Topics in Algebra, Chapter 3 Supplementary Problems, Part 1
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 2](/journal/topics-in-algebra-chapter-3-supplementary-b) (3.10 through 3.18), [part 3](/journal/topics-in-algebra-chapter-3-supplementary-c) (3.19 through 3.24) and [part 4](/journal/topics-in-algebra-chapter-3-supplementary-d) (3.25 through 3.28). 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 3.1: Let $R$ be a commutative ring. Define a *prime ideal* $P$ of $R$ to be an ideal such that if $a,b\in R$ have $ab\in P$, then $a\in P$ or $b\in P$. Show that $P$ is a prime ideal if and only if $R/P$ is an integral domain. Let $a,b\in P$ be such that $$ab+P=(a+P)(b+P)=0+P.$$ $R/P$ is an integral domain is equivalent to the statement that $ab\in P$ implies $a\in P$ or $b\in P$, which is equivalent to the statement that $P$ is a prime ideal. ### Herstein 3.2: Let $R$ be a commutative, unital ring. Prove that a maximal ideal of $R$ is prime. Theorem 3.5.1 states that $M$ is a maximal ideal of commutative, unital ring $R$ if and only if $R/M$ is a field. Therefore if $M$ is a maximal ideal of $R$, we have $R/M$ a field and thus an integral domain. By exercise 3.1, $M$ is a prime ideal. ### Herstein 3.3: Exhibit a ring $R$ where a prime ideal fails to be a maximal ideal. The ring $R$ cannot be a PID, because a prime ideal will always be maximal in that case: if $P=(p)$ is a prime ideal, then an ideal $I=(r)$ containing it will have $r\mid p$ which won't happen in any non-trivial way. Then we avoid PIDs and look for things that are somewhat more exotic, but non-commutative rings are probably too exotic. A natural choice for investigation are polynomial rings. One thought is something like $(x^2+1)$ in $\mathbb{R}[x]$. This is a prime ideal but it's also relatively easy to see that it is maximal. However, if we consider instead $I=(x^2+2)$ in $R=\mathbb{Z}[x]$, we have it. The generator $x^2+2$ is irreducible so, because $\mathbb{Z}[x]$ is a UFD, the ideal is prime. The ideal $$J=(x^2,2)=\{2m+nx^2\mid m,n\in\mathbb{Z}[x]\}$$ properly contains it because it contains elements (e.g. $2$) with degree smaller than $2$. However, $J\ne R$ because $1$ is clearly not in $J$. Therefore $I$ is prime but not maximal in $R=\mathbb{Z}[x]$. Note that $I=(x+2)$ and $J=(x,2)$ would work via the same argument. ### Herstein 3.4: Let $R$ be a finite, commutative, unital ring. Prove that a prime ideal of $R$ is maximal. Let $P$ be a prime ideal of $R$. We have by exercise 3.1 that $R/P$ is an integral domain, and by lemma 3.2.2 that a finite integral domain must be a field. Therefore $R/P$ is a field, so that theorem 3.5.1 gives us that $P$ is a maximal ideal of $R$. ### Herstein 3.5: With $F$ a field, prove that $F[x]$ is isomorphic to $F[t]$. The meaning of this exercise is not clear. It seems that we just want to show that the name of the indeterminate is irrelevant. This can be done by considering the obvious mapping $\phi:F[x]\to F[t]$ with $$\phi(a_0+a_1 x+\cdots+a_n x^n)=a_0+a_1 t+\cdots+a_n t^n.$$ Then $$\phi((fg)(x))=(fg)(t)=f(t)g(t)=\phi(f(x))\phi(g(x))$$ and $$\phi((f+g)(x))=(f+g)(t)=f(t)+g(t)=\phi(f(x))+\phi(g(x)).$$ Therefore $\phi$ is a homomorphism, and it is clear that it is both injective and onto. ### Herstein 3.6: Classify all $\sigma\in{\rm Aut}(F[x])$ which fix the base field $F$. That is, which automorphisms $\sigma$ on $F[x]$ have $\sigma(a)=a$ for all $a\in F$? Let $f\in F[x]$ be given by $f(x)=a_0+\cdots+a_n x^n$. Consider $$\sigma(f(x))=\sigma(a_0+a_1 x+\cdots+a_n x^n)=a_0+a_1\sigma(x)+\cdots+a_n\sigma(x)^n$$ which has simplified because $\sigma$ is a homomorphism and fixes the coefficients. From this we see that the $F$-fixing automorphism $\sigma$ is determined