Topics in Algebra, Chapter 3 Supplementary Problems, Part 2

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 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 3.10: Let $R$ be a ring and let $A,B\subset R$ be ideals with $A\cap B=(0)$. Show that $ab=0$ whenever $a\in A$ and $b\in B$.
Because $A$ is an ideal (and therefore closed under external multiplication), $ab\in A$; because $B$ is an ideal, $ab\in B$. Therefore $ab\in A\cap B=(0)$ so $ab=0$.
### Herstein 3.11: Let $R$ be a ring and let $Z(R)=\{x\in R\mid xy=yx$ for all $y\in R\}$. Prove that $Z(R)$ is a subring of $R$.
This is the analogue of the center of a group, the set of all elements that commute with everything. Let $x_1,x_2\in Z(R)$ and let $y$ be an arbitrary element of $R$. Then
$$x_1 x_2 y=x_1(x_2 y)=x_1(yx_2)=(yx_1)x_2=yx_1x_2$$
showing that $x_1 x_2\in Z(R)$. Similarly, $(x_1+x_2)y=y(x_1+x_2)$ so that $x_1+x_2\in Z(R)$. Of course, if there exists $1\in R$, then $1\in Z(R)$. Therefore $Z(R)$ is a subring of $R$. It is patently a commutative ring.
### Herstein 3.12: Let $R$ be a division ring. Prove that $Z(R)$ is a field.
By exercise 3.11, we know that $Z(R)$ is a commutative subring. If $R$ is a division ring then so too is $Z(R)$. A commutative division ring is a field, by definition.
### Herstein 3.13: Construct a polynomial of degree $3$, irreducible over $F=\mathbb{Z}/3\mathbb{Z}$. Use it to construct a field of $27$ elements.
It seems a difficult problem in general to construct arbitrary irreducible polynomials. However, with small degree and a small field of coefficients, we can force it through. We observe that we may restrict attention to $f\in F[x]$ monic, because $F$ is a field so the highest coefficient is invertible and factoring it out does not affect reducibility. In addition, a reducible degree $3$ polynomial must have a linear factor because the only non-trivial way to partition $3$ is $1+2$. Then if $f$ is monic and reducible, we will be able to write
$$f(x)=(x-\alpha)(x^2+\beta x+\gamma)$$
where the latter polynomial may be further reducible, but that is of no concern. Clearly such a polynomial must map some element $\alpha\in F$ to zero. Returning to the problem at hand, we now know that a degree $3$ polynomial $f\in F[x]$ which has no root in $F$ must be irreducible.
Now a simple brute force search is easy. We restrict to monic degree $3$ polynomials, searching for one which doesn't map any of $0,1,2$ to zero. For simplicity, we keep the $x^2$ term out of it. The first thing to try is $f(x)=x^3+x+1$ but it has $f(1)=0$. Next, $f(x)=x^3+2x+1$, which works: $f(0)=1$, $f(1)=1$ and $f(2)=1$. Therefore one such irreducible polynomial is
$$f(x)=x^3+2x+1,$$
and there are other possibilities.
By exercise 3.9.7, $F[x]/(x^3+2x+1)$ is a field of $3^3=27$ elements.
Note that the connection found here between roots and irreducibility is not of general use. There are sometimes polynomials over fields which have no roots but are nevertheless reducible, such as $(x^2+1)^2=x^4+2x^2+1$ over $\mathbb{R}$. The observation is special for degree $3$, where any reduction involves a degree one term. It does not even extend to higher odd degrees, because one can imagine a fifth degree polynomial that splits into irreducible factors of degree $2$ and $3$, neither of which has a root, e.g. $(x^2-2)(x^3-2)=x^5-2x^3-2x^2+4$ over $\mathbb{Q}$ (both factors are irreducible by exercise 3.10.2).
### Herstein 3.14: Construct a field with $625$ elements.
$625=5^4$ so exercise 3.9.7 suggests that we look for a degree $4$ polynomial $f$ irreducible over $F=\mathbb{Z}/5\mathbb{Z}$. Then $F[x]/(f)$ will be the desired field.
For degree $4$, the methods of the previous problem are not helpful. Therefore we try brute force: writing down the simplest polynomials and manually checking that they are irreducible, hoping to get lucky. A few observations are helpful. 1) If $f\in F[x]$ were reducible, it could be factored into two degree $2$ polynomials, or into a degree $1$ and a degree $3$ polynomial. In the latter case, $f$ must have a root in $F$ due to its linear factor. Therefore we look for candidates which have no roots in $F$, but that is necessary and not sufficient. 2) Fermat's little theorem says that $x^4=1$ in our field. 3) The quadratic residues modulo $5$ are $\{0,1,4\}$.
