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\begin{document}
\section*{Math 332 Winter 2023, Lecture 28: Polynomials}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/23wa}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/23wa}}}

\setcounter{section}{2}

\section{Monoid algebras and polynomials}

Recall: For this entire chapter, we fix a \textbf{commutative} ring $R$.

\setcounter{subsection}{4}

\subsection{Adjoining roots}

\setcounter{subsubsection}{4}

\subsubsection{The general construction}

Last time, we introduced a way to \textquotedblleft adjoin\textquotedblright%
\ a root of a polynomial $b\in R\left[  x\right]  $ to a given commutative
ring $R$: We took the quotient ring $R\left[  x\right]  /b$ (where $S/b$ is
shorthand for $S/bS$).

In particular,%
\begin{align*}
\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)   &  \cong\mathbb{C};\\
\mathbb{Z}\left[  x\right]  /\left(  x^{2}+1\right)   &  \cong\mathbb{Z}%
\left[  i\right]  ;\\
\mathbb{Q}\left[  x\right]  /\left(  x^{2}+1\right)   &  \cong\mathbb{Q}%
\left[  i\right]  ;\\
\mathbb{Q}\left[  x\right]  /\left(  x^{2}-1\right)   &  \cong\mathbb{Q}%
\left[  C_{2}\right]  \cong\mathbb{Q}\times\mathbb{Q}\cong\left\{  \left(
\begin{array}
[c]{cc}%
a & b\\
b & a
\end{array}
\right)  \ \mid\ a,b\in\mathbb{Q}\right\}  ;\\
\mathbb{Z}\left[  x\right]  /m  &  \cong\left(  \mathbb{Z}/m\right)  \left[
x\right]  \ \ \ \ \ \ \ \ \ \ \text{for }m\text{ integer;}\\
\mathbb{Z}\left[  x\right]  /\left(  mx-1\right)   &  \cong R_{m}%
\ \ \ \ \ \ \ \ \ \ \text{for }m\text{ nonzero integer;}\\
R\left[  x\right]  /1  &  \cong\left(  \text{zero ring}\right)  .
\end{align*}


Some of these rings behave nicer than others. Sometimes the quotient ring
$R\left[  x\right]  /b$ contains a copy of the original ring $R$; sometimes it
doesn't. Can we give general criteria for what will happen in what case?

\bigskip

We start with a general fact:

\begin{proposition}
\label{prop.fieldext.proj-morphism}Let $b\in R\left[  x\right]  $ be a polynomial.

\textbf{(a)} The projection map%
\begin{align*}
R\left[  x\right]   &  \rightarrow R\left[  x\right]  /b,\\
p  &  \mapsto\overline{p}%
\end{align*}
is an $R\left[  x\right]  $-algebra morphism, and thus an $R$-algebra morphism.

\textbf{(b)} The map\footnotemark%
\begin{align*}
R  &  \rightarrow R\left[  x\right]  /b,\\
r  &  \mapsto\overline{r}%
\end{align*}
is an $R$-algebra morphism.

\textbf{(c)} For any $p\in R\left[  x\right]  $, we have $p\left[
\overline{x}\right]  =\overline{p}$ in $R\left[  x\right]  /b$.

\textbf{(d)} The element $\overline{x}\in R\left[  x\right]  /b$ is a root of
$b$.
\end{proposition}

\footnotetext{Note the difference between the maps in part \textbf{(a)} and in
part \textbf{(b)}: The map in part \textbf{(a)} takes as input a polynomial
$p\in R\left[  x\right]  $, whereas the map in part \textbf{(b)} takes as
input a scalar $r\in R$ (and treats it as a constant polynomial, i.e., as
$rx^{0}\in R\left[  x\right]  $). If you regard $R$ as a subring of $R\left[
x\right]  $, you can thus view the map in part \textbf{(b)} as a restriction
of the map in part \textbf{(a)}.}

\begin{proof}
Fairly straightforward; see Proposition 4.5.7 in the text.
\end{proof}

So $\overline{x}$ is the root of $b$ that we are pulling out of our hat
through this $R\left[  x\right]  /b$ construction. Now, we come to the
promised criterion for $R\left[  x\right]  /b$ to contain a copy of $R$:

\begin{theorem}
\label{thm.fieldext.basis-monic}Let $m\in\mathbb{N}$. Let $b\in R\left[
x\right]  $ be a polynomial of degree $m$ such that its leading coefficient
$\left[  x^{m}\right]  b$ is a unit of $R$. Then:

\textbf{(a)} Each element of $R\left[  x\right]  /b$ can be uniquely written
in the form%
\[
a_{0}\overline{x^{0}}+a_{1}\overline{x^{1}}+\cdots+a_{m-1}\overline{x^{m-1}%
}\ \ \ \ \ \ \ \ \ \ \text{for }a_{0},a_{1},\ldots,a_{m-1}\in R.
\]


\textbf{(b)} The $m$ vectors $\overline{x^{0}},\overline{x^{1}},\ldots
,\overline{x^{m-1}}$ form a basis of the $R$-module $R\left[  x\right]  /b$.
In particular, this $R$-module is free of rank $m$.

\textbf{(c)} Assume that $m>0$. Then, the $R$-algebra morphism%
\begin{align*}
R  &  \rightarrow R\left[  x\right]  /b,\\
r  &  \mapsto\overline{r}%
\end{align*}
is injective. Therefore, $R$ can be viewed as an $R$-subalgebra of $R\left[
x\right]  /b$ (by identifying each $r\in R$ with its image $\overline{r}\in
R\left[  x\right]  /b$).

\textbf{(d)} Thus, under the assumption that $m>0$, there exists a commutative
ring that contains $R$ as a subring and that contains a root of $b$.
\end{theorem}

\begin{proof}
Again, I refer to the text (Theorem 4.5.9). Part \textbf{(a)} is the
\textquotedblleft hard\textquotedblright\ part, but really a fairly simple
consequence of division-with-remainder for polynomials. Part \textbf{(b)} is
just a restatement of part \textbf{(a)}. Part \textbf{(c)} follows from part
\textbf{(b)} (since $\overline{r}=0$ would mean $r\cdot\overline{x^{0}}%
+0\cdot\overline{x^{1}}+0\cdot\overline{x^{2}}+\cdots+0\cdot\overline{x^{m-1}%
}=0$, but this would violate linear independence of the basis). Part
\textbf{(d)} follows from part \textbf{(c)}.
\end{proof}

Let us summarize: Generalizing Cardano's construction of $\mathbb{C}$, we have
found a way to \textquotedblleft adjoin\textquotedblright\ a (new) root of a
given polynomial $b\in R\left[  x\right]  $ to a given commutative ring $R$.
The resulting commutative ring $R\left[  x\right]  /b$ is always a commutative
$R$-algebra and always contains a root of $b$. When $b$ is non-constant (i.e.,
has positive degree) and has an invertible leading coefficient, this ring
$R\left[  x\right]  /b$ furthermore contains a copy of $R$ as a subring. This
conclusion sometimes remains true even if the leading coefficient of $b$ is
not invertible (e.g., in our $\mathbb{Z}\left[  x\right]  /\left(
mx-1\right)  \cong R_{m}$ example for $m\neq0$), but such claims always
require separate proofs.

\subsection{Field extensions from adjoining roots}

So far so good. But when is $R\left[  x\right]  /b$ a field?

First, we restrict ourselves to the case when $R$ is a field (otherwise, it
happens very rarely). Accordingly, we will call it $F$ instead of $R$.

We begin with two simple facts about polynomials over a field:

\begin{proposition}
\label{prop.fieldext.pol-units}Let $F$ be a field. Then, the units of the
polynomial ring $F\left[  x\right]  $ are the nonzero constant polynomials.
\end{proposition}

\begin{proof}
Easy! (See Proposition 4.6.1 in the text.)
\end{proof}

\begin{proposition}
\label{prop.fieldext.pol-irr}Let $F$ be a field. Let $b\in F\left[  x\right]
$ be a polynomial. Then, $b$ is irreducible (in the sense of Definition
1.14.13 \textbf{(a)} in Lecture 16) if and only if $b$ is an irreducible
polynomial in the classical sense (i.e., if $b$ is non-constant and cannot be
written as a product of two non-constant polynomials).
\end{proposition}

\begin{proof}
Easy! (See Proposition 4.6.2 in the text.)
\end{proof}

There is an easy criterion for polynomials of degree $\leq3$ to be irreducible:

\begin{proposition}
\label{prop.polring.univar.irred123}Let $F$ be a field. Let $b\in F\left[
x\right]  $ be a polynomial such that $2\leq\deg b\leq3$. Then, $b$ is
irreducible if and only if $b$ has no root in $F$.
\end{proposition}

\begin{proof}
$\Longrightarrow:$ Assume that $b$ is irreducible. If $b$ has a root $r$ in
$F$, then $x-r\mid b$ in $F\left[  x\right]  $ (by Proposition 3.3.11 in
Lecture 26), which entails that $b=\left(  x-r\right)  \cdot q$ for some $q\in
F\left[  x\right]  $, and thus $b$ is not irreducible (because degree
considerations show that $\deg q=\underbrace{\deg b}_{\geq2}-\,1\geq2-1=1$, so
that $q$ is not a unit, and of course $x-r$ is not a unit either). But we
assumed that $b$ is irreducible. Thus, $b$ cannot have a root in $F$. The
\textquotedblleft$\Longrightarrow$\textquotedblright\ direction of Proposition
\ref{prop.polring.univar.irred123} is thus proved.

$\Longleftarrow:$ Assume that $b$ has no root in $F$. We must show that $b$ is
irreducible. In other words, we must prove that $b$ cannot be written as a
product of two non-constant polynomials (since $\deg b\geq2$ shows that $b$ is
not constant).

Assume the contrary. Thus, $b=uv$ for two non-constant polynomials $u,v\in
F\left[  x\right]  $. Consider these polynomials $u$ and $v$. They satisfy
$\deg u\geq1$ and $\deg v\geq1$ (since they are non-constant). However, from
$b=uv$, we obtain $\deg b=\deg\left(  uv\right)  =\deg u+\deg v$ (by
Proposition 3.3.5 \textbf{(c)} in Lecture 15, since $F$ is a field and thus an
integral domain). If both $\deg u$ and $\deg v$ were $\geq2$, then this would
entail $\deg b=\underbrace{\deg u}_{\geq2}+\underbrace{\deg v}_{\geq2}%
\geq2+2=4$, which would contradict $\deg b\leq3<4$. Thus, $\deg u$ and $\deg
v$ cannot both be $\geq2$. Hence, at least one of $\deg u$ and $\deg v$ is at
most $1$. We WLOG assume that $\deg u\leq1$ (otherwise, swap $u$ with $v$).
Then, $\deg u=1$ (since $\deg u\leq1$ and $\deg u\geq1$). In other words,
$u=\alpha x+\beta$ for some $\alpha,\beta\in F$ with $\alpha\neq0$. Consider
these $\alpha,\beta$. Since $F$ is a field, $\alpha\in F$ is a unit (since
$\alpha\neq0$), so that $-\dfrac{\beta}{\alpha}$ is a well-defined element of
$F$. From $u=\alpha x+\beta$, we obtain $u\left[  -\dfrac{\beta}{\alpha
}\right]  =\alpha\left(  -\dfrac{\beta}{\alpha}\right)  +\beta=0$. Now,
substituting $-\dfrac{\beta}{\alpha}$ for $x$ into $b=uv$, we obtain%
\[
b\left[  -\dfrac{\beta}{\alpha}\right]  =\underbrace{u\left[  -\dfrac{\beta
}{\alpha}\right]  }_{=0}\cdot\,v\left[  -\dfrac{\beta}{\alpha}\right]  =0.
\]
This shows that $-\dfrac{\beta}{\alpha}$ is a root of $b$. Hence, $b$ has a
root in $F$, contradicting our assumption that $b$ has no root in $F$. This
contradiction shows that our last assumption was false. This completes the
proof of the \textquotedblleft$\Longleftarrow$\textquotedblright\ direction of
Proposition \ref{prop.polring.univar.irred123}.
\end{proof}

Note that Proposition \ref{prop.polring.univar.irred123} would fail if $\deg
b$ was allowed to be $4$. For instance, the degree-$4$ polynomial $x^{4}%
+4\in\mathbb{Q}\left[  x\right]  $ has no roots in $\mathbb{Q}$ (or in
$\mathbb{R}$, for that matter), but it is not irreducible\footnote{In fact,
$x^{4}+4=\left(  x^{2}+2x+2\right)  \left(  x^{2}-2x+2\right)  $.}.