entirely by where it maps the polynomial $x=0+1x+0x^2+\cdots$. If ${\rm deg}(\sigma(x))=0$, then the map can not be surjective, because only constants are in the image of $\sigma$. If ${\rm deg}(\sigma(x))>1$, then again the map $\sigma$ can not be surjective: the degree of a non-constant polynomial $a_0+a_1\sigma(x)+\cdots+a_n\sigma(x)^n$ is $n\cdot{\rm deg}(\sigma(x))\ge{\rm deg}(\sigma(x))>1$. Hence no polynomials of degree $1$ would be in the image of $\sigma$. Therefore $\sigma(x)=\alpha x+\beta$ for some $\alpha,\beta\in F$ is the only remaining possibility. The question is whether $\sigma(x)$ so chosen respects the fact that the map must be an automorphism. We have $$\sigma(f(x))=\sigma(a_0+a_1 x+\cdots+a_n x^n)=a_0+a_1(\alpha x+\beta)+\cdots+a_n(\alpha x+\beta)^n=f(\alpha x+\beta).$$ This is simply a composition and we easily see that $\sigma$ is still a homomorphism: $$\sigma((f+g)(x))=(f+g)(\alpha x+\beta)=f(\alpha x+\beta)+g(\alpha x+\beta)=\sigma(f(x))+\sigma(g(x))$$ and $$\sigma((fg)(x))=(fg)(\alpha x+\beta)=f(\alpha x+\beta)g(\alpha x+\beta)=\sigma(f(x))\sigma(g(x)).$$ Suppose $f,g\in F[x]$ are such that $\sigma(f)=\sigma(g)$, and suppose $\alpha\ne 0$. Then $$0=(f-g)(\alpha x+\beta)=b_0+b_1(\alpha x+\beta)+\cdots+b_m(\alpha x+\beta)^m$$ where $b_i$ are the differences of the coefficients of $f$ and $g$, and $m$ is the maximum of the two degrees. Starting from the degree $m$ term, it is clear that $b_m=0$ because there is no way to cancel the $x^m$ term otherwise. Next the degree $m-1$ term suffers the same fate, and so on, down the chain. Thus all of the coefficients must vanish identically, so that $\sigma(f)=\sigma(g)$ implies $f=g$, i.e. that $\sigma$ is injective. This argument would fail if $\alpha=0$ because there are no powers of $x$ to speak of. However, $\sigma$ is not injective in the $\alpha=0$ case: $\sigma(x)=\sigma(\beta)$ and $x\ne\beta$. Therefore we must restrict $\alpha\ne 0$. Observe that, if $\sigma(f(x))=f(\alpha x+\beta)$, then $$\sigma(\alpha^{-1}x-\alpha^{-1}\beta)=\alpha^{-1}(\alpha x+\beta)-\alpha^{-1}\beta=x.$$ Therefore $\sigma$ is surjective because $$a_0+a_1 x+\cdots+a_n x^n$$ is the image under $\sigma$ of $$a_0+a_1(\alpha^{-1}x-\alpha^{-1}\beta)+\cdots+a_n(\alpha^{-1}x-\alpha^{-1}\beta)^n.$$ To summarize, we have shown that if $\sigma$ is an automorphism of $F[x]$ which fixes the coefficient field $F$, then $\sigma$ must be of the form $$\sigma(f(x))=f(\alpha x+\beta)$$ where $\alpha,\beta\in F$ and $\alpha\ne 0$. If we wanted to replace $F$ by a ring $R$, we note that if $R$ is not an integral domain, then it is conceivable that ${\rm deg}(\sigma(x))>1$ would be an acceptable situation, so there might be much more to study here. However, if $R$ is an integral domain, the argument carries through but $\alpha$ must be a unit rather than simply non-zero. ### Herstein 3.7: Let $R$ be a commutative ring and let $N=\{r\in R\mid r^m=0$ for some $m\in\mathbb{Z}\}$. ### Prove that ### (a) $N$ is an ideal of $R$. ### (b) If $(r+N)^m=N$ in $R/N$ for some $m\in\mathbb{Z}$, then $r\in N$. ### $N$ is called the *nilradical* of $R$. **(a)** If $r,s\in N$ with $r^m=0$ and $s^n=0$, then $$(r+s)^{m+n}=\sum_{k=0}^{m+n}\binom{m+n}{k}r^k s^{m+n-k}$$ by the binomial theorem. The idea here is that every term in this sum is zero because either $r$ is raised to a high enough power or $s$ is raised to a high enough power. This is most clear if we change the variable of summation to $\ell=k-m$ so that $$(r+s)^{m+n}=\sum_{\ell=-m}^n\binom{m+n}{m+\ell}r^{m+\ell}s^{n-\ell}.