$f(x)=x^4+1$ has no root because $x^4+1=2$ for $x\in F$. We then try to factor it as
$$x^4+1=(x^2+\alpha x+\beta)(x^2+\gamma x+\delta)$$
and find that the equations for each coefficient have a consistent solution: $\alpha=\gamma=0$, $\beta=2$, $\delta=3$. Thus $x^4+1=(x^2+2)(x^2+3)$ is reducible.
$f(x)=x^4+x+1$ has a root at $x=3$. More generally, $x^4+kx+1$ takes the values $2+kx$, and quick inspection shows that any non-zero $k$ gives a polynomial with a root, which is reducible.
$f(x)=x^4+x^2+1$ takes the values $2+x^2\in\{1,2,3\}$ on $F$, so it has no roots. However, again writing down the equations for the coefficients in a product of quadratics, we find that $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ is reducible.
$f(x)=x^4+x^2+x+1$ takes the values $2+x+x^2\in\{2,3,4\}$ on $F$, so it has no roots. The equations for the coefficients are $\gamma=-\alpha$, $\delta=\beta^{-1}$, $\beta+\delta+\alpha\gamma=1$ and $\alpha\delta+\beta\gamma=1$. Combining these four equations, we see that
$$\beta(\beta-\alpha^2)=0\qquad{\rm and}\qquad\alpha(1-\beta^2)=1.$$
From the first of these new equations, we see that $\beta=\alpha^2$, because $\beta\delta=1$ precludes the possibility that $\beta=0$. Next, we see that $\alpha(1-\alpha^4)=1$. However, we know that $\alpha^4=1$ by little Fermat, so we have our desired contradiction at long last. Thus $f(x)=x^4+x^2+x+1$ is irreducible over $F$ and $F[x]/(f)$ is a field of order $625$.
### Herstein 3.15: Let $F$ be a field, let $p\in F[x]$, and let $R=F[x]/(p)$. Show that the nilradical $N$ of $R$ is $(0)$ if and only if $p$ is not divisible by the square of any polynomial.
For an element $f\in F[x]$ to be in the nilradical $N$ of $R$, it means there exists an integer $n$ such that $p(x)$ divides $f(x)^n$. First we show that $N$ being trivial implies that $p$ cannot be divisible by a square. Consider the contrapositive statement: if $p$ is divisible by a square, say $p=f^2 g$ with $f,g\in F[x]$, then $(fg)^2=pg\in(p)$ so that $N$ is non-trivial because $fg$ is in it. Hence if $N$ is trivial then $p$ must not be divisible by a square.
Now suppose that $p$ is not divisible by any square. Using the fact that $F[x]$ is a UFD, we can write $p=\pi_1\cdots\pi_m$ with the $\pi_i\in F[x]$ irreducible and all being unique, $\pi_i\ne\pi_j$ if $i\ne j$. If $f\in N$, then there is $n\in\mathbb{Z}$ with $p\mid f^n$. Then by lemma 3.7.6 every $\pi_i\mid f$ and therefore $p\mid f$. Viewed in the quotient ring $R$, $f=0$. Hence if $p$ is not divisible by a square, the nilradical of $R$ is trivial.
Note that the squarefree property is truly necessary in the second paragraph. Take this example in the integers: $p=18=2\cdot 3^2$ and $f=6$. We have $p\mid f^2$ but of course $p\nmid f$.
### Herstein 3.16: Prove that $f(x)=x^4+x^3+x+1$ is not irreducible over any field (note no $x^2$ term).
We observe that $-1$, which belongs to any field, is a root of $f$. Hence $f(x)$ is divisible by the polynomial $(x+1)$ and is therefore irreducible.
### Herstein 3.17: Prove that $f(x)=x^4+2x+2$ is irreducible over $\mathbb{Q}$.
This is immediate from the Eisenstein criterion with $p=2$.
### Herstein 3.18: Let $F$ be a finite field and let the characteristic of $F$ be $p$. Prove that $p$ is prime and that $|F|=p^n$ for some $n\in\mathbb{Z}$. Also prove that if $a\in F$ then $a^{p^n}=a$.