Now we come to the actually crucial result of this section:

\begin{theorem}
\label{thm.fieldext.field-irr}Let $F$ be a field. Let $b\in F\left[  x\right]
$ be a nonzero polynomial. Then, the ring $F\left[  x\right]  /b$ is a field
if and only if $b$ is irreducible.
\end{theorem}

\begin{proof}
This is the polynomial analogue of the fact that $\mathbb{Z}/n$ is a field
(for a given positive integer $n$) if and only if $n$ is prime (part of
Corollary 1.6.2 in Lecture 5). Our proof of the latter fact can be adapted to
Theorem \ref{thm.fieldext.field-irr} (with some changes: most importantly,
\textquotedblleft the numbers $1,2,\ldots,n-1$\textquotedblright\ must be
replaced by \textquotedblleft the nonzero polynomials of degree $<\deg
b$\textquotedblright, and we need Bezout's theorem for polynomials instead of
Bezout's theorem for integers\footnote{\textbf{Bezout's theorem for
polynomials} says that if $a$ and $b$ are two polynomials in $F\left[
x\right]  $, then $\gcd\left(  a,b\right)  $ (more precisely, any gcd of $a$
and $b$) can be written in the form $ua+vb$ for some polynomials $u,v\in
F\left[  x\right]  $. This follows easily from the fact that the polynomial
ring $F\left[  x\right]  $ is a PID (which, in turn, follows from Theorem
3.3.16 in Lecture 26).}). Alternatively, you can find the proof of the
\textquotedblleft$\Longleftarrow$\textquotedblright\ direction in the text
(Theorem 4.6.4), and that is the only direction we will need.
\end{proof}

Here is an instructive example:

\begin{itemize}
\item The polynomial $x^{2}+\overline{1}\in\left(  \mathbb{Z}/3\right)
\left[  x\right]  $ is irreducible. (This is easily checked using Proposition
\ref{prop.polring.univar.irred123}.)

Thus, Theorem \ref{thm.fieldext.field-irr} yields that the ring
\[
\left(  \mathbb{Z}/3\right)  \left[  x\right]  /\left(  x^{2}+\overline
{1}\right)
\]
is a field. Theorem \ref{thm.fieldext.basis-monic} \textbf{(b)} tells us that
this field is a free $\mathbb{Z}/3$-module of rank $2$ (with basis $\left(
\overline{x^{0}},\overline{x^{1}}\right)  $). In other words, it is isomorphic
to $\left(  \mathbb{Z}/3\right)  ^{2}$ as a $\mathbb{Z}/3$-module. So its size
is%
\[
\left\vert \left(  \mathbb{Z}/3\right)  ^{2}\right\vert =3^{2}=9.
\]


Thus, we have found a field with $9$ elements. (Aside: This field is also
isomorphic to the quotient ring $\mathbb{Z}\left[  i\right]  /3$, so there is
a second way to get it. In a sense, this is not surprising: $\mathbb{Z}\left[
i\right]  /3$ is obtained from $\mathbb{Z}$ by first adjoining a square root
of $-1$ and then quotienting out all multiples of $3$, whereas $\left(
\mathbb{Z}/3\right)  \left[  x\right]  /\left(  x^{2}+\overline{1}\right)  $
is obtained from $\mathbb{Z}$ by first quotienting out all multiples of $3$
and then adjoining a square root of $-1$. It stands to reason that the results
should be the same both times, although of course this should be rigorously proved.)
\end{itemize}

\section{Finite fields}

So we have found a field with $9$ elements (in addition to all our finite
fields $\mathbb{Z}/p$ with $p$ elements for prime numbers $p$). What other
finite fields can we find?

A moment of thought reveals that we can find a finite field with $p^{2}$
elements for any prime $p\equiv3\operatorname{mod}4$ using the same
construction as our $9$-element field (since $p\equiv3\operatorname{mod}4$
entails that $\overline{-1}\in\mathbb{Z}/p$ is not a square\footnote{This is
implicit in the solution to Exercise 1 \textbf{(b)} on midterm \#2.}, and thus
the polynomial $x^{2}+\overline{1}\in\left(  \mathbb{Z}/p\right)  \left[
x\right]  $ is irreducible\footnote{again using Proposition
\ref{prop.polring.univar.irred123}}). What else can we find?

