Topics in Algebra, Chapter 3 Supplementary Problems, Part 4
This page covers the supplementary problems at the end of chapter 3. This page is split into parts so as to not get excessively long. **See also**: [part 1](/journal/topics-in-algebra-chapter-3-supplementary-a) (3.1 through 3.9), [part 2](/journal/topics-in-algebra-chapter-3-supplementary-b) (3.10 through 3.18) and [part 3](/journal/topics-in-algebra-chapter-3-supplementary-d) (3.19 through 3.24). The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book. ### Herstein 3.25: Let $R$ be a ring with no non-zero nilpotent elements such that $(ab)^2=a^2b^2$ for all $a,b\in R$. Prove that $R$ is commutative. After struggling with this problem for a while, I [posted it on]( where I was directed to a [paper by John Wavrik]( The paper takes up several problems of the form "supposing $R$ has thus and such a property, prove $R$ is commutative"; it's a nice companion to this stretch of problems in Topics in Algebra. Theorem 3 deals specifically with this problem. A rational argument was given ex post facto by Andreas Blass in the thread, and I combine the ideas here. To begin, and as a hint to anybody who might be trying this problem, you might say that the key observation in this problem is the following: it can be shown that $(ab-ba)^3=0$ for any $a,b\in R$. Therefore the condition of no non-zero nilpotents is stronger than it needs to be. The condition of the problem can be rewritten $$a(ab-ba)b=a[a,b]b=0,$$ showing that the commutator $[a,b]=ab-ba$ is killed when sandwiched on the left by $a$ and on the right by $b$. Now, with the hint from above, we write out $$ \begin{multline} (ab-ba)^3=(ab-ba)(ab-ba)(ab-ba)\\ =ab(ab-ba)ab-ab(ab-ba)ba-ba(ab-ba)ab+ba(ab-ba)ba. \end{multline} $$ The final term we already know to vanish, and the first term vanishes in the same way because $$0=(ba)^2-b^2a^2=b(ab-ba)a.$$ One way to proceed with the argument is to show that $a(ab-ba)a=b(ab-ba)b=0$. We use a nice trick pointed out by Andreas Blass. Note that $[a+b,b]=[a,b]$ so use the first result to write $$0=(a+b)[a+b,b]b=(a+b)(ab-ba)b.$$ Subtracting $a(ab-ba)b=0$ from this result, we find that $$b(ab-ba)b=0.$$ Thus the commutator of $[a,b]$ is annihilated when sandwiched on both sides by $b$. By symmetry (same argument applied to $[b,a]=-[a,b]$), left and right multiplying by $a$ also sends the commutator to zero. Hence the second and third terms of the expansion are also zero, and we have $(ab-ba)^3=0$. By the stipulation of the problem, this implies that $ab-ba=0$ for arbitrary $a,b\in R$, so that $R$ is commutative. ### Herstein 3.26: Let $R$ be a ring with no non-zero nilpotent elements such that $(ab)^2=(ba)^2$ for all $a,b\in R$. Prove that $R$ is commutative. I don't have a good solution to this problem. It is treated as theorem 5 in the above-mentioned [paper by John Wavrik]( As in 3.25, the assumption of no nilpotents is stronger than necessary — apparently, it can be shown that $(ab-ba)^5=0$ for all $a,b\in R$. I would be interested to hear of an elegant way of showing it. ### Herstein 3.27: Let $p_1,\ldots,p_k\in\mathbb{Z}$ be distinct primes, $n=p_1\cdots p_k$, and $R=\mathbb{Z}/n\mathbb{Z}$. Prove there exist exactly $2^k$ elements $x\in R$ with $x^2=x$. First, observe that $x^2-x=0$ modulo $n$ means that $$n=p_1\cdots p_k\mid x(x-1).$$ Every prime in the list must individually divide either $x$ or $x-1$ (these scenarios are mutually exclusive, or else we would have a prime dividing $x-(x-1)=1$). Hence we have $$x\equiv 0\bmod p_i\qquad{\rm or}\qquad x\equiv 1\bmod p_i$$ for every $i\in\{1,\ldots,k\}$. Each of the $k$ primes has two choices, and there are therefore $2^k$ potential solutions here. We must show that each choice gives rise to exactly one solution. We can write a "choice tuple" $C=\{a_1,\ldots,a_k\}$, representing the fact that $x=a_i\bmod p_i$ for each $i$. It is clear that two distinct choices $C$ and $C'$ will never give rise to the same solution, $x$: for the choices to be distinct, they must differ in the remainder upon division by some $p_i$. Therefore we need only show (i) that a solution exists for any choice $C$, and (ii) that the solution is unique, modulo $n$. First, uniqueness: suppose $x$ and $x'$ are two solutions for the same choice tuple $C$. Then they agree in every remainder upon division by the $\{p_i\}$ and thus $$x-x'\equiv 0\bmod p_i$$ for all $i\in\{1,\ldots,k\}$. As a result, the product $n=p_1\cdots p_k$ must also divide $x-x'$. In other words, $x'\equiv x\bmod n$, which is the desired statement of uniqueness. Now fix $i\in\{1,\ldots,k\}$ and consider $p_i$ and $\frac{n}{p_i}=\prod_{j\ne i}p_j$. These two integers are coprime, so there exist $r_i,s_i\in\mathbb{Z}$ with $$r_i p_i+s_i\cdot\frac{n}{p_i}=1.