$$ Now if $\ell\ge0$ we have $r^{m+\ell}=0$, while $\ell<0$ gives $s^{n-\ell}=s^{n+|\ell|}=0$. Hence $(r+s)^{m+n}=0$ and so $r+s\in N$ whenever $r,s\in N$. In addition, if $r\in N$ with $r^m=0$ and $\alpha\in R$, then $(\alpha r)^m=\alpha^m r^m=0$ so $\alpha r\in N$. Therefore, $N$ is an ideal of $R$. **(b)** If $(r+N)^m=N$, then $r^m\in N$. Thus there exists $n$ such that $0=(r^m)^n=r^{mn}$, which shows that $r$ also belongs to $N$. This is reminiscent of $R/N$ being an integral domain. It's not that we necessarily have no zero divisors in $R/N$, but we have no non-trivial nilpotent elements in $R/N$. Of course, nilpotent elements are a particular class of zero divisor. If $R/N$ were an integral domain, then we could say by exercise 3.1 that $N$ is a prime ideal. While this might not be true, it is indeed true that the nilradical is related to prime ideals. It turns out that $N$ is the intersection of all prime ideals of $R$ ([proof]( ### Herstein 3.8: Let $R$ be a commutative ring and let $A$ be an ideal of $R$. Let $N(A)=\{r\in R\mid r^m\in A$ for some $m\in\mathbb{Z}\}$. ### Prove that ### (a) $N(A)$ is an ideal of $R$ containing $A$. ### (b) $N(N(A))=N(A)$. ### $N(A)$ is called the *radical* of $A$. **(a)** If $r\in A$ then $r$ is immediately a member of $N(A)$ because $r^1\in A$. Thus $A\subset N(A)$. If $r,s\in N(A)$ then the proof of exercise 3.7a carries over almost verbatim to show that $r+s\in N(A)$. The various terms in the binomial expansion do not vanish, but they all belong to $A$, which is an ideal. Hence $r+s\in N(A)$. If $r\in N(A)$ with $r^m\in A$ and $\alpha\in R$, then $(\alpha r)^m=\alpha^m r^m\in A$ so that $\alpha r\in N(A)$. Therefore $N(A)$ is an ideal containing $A$. **(b)** Let $r\in N(N(A))$ so that $r^m\in N(A)$ for some $m\in\mathbb{Z}$. Then there exists $n\in\mathbb{Z}$ with $(r^m)^n=r^{mn}\in A$ and hence $r\in N(A)$. This shows that $N(N(A))\subset N(A)$. By part (a), we already know that $N(A)\subset N(N(A))$. Therefore $N(N(A))=N(A)$. The nilradical is the radical of the zero ideal. Note that the parts of exercise 3.7 are the respective special cases of the parts of exercise 3.8, despite the fact that part (b) of exercise 3.7 is expressed somewhat differently. ### Herstein 3.9: Describe the nilradical of $\mathbb{Z}/n\mathbb{Z}$ in terms of $n$. If $r\in\mathbb{Z}/n\mathbb{Z}$ is an element of the nilradical, then there exists $m,a\in\mathbb{Z}$ with $r^m=an$. The prime factors of the left hand side will always be exactly those of $r$. If one of the prime factors of $n$ is missing from $r$, then this equation will have no solution. To be more explicit, say that $$n=p_1^{i_1}\cdots p_k^{i_k};$$ then we claim that $r$ belongs to the nilradical if and only if $r$ is a multiple of $p_1\cdots p_k$. This statement correctly describes even the trivial case of $r=0$, but we exclude that case in what follows. If $r$ is non-zero and of this form, then take the power $m$ large enough that each prime $p_l$ is raised to a power greater than its power in $n$, and then choose the coefficient $a$ to make up the deficit. On the other hand, if non-zero $r$ is known to be in the nilradical, then we have $r^m=an$ for some $m,a$. If $p$ is a prime dividing $n$, then it also must divide the left hand side, $p\mid r^m$. As $p$ is a prime, we have further that $p\mid r$. This holds for each prime, so every prime dividing $n$ must also appear, with at least one power, in $r$. This proves the claim. As an example, consider $\mathbb{Z}/24\mathbb{Z}$ where $24=2^3\cdot 3$. The prime product is $2\cdot 3=6$. The elements of the nilradical of $\mathbb{Z}/24\mathbb{Z}$ are $6$ ($6^3=24\cdot9$), $2\cdot 6$ ($12^2=24\cdot6$), $3\cdot 6$ ($18^3=24\cdot 243$) and of course $0$. On the other hand, something like $2^1\cdot 3^0$ is not in the nilradical because $24a$ is always divisible by $3$ whereas $2^m$ never is.
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