We can be more flexible and replace our polynomial $x^{2}+\overline{1}$ by
other degree-$2$ polynomials. It is not hard to show that for each prime $p$,
there exists an irreducible degree-$2$ polynomial in $\left(  \mathbb{Z}%
/p\right)  \left[  x\right]  $: Indeed, for $p=2$, you can take the polynomial
$x^{2}+x+\overline{1}$. For odd $p$, you can find some element $u$ of
$\mathbb{Z}/p$ that is not a square (exercise!), and use the polynomial
$x^{2}-u$. Thus, you find an irreducible degree-$2$ polynomial $b\in\left(
\mathbb{Z}/p\right)  \left[  x\right]  $, and then the quotient ring $\left(
\mathbb{Z}/p\right)  \left[  x\right]  /b$ is a finite field with $p^{2}$
elements. (See Proposition 5.1.1 in the text for the details of this construction.)

Thus we found finite fields of size $p^{2}$ for any prime $p$. What about
other sizes? What about a finite field of size $6$ or $8$ or $24$ ?

\subsection{The characteristic of a field}

Let us first discuss finite fields whose size is not a prime power (but a
number such as $6$ or $10$). Do such fields exist?

The answer is \textquotedblleft no\textquotedblright. Each finite field has
its \textquotedblleft favorite prime\textquotedblright, and its size is a
power of that prime. This prime is the so-called \textquotedblleft
characteristic\textquotedblright\ of the field, and is defined as follows (not
just for finite fields):

\begin{definition}
\label{def.finfield.char}Let $F$ be a field. The \textbf{characteristic} of
$F$ is an integer called $\operatorname*{char}F$, defined as follows:

\begin{itemize}
\item If there exists a positive integer $n$ such that $n\cdot1_{F}=0_{F}$,
then $\operatorname*{char}F$ is the \textbf{smallest} such $n$.

\item If such an $n$ does not exist, then $\operatorname*{char}F$ is defined
to be $0$.
\end{itemize}
\end{definition}

Roughly speaking, $\operatorname*{char}F$ is \textquotedblleft how often you
have to add $1_{F}$ to itself to circle back to $0_{F}$\textquotedblright%
\ (except that you declare it to be $0$ if this never happens). Examples:

\begin{itemize}
\item We have $\operatorname*{char}\left(  \mathbb{Z}/p\right)  =p$ for any
prime $p$.

\item Our finite field of size $9$ has characteristic $3$.

\item We have $\operatorname*{char}\mathbb{Q}=0$.
\end{itemize}

The characteristic of a field $F$ \textquotedblleft knows a
lot\textquotedblright\ about $F$. Here is some of that:

\begin{theorem}
\label{thm.finfield.char.basics}Let $F$ be a field. Let
$p=\operatorname*{char}F$. Then:

\textbf{(a)} We have $pa=0$ for each $a\in F$.

\textbf{(b)} The field $F$ is a $\mathbb{Z}/p$-algebra (with action given by
$\overline{n}\cdot a=na$ for all $n\in\mathbb{Z}$ and $a\in F$).

\textbf{(c)} The number $p$ is either prime or $0$.

\textbf{(d)} If $F$ is finite, then $p$ is a prime.

\textbf{(e)} If $F$ is finite, then $\left\vert F\right\vert =p^{m}$ for some
positive integer $m$.

\textbf{(f)} If $p$ is a prime, then $F$ contains \textquotedblleft a copy of
$\mathbb{Z}/p$\textquotedblright\ (that is, a subring isomorphic to
$\mathbb{Z}/p$).

\textbf{(g)} If $p=0$, then $F$ contains \textquotedblleft a copy of
$\mathbb{Q}$\textquotedblright\ (that is, a subring isomorphic to $\mathbb{Q}%
$): namely, the map%
\begin{align*}
\mathbb{Q}  &  \rightarrow F,\\
\dfrac{a}{b}  &  \mapsto\dfrac{a\cdot1_{F}}{b\cdot1_{F}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{for }a,b\in\mathbb{Z}\text{ with }%
b\neq0\right)
\end{align*}
is an injective ring morphism.
\end{theorem}

\begin{proof}
See Theorem 5.2.2 in the text.

