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\begin{document}
\section*{Math 332 Winter 2023, Lecture 27: 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}{3}

\subsection{Intermezzo: Quotients of $R$-algebras}

A ring can be quotiented modulo an ideal.\footnote{That is: If $S$ is a ring
and $I$ is an ideal of $S$, then there is a quotient ring $S/I$.}

An $R$-module can be quotiented modulo a submodule.\footnote{That is: If $M$
is an $R$-module and $I$ is an $R$-submodule of $M$, then there is a quotient
$R$-module $M/I$.}

Thus, you would expect that an $R$-algebra (which, by its very definition, is
always a ring and an $R$-module at the same) can be quotiented modulo an
appropriate \textquotedblleft ideal-submodule\textquotedblright\ (i.e., an
ideal that happens to be an $R$-submodule as well).

This is indeed the case, and it's even simpler than that, because we don't
even need to define these \textquotedblleft ideal-submodules\textquotedblright%
; instead, any ideal will do, since any ideal of an $R$-algebra is
automatically an $R$-submodule. Let us state this all as a theorem:

\begin{theorem}
\label{thm.algs.quotients}Let $A$ be an $R$-algebra. Let $I$ be an ideal of
$A$. Then:

\textbf{(a)} The ideal $I$ is also an $R$-submodule of $A$.

\textbf{(b)} The quotient ring $A/I$ and the quotient $R$-module $A/I$ fit
together to form an $R$-algebra.

\textbf{(c)} The canonical projection $\pi:A\rightarrow A/I$ (which sends each
$a\in A$ to its residue class $\overline{a}=a+I$) is an $R$-algebra morphism
(from the original $R$-algebra $A$ to the $R$-algebra $A/I$ that we just
constructed in part \textbf{(b)}).
\end{theorem}

\begin{proof}
\textbf{(a)} We need to prove that $I$ is closed under scaling. But this
follows from the fact that
\[
r\underbrace{i}_{=1_{A}\cdot i}=r\left(  1_{A}\cdot i\right)  =\left(
r\cdot1_{A}\right)  i\ \ \ \ \ \ \ \ \ \ \text{for all }r\in R\text{ and }i\in
I
\]
(and from the fact that $I$ is an ideal of $A$, so that $\left(  r\cdot
1_{A}\right)  i\in I$).

\textbf{(b)}, \textbf{(c)} follow using standard techniques.
\end{proof}

Quotient $R$-algebras also have a universal property:

\begin{theorem}
[Universal property of quotient algebras, elementwise form]%
\label{thm.quotalg.uniprop1}Let $A$ be an $R$-algebra. Let $I$ be an ideal of
$A$.

Let $B$ be an $R$-algebra. Let $f:A\rightarrow B$ be an $R$-algebra morphism.
Assume that $f\left(  I\right)  =0$ (by this we mean that $f\left(  i\right)
=0$ for all $i\in I$). Then, the map%
\begin{align*}
f^{\prime}:A/I  &  \rightarrow B,\\
\overline{a}  &  \mapsto f\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{for all }a\in A\right)
\end{align*}
is well-defined (i.e., the value $f\left(  a\right)  $ depends only on the
residue class $\overline{a}$, not on $a$ itself) and is an $R$-algebra morphism.
\end{theorem}

\begin{proof}
This is analogous to our proof of the universal property of quotient rings
(Theorem 1.9.6 in Lecture 10). This time, we just need to discuss the scaling
along with the addition and the multiplication (but it is just as easy to handle).
\end{proof}

\subsection{Adjoining roots}

\subsubsection{A notation for quotients by principal ideals}

We introduce a convenient shorthand notation for quotient rings modulo
principal ideals:

\begin{convention}
\label{conv.quotring.S/a}Let $S$ be any commutative ring, and let $a\in S$.
Then, $S/a$ shall mean the quotient ring $S/aS$.
\end{convention}

This generalizes our shorthand notation $\mathbb{Z}/n$ for the quotient ring
$\mathbb{Z}/n\mathbb{Z}$ (when $n$ is an integer). This shorthand becomes
particularly useful when $S$ is not $\mathbb{Z}$ but something more
complicated like a polynomial ring, so that you really don't want to write $S$ twice.

