Solutions to some problems from Chapter 1 in Atiyah-Macdonald. And more importantly, insights and generalizations

If $x\in J$, where $J$ is the Jacobian ideal of a ring $R$, then $1-xy$ is a unit for all $y\in R$.

$1-xy$ looks like a particularly arbitrary expression. Let us break it down. $xy$ for all $y\in R$ is an alternate representation for $xR$. All maximal ideals contain $xR$. Now let us suppose that $1-xy$ is not a unit. Then $(1-xy)$ is contained in SOME maximal ideal (possibly not all). Let one of the maximal ideals $1-xy$ is contained in be $M$. Now $M$ also contains $xR$! Hence, that maximal ideal contains $1$, which is a contradiction.

Generalizing this, $a+xy$ should be a unit for any unit $a$ and for all $y\in R$.

Now the converse. Prove that $(1-xy)$ is a unit- $x$ belongs to the Jacobson radical. Suppose there is a maximal ideal that $x$ does not belong to. Let that ideal be $T$. Then $\exists t\in T$ such that $xm+tr=1$. There $1-xm=tr$. Now according to the condition stated above, $tr$ should be a unit. However, if that were true, $T$ wouldn’t be a maximal ideal. Hence contradiction. Actually this is true for the general expression $a-xy$, where $a$ is any unit in commutative ring $A$.

Proofs are for the verification of mathematical fact. They are not necessarily intuitive. They rarely offer deep glimpses into mathematical structures. How to “see” that $1-xy$ is a unit $\implies x\in J$? Assume that $I$ is an ideal of $R$ which does not contain $x$. Then $x$ can be added to it to create a bigger ideal. Hence it is contained in all maximal ideals. How can one be sure that the addition of $x$ will not create a unit within the ideal? It is not like the addition of $1$ is the only thing that can create a unit. There are very many ways of creating a unit- by the addition of ANY element in the ideal. Hence, the fact that $1-xy$ is a unit for ALL $y$ is an important condition for $x$ to be present in all maximal ideals. I unfortunately cannot offer a way to “see” this fact as of now. I can only stress on relevant facts.

Now I move on to solving problems.

1. It is easy to show that $(1+x)$ is a unit if $x$ is nilpotent. Is there a generalization? Yes. That is the next question. Any power of a unit is also a unit, like any power of a nilpotent is nilpotent. What about the product of a unit and nilpotent? In a commutative ring, it will obviously be nilpotent.

Let us generalize this. Any power of a nilpotent is a nilpotent. Any power of a unit is a unit. The product of a unit and nilpotent is nilpotent. The sum of a unit and nilpotent is a unit. Looking at things in some generality always makes me feel better.

2. i) Let $f(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$. If it is a unit, then there exists a polynomial $g(x)=b_0+\dots + b_mx^m$ such that $f(x)g(x)=1$ for all $x$. The condition for this is $a_0b_0=1$ and $a_nb_m=0,a_{n-1}b_m+a_nb_{m-1}=0$, etc. I’ve tried hard to prove it through a method other than the one mentioned in the book, but failed. I am used to only solving linear equations through substitution. Is this not substitution? No. Assuming that the linear equations are accurate, we can do a variety of things. We could have added them up and equated the sum to $1$. We could have multiplied all of the equations and equated the product to $0$. Substitution is just one of the many things we could have done. I need a broader perspective on solving linear equations.

Also note that $b_0$ is a unit too and $b_1,b_2,\dots b_m$ are nilpotent.

Another thing to note is that if $a+b$ is a unit, and $b$ is nilpotent, then $a$ is a unit. This is easy to prove. Let $a+b=c$, where $c$ is a unit. Then $a=c-b$, which is a unit by problem 1.

Proving the converse is easy. Use multinomial theorem.

ii) Similar to i.

iii) and iv) Done

3. Assume that all the polynomials are in $x_1$. If $f$ is a unit, all coefficients except for the constant are nilpotent. All those coefficients are polynomials in $\{x_2,x_3,\dots,x_n\}$. Proceed in this manner. This stuff is easy!