$$ Multiply through by $a_i$ and define $$x_i=a_i(1-r_i p_i)=a_i s_i\prod_{j\ne i}p_j.$$ From the first equality, we see that $x_i\equiv a_i\bmod p_i$. From the second equality, we see that $x_i\equiv 0\bmod p_j$ for all $j\ne i$. Now, if we define $x_i$ in this way for each $i$ and construct $$x=\sum_{i=1}^k x_i,$$ it must solve our system of simultaneous congruences. Now we have shown that, given a choice tuple $C$, a solution exists and is unique modulo $n$. Because there are $2^k$ choice tuples, this proves that there are $2^k$ solutions of $x^2=x$ in $R$. This problem boils down to a special case of the [Chinese Remainder Theorem](, whose proof is essentially the same as the proof of existence presented above. A related problem is Herstein's exercise 1.3.15. ### Herstein 3.28: Construct a non-zero polynomial $q\in\mathbb{Z}[x]$ which has no rational roots but for which there exists a solution $x\in\mathbb{Z}$ to $q(x)\equiv 0\bmod p$ for every prime $p$. My first instinct here was to try the usual polynomials which have no roots over $\mathbb{Q}$ or $\mathbb{R}$, such as $x^2-2$ and $x^2+1$. Recall the results of chapter 3.8, where Lemma 3.8.2 and exercise 3.8.4 combine to state that, if $p$ is a prime, then $$x^2+1\equiv 0\bmod p$$ has a solution if and only if $p=4k+1$ for some $k\in\mathbb{Z}$. Then our starting point is $$q_0(x)=x^2+1$$ which has no rational roots but does have a solution $x$ to $q_0(x)\equiv 0\bmod p$ for any $4k+1$ prime $p$. Proceeding from here, the plan is to tack on additional factors to cover the cases of other sorts of primes. One might hope that there is another common non-trivial (i.e. not a perfect square in $\mathbb{Z}$) quadratic residue among all $4k+3$ primes, as $-1$ was common to all $4k+1$ primes. This is not the case. However, we will prove below that if $p$ is of the form $4k+3$ with $k\in\mathbb{Z}$, then one of $2$ or $-2$ is always a quadratic residue modulo $p$ ([hint from JavaMan on]( Therefore, if we augment: $$q_1(x)=(x^2+1)(x^2+2)(x^2-2),$$ then it still has no rational roots, but it has a solution $x$ to $q_1(x)\equiv 0\bmod p$ for any odd $p$. The only remaining concern is the prime $2$, but we see that $x=0$ is already a root modulo $2$. In summary, the polynomial $$q(x)=(x^2+1)(x^2+2)(x^2-2)\in\mathbb{Z}[x]$$ has no rational roots but has a solution $x$ to $q(x)\equiv 0\bmod p$ for any prime $p$. **Lemma**: If $p$ is an odd prime, then there are as many quadratic residues as quadratic non-residues modulo $p$. **Proof**: Note: for purposes of symmetry, we do not consider $0$ to be a quadratic residue. The candidates for quadratic residues modulo $p$ are $\{1,\ldots,p-1\}$; there are an even number of them. We can enumerate the actual residues by simply considering the set $$\{x^2\mid x\in\{1,\ldots,p-1\}\}\subset\mathbb{Z}/p\mathbb{Z}.$$ In doing so, we note that $(p-x)^2=p^2-2px+x^2=x^2\bmod p$, so fully half of those squares do not contribute distinct residues to the set. This puts an upper bound of $(p-1)/2$ on the number of quadratic residues modulo odd prime $p$. Are there any other duplicates? No. Suppose $x$ and $y$ square to the same residue: $x^2\equiv y^2\bmod p$. Then $(x+y)(x-y)\equiv 0\bmod p$ so that $p\mid(x-y)$ or $p\mid(x+y)$. Restricting $x,y\in\{1,\ldots,p-1\}$, the first case gives $y=x$ and the second case gives $y=p-x$. Therefore, there are no other duplicates: exactly half, $(p-1)/2$, of the values $\{1,\ldots,p-1\}$ are quadratic residues modulo $p$. This leaves the other half as quadratic non-residues. **Lemma**: If $p=4k+3$ is a prime, with $k\in\mathbb{Z}$, and $a\in\{1,\ldots,p-1\}$, then exactly one of $a$ or $-a$ is a quadratic residue modulo $p$. **Proof**: By the lemma above, there are $(p-1)/2$ quadratic residues modulo $p$. For each one, we will exhibit a complementary, unique quadratic non-residue, thus accounting for all $p-1$ values in $\{1,\ldots,p-1\}$. Let $m\in\{1,\ldots,p-1\}$ be a quadratic residue modulo $p$, with $x$ such that $x^2\equiv m\bmod p$. Suppose that $-m=p-m$ is also a quadratic residue, with $y$ such that $y^2\equiv -m\bmod p$. Then, because $(m,p)=1$, we can invert, writing $$y^{-2}\equiv -m^{-1}\bmod p$$ and hence $$(xy^{-1})^2\equiv -1\bmod p.$$ This is impossible by exercise 3.8.4 because $p\equiv 3\bmod 4$. Therefore it must be that $-m$ is a non-residue. To each of the $(p-1)/2$ quadratic residues $m$, we associate a quadratic non-residue $-m=p-m$. In this way, we enumerate the whole set $\{1,\ldots,p-1\}$. Every element is therefore a quadratic residue or its complementary non-residue. **Corollary**: If $p=4k+3$ is a prime, with $k\in\mathbb{Z}$, then one of $2$ or $-2$ is a quadratic residue modulo $p$.
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