### 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 $\langle x,y\rangle=\langle x+y+(n-1)xy\rangle$

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