4. The general trick is to take an element $f$ of the Jacobson ideal, assume it is not nilpotent, take the unit $1-fy$, and put $y=x$. It is easy to see that $y$ can be ANY multiple of $x$. Also, $f$ need not be a polynomial. It can also be a constant element of the ring. This trick works even then.
In what kinds of rings are the Jacobson radical and the nilradical equal? It is known that the nilradical is always a subset of the Jacobson radical. It is difficult to comment as of now when the Jacobson ideal is in general equal to the nilradical. Here $A[x]$ is a particular ring with very specific properties for nilpotents. Other rings may have similar properties for nilpotents. What can be said is that for the properties that $1-fy$ must possess to be a unit for all $y$, those properties must make $f$ a nilpotent element, amongst other things that different values of $y$ may signify. Hence the Jacobson radial is likely to be a proper superset of the nilradical in most rings, as these are pretty specialized conditions.

5. i) Let the inverse of $f$ be $g$. Equate coefficients to $0$, and find suitable coefficients for $g$. Now you may wonder why this does not work for the finite degree case (problem 2). This is because after a finite number of steps (determining coefficients of $g$), you arrive upon a contradiction. Here, you can construct the power series $g$ from the coefficients determined, and see that multiplication happens perfectly, giving $1$. In the case of finite degree polynomials, if you construct the polynomial with coefficients derived in this manner, you will meet a contradiction.

ii) Mostly hinges on the fact that if $a,b$ are nilpotents, then so is $b-a$. There’s not a lot of coefficient hunting here. Only inductive reasoning. The converse does not HAVE to be true, going by the argument I put forward in problem 2. But I will still have to check Chapter 7 Exercise 2.

iii) Easy

6. Well I owe my solution to this link- http://asgarli.wordpress.com/2013/04/22/idempotent-elements-outside-the-nilradical/. However, I think my solution is slightly more straightforward. Let us assume that $J\neq N$, where $J$ is the Jacobson radical and $N$ is the nilradical. Then there exists an idempotent element in $J$. Let that element be $e$. Then $1-ey$ is a unit for all $y\in R$. In particular, $1-e^2=(1+e)(1-e)=1-e$ is a unit. We get $1+e=1$. Hence $e=0$, which is a contradiction as $e$ is assumed to be non-zero. Hence the Jacobson radical cannot contain any element apart from nilpotent elements.

7. Let $P$ be a prime ideal. Assume $P$ is not maximal. Then there exists $a\in R$ such that $(P,a)$ is not equal to $R$. Now $a^n=a$ for some $n\in\Bbb{N}$. Hence $a^n-a=0$. Therefore $a(a^{n-1}-a)=0$. If we take $R/P$, we still have $(a+P)(a^{n-1}-1+P)=0$. Seeing as this is an integral domain, and knowing that $a\neq 0$, we get $a^{n-1}=1+p$, where $p\in P$. Now follow closely: when we add $a$ to $P$, we also add $a^{n-1}$ to it. Hence, we’re effectively adding $1$ to $P$. This is a contradiction.

Alternate proofs can be found elsewhere. A common proof is to say that $R/P$ is maximal, as $x$ has $x^{n-2}$ is the multiplicative inverse of $x$. I just wanted to add a slightly different flavour to it.

8. Zorn’s lemma. A clever chain argument is used here. Try for yourself.

9. This is equivalent to proving that the intersection of all prime ideals containing $a$ is $r(a)$. Atiyah-Macdonald gives a concise proof: Consider the ring $R/a$. The inverses of all prime ideals containing $a$ will be prime ideals containing $a$ (inverse obviously of a mapping from $R$ to $R/a$). The nilradical will be the intersection of all prime ideals in $R/a$. Check that the inverse of this nilradical will indeed include the whole of $r(a)$.

10. $i)\to ii)$ Solve for yourself. Remember that every maximal ideal is prime.

$i)\to iii)$ The lone prime ideal is the maximal ideal, which is also equal to the nilradical.

$ii)\to i)$ Let us take the nilradical as the starting set. Every prime ideal can be built on it. Every prime ideal has to contain it. Draw pictures to give yourself a visual understanding of this!! However, we cannot build a bigger prime ideal, starting out from the nilradical, as all external elements are units! Hence there is only one prime ideal.