(In a nutshell: Part \textbf{(a)} follows from the $pa=\underbrace{p\cdot
1_{F}}_{=0_{F}}\cdot\,a=0_{F}\cdot a=0$ trick. Part \textbf{(b)} follows
easily using part \textbf{(a)} -- you just need to show that $na$ depends only
on $\overline{n}\in\mathbb{Z}/p$ and not on $n\in\mathbb{Z}$. Part
\textbf{(c)} uses the fact that $F$ is an integral domain to argue that
$\operatorname*{char}F$ cannot be composite (e.g., if $\operatorname*{char}F$
was $6$, then we would have $\left(  2\cdot1_{F}\right)  \cdot\left(
3\cdot1_{F}\right)  =6\cdot1_{F}=0_{F}$, which would lead to $2\cdot1_{F}$ or
$3\cdot1_{F}$ being zero, but this would mean that $\operatorname*{char}F$ is
actually $2$ or $3$ rather than $6$). Part \textbf{(d)} follows from a
pigeonhole argument or from Lagrange's theorem from group theory. Part
\textbf{(e)} follows from parts \textbf{(b)} and \textbf{(d)}, since a
$\mathbb{Z}/p$-algebra is always a $\mathbb{Z}/p$-module, and every
$\mathbb{Z}/p$-module is free (because $p$ is prime) and thus isomorphic to
$\left(  \mathbb{Z}/p\right)  ^{m}$ for some $m\in\mathbb{N}$. Part
\textbf{(f)} follows easily from part \textbf{(b)} and the definition of a
characteristic. Part \textbf{(g)} is a bit trickier, but will not be used.)
\end{proof}

Parts \textbf{(e)} and \textbf{(f)} of Theorem \ref{thm.finfield.char.basics}
entail that the size of any finite field is a power of a prime (and said prime
is the characteristic of the field). Thus, a finite field cannot have size $6$
or $10$ or $15$ for example.

This leaves the question of prime powers. We know that finite fields of sizes
$p$ and $p^{2}$ exist for any prime $p$, but what about $p^{3}$ or $p^{4}$ or
$p^{29}$ ?

In general, if we have a prime $p$ and a positive integer $m$, and if we know
any irreducible polynomial of degree $m$ over $\mathbb{Z}/p$, then we easily
obtain a finite field of size $p^{m}$ (by adjoining a root of this
polynomial). But is there such an irreducible polynomial?

Fortunately, such irreducible polynomials do always exist. Proving this is not
at all easy, but can be done. Alternatively, there is another way to construct
finite fields of any given prime-power size.

Let me just state the result:

\begin{theorem}
\label{thm.finfield.ex-uni}\textbf{(a)} For any prime $p$ and any positive
integer $m$, there exists a finite field of size $p^{m}$.

\textbf{(b)} Any two finite fields of the same size are isomorphic.
\end{theorem}

How do these finite fields look like? Theoretically, the answer is clear: Pick
an irreducible polynomial $b$ of degree $m$ over $\mathbb{Z}/p$, and adjoin
its root to $\mathbb{Z}/p$ (that is, form the quotient ring $\left(
\mathbb{Z}/p\right)  \left[  x\right]  /b$). No explicit formula for such a
polynomial is known (except in specific cases like $m=p$). Finding a nice
explicit construction of a finite field of size $p^{m}$ is an open problem!
The brute-force approach (try every degree-$m$ polynomial until you find an
irreducible one) might sound stupid, but is not too far from the best known algorithm.

\begin{proof}
[Proof of Theorem \ref{thm.finfield.ex-uni}.]See the text (\S 5.2 till \S 5.4
for part \textbf{(a)}; \S 5.5 for part \textbf{(b)}).
\end{proof}

Sadly, our treatment of finite fields has to stop at this strange and
unsatisfactory place. More can be found in Chapter 5 of the text, and much
more in dedicated books on the subject, such as \cite{MulMum07} (an
introduction) and \cite{LidNie00} (a comprehensive reference).

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[LidNie00]{LidNie00}%
\href{https://doi.org/10.1017/CBO9780511525926}{Rudolf Lidl, Harald
Niederreiter, \textit{Finite fields}, Encyclopedia of Mathematics and its
Applications \textbf{20}, Cambridge University Press, 2nd edition 1997.}

\bibitem[MulMum07]{MulMum07}%
\href{https://bookstore.ams.org/view?ProductCode=STML/41}{Gary L. Mullen, Carl
Mummert, \textit{Finite Fields and Applications}, Student Mathematical Library
\textbf{41}, AMS 2007.}
\end{thebibliography}


\end{document}