First, the notion of characteristic (as defined by Herstein) is relevant here because a field is an integral domain (exercise 3.2.12). The characteristic of $F$ is finite by the pigeonhole principle applied to the set $\{1,2\cdot1,3\cdot1,\ldots\}$: the list must repeat because $F$ is finite, so there exist $\alpha,\beta\in\mathbb{Z}$ such that $\alpha1=\beta1$; therefore $|\alpha-\beta|1=0$. By exercise 3.2.6, the characteristic of $F$ is prime because it is finite.
Because $F$ has characteristic $p$, we know that
$$F_0=\{0,1,2,\ldots,p-1\}\subset F.$$
Now, we are familiar with $F_0\cong\mathbb{Z}/p\mathbb{Z}$; it is a field containing $p$ elements. If $F=F_0$ then we have shown that $|F|=p^1$ and we are done. However, suppose there exists $x\in F$ with $x\not\in F_0$, and consider the set
$$F_1=\{\alpha_1 x+\alpha_0\mid\alpha_0,\alpha_1\in F_0\}.$$
Clearly $F_0\subset F_1\subset F$, and we will show in a moment that it contains $p^2$ elements. Note that we will **not** claim that $F_1$ is a field or even closed under multiplication. For instance, it is not clear at all at this point whether $x^2$ would belong to $F_1$. Nevertheless, we will find this construction useful. The sketch of the proof from this point is as follows: we repeat this procedure for as long as $F$ contains an element outside our constructed subsets. The procedure surely terminates because each step generates a proper superset of the preceding step's set, and $F$ is finite. Moreover, the set generated in any step is $p$ times as large as the preceding set. Hence, when the procedure terminates, we realize that the size of $F$ must be a power of $p$.
First we show that $F_1$ contains $p^2$ elements. It is clear that we can enumerate $p^2$ elements in $F_1$ because there are $p$ choices for each of the coefficients $\alpha_0$ and $\alpha_1$. However, must they all be unique? Yes. If $\alpha'_1 x+\alpha'_0=\alpha_1 x+\alpha_0$, then we see that
$$(\alpha'_1-\alpha_1)x=\alpha_0-\alpha'_0.$$
If $\alpha'_1-\alpha_1$ is non-zero, then it is an invertible element of $F_0$, so that we have
$$x=(\alpha'_1-\alpha_1)^{-1}(\alpha_0-\alpha'_0)\in F_0,$$
a contradiction. Therefore $\alpha'_1=\alpha_1$ and consequently $\alpha'_0=\alpha_0$, the element is unique. This proves that there are $p^2$ elements in $F_1\subset F$.
Now we would like to show the validity of the procedure in general. Suppose $F_{k-1}\subset F$ with $|F_{k-1}|=p^k$, and suppose there exists $y\in F$ with $y\not\in F_{k-1}$. Then construct
$$F_k=\{\alpha_k y+\beta\mid \alpha_k\in F_0,\ \beta\in F_{k-1}\}.$$
We can enumerate $p^k\cdot p=p^{k+1}$ elements in $F_k$, and clearly $F_k\subset F$. Are any of the $p^{k+1}$ elements duplicates? No. If
$$\alpha'_k y+\beta'=\alpha_k y+\beta,$$
then $(\alpha'_k-\alpha_k)y=\beta-\beta'$ and the argument from above applies again: if $\alpha'_k-\alpha_k$ is non-zero, then it is an invertible element of $F_0$ and
$$y=(\alpha'_k-\alpha_k)^{-1}(\beta-\beta').$$
We know, however, that $F_{k-1}$ is closed under multiplication by elements of $F_0$, because $F_0$ is a field. Thus, this line of reasoning has us conclude that $y\in F_{k-1}$, a contradiction. Therefore we have $\alpha'_k=\alpha_k$ and consequently $\beta'=\beta$, the two elements are identical. Now we have shown the recurrence $|F_k|=p|F_{k-1}|$.
Because $F$ is finite, it will eventually be exhausted of its elements and we will have some maximal $n$ such that $F_n\subset F$, but there does not exist another $z\in F$ with $z\not\in F_n$. But this means that $F\subset F_n$, so that $F=F_n$ and $|F|=p^{n+1}$. This is the desired result: the number of elements in a finite field must be a power of a prime.
Finally, suppose that $|F|=p^n$. We have that the non-zero elements of $F$ form a group under multiplication, and there are $p^n-1$ of them. By Lagrange's theorem, the order of any element divides $p^n-1$. Then if $a\in F$, we surely have
$$a^{p^n-1}=1$$
or, what is the same,
$$a^{p^n}=a.$$