11. i) $2x=x+x=2x^2=2x^2+2x$. Hence, $2x=0$. How many ways are there of solving such a question? What are the kind of strategies one should opt for? Is hit and miss the only way? In general, we can construct $nx=(x+x+\dots+x)=(x^2+x^2+\dots x^2)=(x+x)^2+x^2+x^2+\dots+x^2=\dots=(x+x+\dots+x)^2$

ii) Proving every prime ideal is maximal is easy. Refer to problem 7. However, we notice that in a lot of the last problems we’ve had prime ideals equal to maximal ideals. What is the general case? When are prime ideals equal to maximal ideals? Problem 7 says one possibility is when for all $x\in R$, $x^n=x$ for some $n\in\Bbb{N}$. The $n$ doesn’t have to be the same for different elements. But what if $x^n=2x$? or $3x$? Does it still work? It does not always work. BUT IT WORKS FOR $x^n=ax$, WHERE $a$ IS A UNIT!! Check if for yourself. Here is the generalization we were waiting for!!

Proving that $R/p$ is a field with two elements is easy if you walk down the same path as mentioned above. For $a\in R/p$, we have $a^2-a=0$. Hence $a(a-1)=0$. This shows that either $a\in p$ or $a\in a+p$.

Note: Constructing ring modulo prime ideal seems to be a source of many useful discoveries. When is this technique useful? It is useful when we’re considering elements with respect to the prime ideal- namely whether they’re inside or outside it. It is unlikely to be useful in a lot of other places.

iii) This involves the creation of a generator. One option is $x+y+xy$. Are there other generators?

More importantly, in what kinds of cases is every finitely generated ideal principal? One possibility is when the Euclidean algorithm is valid, and it is possible to determine the gcd of the generators. Another is when rings have special relations, which allow the creation of generators by multiplication with a single element. For example, if the ring had the relation that $xy=0$, and $x^2=x,y^2=y$, then $\langle x,y\rangle=\langle x+y\rangle$. If the ring had a relation $x^2=x,y^2=y,nxy=0$, then $\langle x,y\rangle=\langle x+y+(n-1)xy\rangle$

### Localization of a ring

So what exactly is the localization of a ring?

It is creating a field-imitation (and NOT necessarily a field with multiplicative inverses) for every ring. It includes the creation of multiplicative inverse- imitations (for elements that are not zero-divisors).

How does it do this? It apes the steps taken to create a field of fractions from an integral domain. A good description can be found on pg. 60 of Watkins’ “Topics in Commutative Ring Theory”, First Edition.

I would like to concentrate on one small point. Why is it that $\frac{a}{b}=\frac{c}{d}\implies t(ad-bc)=0$ where $b,d,t$ belong to the same multiplicative system? This description does not exactly ape the condition for integral domains: $\frac{a}{b}=\frac{c}{d}\implies (ad-bc)=0$. Simply put, why the extra $t$??

Because we have no means of isolating a regular element that is a common factor!! In every case, including that of integral domains, we get something like $h(ad-bc)=0$, where $h\in R$. Then we say $h$ is not a zero divisor. Hence, $ad-bc=0$. Here, we can’t say anything about $h$. Hence, the condition here is that $\frac{a}{b}=\frac{c}{d}\implies h(ad-bc)=0$.

Now where does the “multiplicative system” part come in?? That part seems fairly arbitrary, doesn’t it? How does it help prove transitivity? Because you’re going to get something of the form $h(af-be)$ anyway! Hence, if you didn’t add the $t$, how would you justify the $h$? Simple?

But why “multiplicative system”? Why not a simple $\frac{a}{b}=\frac{c}{d}\implies t(ad-bc)=0$ for some $t\in R$? Mainly because this allows us to localize $R$ with respect to different multiplicative systems within the ring. It helps us to generalize. We can choose the denominators and $t's$ (as in $t(ad-bc)$) from subsets of the ring, or the whole ring itself. It is the properties of these subsets that we will explore in the coming paragraphs.

Say we localize with respect to multiplicative system  $E$ of ring $R$. Then we choose all denominators from $E$. We need to choose denominators from the same multiplicative system because the way fraction multiplication is defined: we want multiplicative closure.

Why choose $t$ in $t(ad-bc)$ from the same multiplicative system as the denominators too? Why choose $t$ from a multiplicative system at all? We have to choose $t$ from the same multiplicative system because soon we start having products of $t's$, and we don’t want them to be from outside of the set we’re choosing the $t's$ from. This multiplicative system is the same as that of the denominators because denominators are also multiplied to the $t's$. Hence, we want algebraic closure.