We recall briefly the intuition behind quotient rings: If $I$ is an ideal of a
ring $R$, then $R/I$ is \textquotedblleft what becomes of $R$ if we equate all
elements of $I$ to zero\textquotedblright. Thus, in particular, the ring $S/a$
in Convention \ref{conv.quotring.S/a} is \textquotedblleft what becomes of $S$
if we equate all multiples of $a$ to zero\textquotedblright, or, even more
simply, \textquotedblleft what becomes of $S$ if we equate $a$ to
zero\textquotedblright\ (since equating $a$ to zero automatically forces all
multiples of $a$ to become $0$ as well). For example, $\mathbb{Z}/7$ is what
becomes of $\mathbb{Z}$ if we equate $7$ to zero.

\subsubsection{Why adjoin roots}

We now come to one of the most important applications of polynomials to
algebra itself: a way to \textquotedblleft adjoin\textquotedblright\ roots of
a polynomial to a given commutative ring (i.e., to \textquotedblleft
create\textquotedblright\ roots out of thin air).

The classical example for this is the invention of the complex numbers. This
idea is somewhat obscured by the fact that nowadays, complex numbers are
usually defined in a down-to-earth, explicit way (viz., as pairs of real
numbers\footnote{with addition being entrywise, multiplication being defined
by the rule $\left(  a,b\right)  \left(  c,d\right)  =\left(
ac-bd,\ ad+bc\right)  $, and so on}) that eschews any use of
polynomials\footnote{I think this modern definition is due to Hamilton in the
19th century.}. However, this was not always the case. Gerolamo Cardano, back
in the 16th century, introduced complex numbers as a tool for solving cubic
equations. Specifically, he had a formula (due to Scipione del Ferro) for
solving a cubic equation ($ax^{3}+bx^{2}+cx+d=0$) that involved square roots.
Algebraically, the formula worked well, but quite forbiddingly, a negative
number would often appear under these square roots, even though the original
cubic equation had three real solutions! It was this vexing situation (and not
some abstract curiosity along the lines of \textquotedblleft what kind of
other numbers could we cook up?\textquotedblright) that prompted Cardano to
ponder the possibility of imaginary (and thus complex) numbers.

Specifically, Cardano imagined an \textquotedblleft imaginary
unit\textquotedblright\ $i$ that satisfies $i^{2}=-1$. This new
\textquotedblleft number\textquotedblright\ would then entail further
\textquotedblleft numbers\textquotedblright\ (by forming sums and products of
this \textquotedblleft number\textquotedblright\ with existing real numbers).
Since $i^{2}=-1$, all these new \textquotedblleft numbers\textquotedblright%
\ can actually be simplified to the form $a+bi$ with $a,b\in\mathbb{R}$, so it
is like an extra degree of freedom is getting added to the real numbers.

Such flights of fancy are often dangerous; after all, one could equally well
imagine an \textquotedblleft infinite number\textquotedblright\ $\infty$
satisfying $0\cdot\infty=1$, but this new \textquotedblleft
number\textquotedblright\ would cause the number system to collapse (i.e., all
numbers to become equal), since it would entail $1+1=0\cdot\infty+0\cdot
\infty=\underbrace{\left(  0+0\right)  }_{=0}\cdot\,\infty=0\cdot\infty=1$ and
therefore $1=0$ and so on. There is a good reason why the modern definition of
complex numbers (as pairs of real numbers) is preferrable.

\subsubsection{How to adjoin roots}

Yet, the ability to invent new numbers satisfying desired equalities is a good
power to have, and it would be great if we could tell when such an invention
is harmless (as opposed to collapsing some of the existing numbers). So let us
try to put it on a rigorous footing. What does it mean to introduce a new number?

The simplest case is when we want to introduce a new number $x$ that satisfies
no relations (other than the ring axioms). That just means we work in the
polynomial ring $\mathbb{R}\left[  x\right]  $. So our \textquotedblleft new
number\textquotedblright\ is just the indeterminate of a polynomial ring.

\begin{noncompile}
More generally, if we start with an arbitrary commutative ring $R$,
introducing our new \textquotedblleft number\textquotedblright\ $x$ takes us
into the polynomial ring $R\left[  x\right]  $.
\end{noncompile}

However, in more interesting cases (such as Cardano's), we also want our new
number to satisfy some equalities, such as $i^{2}=-1$ (in the case of the
imaginary unit). The indeterminate $x$ in the ring $\mathbb{R}\left[
x\right]  $ does not satisfy any such equalities (beyond the ones that follow
from the ring axioms). But we can use quotient rings to make it satisfy
whatever we want! Recall that a quotient ring $S/a$ (where $S$ is a
commutative ring and $a\in S$ is an element) is essentially \textquotedblleft
what becomes of $S$ if we equate $a$ to $0$\textquotedblright. Thus, if we
want our indeterminate $x$ to satisfy $x^{2}=-1$, then we just have to take
the quotient ring $\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)  $ of
our polynomial ring $\mathbb{R}\left[  x\right]  $ (thus equating $x^{2}+1$
with $0$). The residue class $\overline{x}$ of $x$ in this quotient ring
$\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)  $ will then satisfy
$\overline{x}^{2}+\overline{1}=\overline{x^{2}+1}=0$, so that $\overline
{x}^{2}=-\overline{1}$, which means that it is an \textquotedblleft imaginary
unit\textquotedblright. So the quotient ring $\mathbb{R}\left[  x\right]
/\left(  x^{2}+1\right)  $ is a rigorous interpretation of Cardano's
suggestion to introduce a new \textquotedblleft number\textquotedblright\ $i$
satisfying $i^{2}=-1$.

This method generalizes to an arbitrary commutative ring $R$ instead of
$\mathbb{R}$, and to an arbitrary polynomial $b\in R\left[  x\right]  $. In
general, if we start with any commutative ring $R$ and any polynomial $b\in
R\left[  x\right]  $, then the quotient ring $R\left[  x\right]  /b$ has an
element $\overline{x}$ (the residue class of the polynomial $x$) that is a
root of $b$ (we will give the easy proof of this soon). This quotient ring
$R\left[  x\right]  /b$ is not just a ring, but actually a commutative
$R$-algebra (by Theorem \ref{thm.algs.quotients} \textbf{(b)}); each element
$r\in R$ gives rise to an element $\overline{r}\in R\left[  x\right]  /b$ (the
residue class of the constant polynomial $r\in R\subseteq R\left[  x\right]
$). As we said, it may happen that some of the existing elements of $R$
\textquotedblleft collapse\textquotedblright\ in this quotient ring $R\left[
x\right]  /b$ (that is, we might have $\overline{r}=\overline{s}$ in $R\left[
x\right]  /b$ for two distinct elements $r$ and $s$ of $R$), but such
\textquotedblleft collapses\textquotedblright\ are natural and cannot be
prevented. In the next lecture, we will see a sufficient criterion for when
such collapses don't happen.

\subsubsection{Some examples}

Let us first see some examples of this construction.

As we just said, Cardano's complex numbers are the elements of the quotient
ring $\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)  $, whereas the
modern complex numbers are the elements of the ring $\mathbb{C}$ (defined as
pairs of real numbers). Let us now show that these two rings are isomorphic
(better yet, that these two $\mathbb{C}$-algebras are isomorphic):

\begin{proposition}
\label{prop.fieldext.C}We have
\[
\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)  \cong\mathbb{C}%
\ \ \ \ \ \ \ \ \ \ \text{as }\mathbb{R}\text{-algebras.}%
\]
Concretely: There is an $\mathbb{R}$-algebra isomorphism%
\begin{align*}
\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)   &  \rightarrow
\mathbb{C},\\
\overline{p}  &  \mapsto p\left[  i\right]  .
\end{align*}

\end{proposition}

\begin{proof}
[Proof sketch.]Here is a six-step procedure to prove this claim (and, more
generally, to prove any claim like this):

\begin{enumerate}
\item Give a putative definition of the alleged isomorphism.

\item Prove that this definition actually defines a map (or, to put it
colloquially, that \textquotedblleft the map is well-defined\textquotedblright).

\item Prove that this map is an $\mathbb{R}$-algebra morphism.

\item Prove that this map is injective.

\item Prove that this map is surjective.

\item Conclude that this map is an $\mathbb{R}$-algebra isomorphism (since
every invertible $\mathbb{R}$-algebra morphism is an $\mathbb{R}$-algebra isomorphism).
\end{enumerate}

Let us just say a few words about these six steps.

Step 1 has already been done in the statement of Proposition
\ref{prop.fieldext.C}: Our map is defined to be the map%
\begin{align*}
\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)   &  \rightarrow
\mathbb{C},\\
\overline{p}  &  \mapsto p\left[  i\right]  .
\end{align*}


For Step 2, we need to prove that if two polynomials $p,q\in\mathbb{R}\left[
x\right]  $ satisfy $\overline{p}=\overline{q}$, then $p\left[  i\right]
=q\left[  i\right]  $. Let us do this. Let $p,q\in\mathbb{R}\left[  x\right]
$ be two polynomials that satisfy $\overline{p}=\overline{q}$. Then,
$p-q\in\left(  x^{2}+1\right)  \mathbb{R}\left[  x\right]  $, so that
$p-q=\left(  x^{2}+1\right)  r$ for some $r\in\mathbb{R}\left[  x\right]  $.
Consider this $r$. Substituting $i$ for $x$ in the equality $p-q=\left(
x^{2}+1\right)  r$, we obtain%
\[
p\left[  i\right]  -q\left[  i\right]  =\underbrace{\left(  i^{2}+1\right)
}_{\substack{=0\\\text{(since }i^{2}=-1\text{)}}}r\left[  i\right]  =0,
\]
so that $p\left[  i\right]  =q\left[  i\right]  $. Thus, Step 2 has been completed.

Step 3 is straightforward (since addition, multiplication and scaling on
$\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)  $ are defined by
$\overline{a}+\overline{b}=\overline{a+b}$ and $\overline{a}\cdot\overline
{b}=\overline{ab}$ and $r\overline{a}=\overline{ra}$).

Alternatively, Steps 2 and 3 can be handled in one swoop using the universal
property of quotient $R$-algebras (Theorem \ref{thm.quotalg.uniprop1}). This
is how I do it in the text (proof of Proposition 4.5.2).

Now we come to Step 4 (injectivity): We need to prove that the map%
\begin{align*}
\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)   &  \rightarrow
\mathbb{C},\\
\overline{p}  &  \mapsto p\left[  i\right]
\end{align*}
is injective. In other words, we need to prove that its kernel is $\left\{
0\right\}  $ (since it is an $\mathbb{R}$-linear map, and we
know\footnote{from Lemma 2.4.9 in Lecture 20} that a linear map is injective
if its kernel is $\left\{  0\right\}  $). In other words, we need to prove
that if a polynomial $p\in\mathbb{R}\left[  x\right]  $ satisfies $p\left[
i\right]  =0$, then $\overline{p}=0$ (that is, $p$ is a multiple of $x^{2}+1$).

So let $p\in\mathbb{R}\left[  x\right]  $ be a polynomial that satisfies
$p\left[  i\right]  =0$. Let us apply the division-with-remainder theorem for
polynomials (Theorem 3.3.8 \textbf{(a)} in Lecture 25) to $R=\mathbb{R}$ and
$b=x^{2}+1$ and $a=p$ (we can do this, since the leading coefficient of
$x^{2}+1$ is a unit). We conclude that there is a \textbf{unique} pair
$\left(  q,r\right)  $ of polynomials in $\mathbb{R}\left[  x\right]  $ such
that%
\[
p=q\cdot\left(  x^{2}+1\right)  +r\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \deg r<\deg\left(  x^{2}+1\right)  .
\]
Consider this pair $\left(  q,r\right)  $. From $p=q\cdot\left(
x^{2}+1\right)  +r$, we obtain $p-r=q\cdot\left(  x^{2}+1\right)  \in\left(
x^{2}+1\right)  \cdot\mathbb{R}\left[  x\right]  $, so that $\overline
{p}=\overline{r}$ in $\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)  $.
Hence, $p\left[  i\right]  =r\left[  i\right]  $ (by the same argument that we
used in Step 2 above, but now with $r$ instead of $q$). Hence, $r\left[
i\right]  =p\left[  i\right]  =0$.

From $\deg r<\deg\left(  x^{2}+1\right)  =2$, we see that $\deg r\leq1$, so
that $r$ is a polynomial of degree $\leq1$. In other words, $r=a+bx$ for some
$a,b\in\mathbb{R}$. Consider these $a,b$. Now, our map (whose injectivity we
are currently proving) sends $\overline{r}$ to $\underbrace{r}_{=a+bx}\left[
i\right]  =\left(  a+bx\right)  \left[  i\right]  =a+bi$. Thus, we find
$a+bi=r\left[  i\right]  =0$. Therefore, $a=b=0$ (since $a,b\in\mathbb{R}$).
Thus, $r=0$ (since $r=a+bx$). Hence, $\overline{p}=\overline{r}=\overline
{0}=0$, as we desired to prove. Thus, we are done with Step 4.

Step 5 is again almost trivial: Every complex number $\left(  a,b\right)
\in\mathbb{C}$ can be written as%
\[
\left(  a,b\right)  =a+bi=\left(  a+bx\right)  \left[  i\right]  ,
\]
which is the image of the residue class $\overline{a+bx}$ under our map. Thus,
our map is surjective.

Finally, Step 6 is automatic: Since our map is injective and surjective, it is
bijective, i.e., invertible. But an invertible $\mathbb{R}$-algebra morphism
is automatically an $\mathbb{R}$-algebra isomorphism (by a result from
\S 2.9.6 in Lecture 22).

Thus, all six steps have been made, so that Proposition \ref{prop.fieldext.C}
is proved.
\end{proof}

There are many other results like Proposition \ref{prop.fieldext.C}, revealing
some known and unknown rings as quotients of polynomial rings. Here is a
selection (see \S 4.5.1 in the text for proofs):

\begin{proposition}
\textbf{(a)} Recall the ring $\mathbb{Z}\left[  i\right]  $ of Gaussian
integers. Then,%
\[
\mathbb{Z}\left[  x\right]  /\left(  x^{2}+1\right)  \cong\mathbb{Z}\left[
i\right]  \ \ \ \ \ \ \ \ \ \ \text{as }\mathbb{Z}\text{-algebras.}%
\]
More concretely, the map%
\begin{align*}
\mathbb{Z}\left[  x\right]  /\left(  x^{2}+1\right)   &  \rightarrow
\mathbb{Z}\left[  i\right]  ,\\
\overline{p}  &  \mapsto p\left[  i\right]
\end{align*}
is a $\mathbb{Z}$-algebra isomorphism.

\textbf{(b)} Recall the ring $\mathbb{Q}\left[  i\right]  $ of Gaussian
rationals. Then,%
\[
\mathbb{Q}\left[  x\right]  /\left(  x^{2}+1\right)  \cong\mathbb{Q}\left[
i\right]  \ \ \ \ \ \ \ \ \ \ \text{as }\mathbb{Q}\text{-algebras.}%
\]
More concretely, the map%
\begin{align*}
\mathbb{Q}\left[  x\right]  /\left(  x^{2}+1\right)   &  \rightarrow
\mathbb{Q}\left[  i\right]  ,\\
\overline{p}  &  \mapsto p\left[  i\right]
\end{align*}
is a $\mathbb{Q}$-algebra isomorphism.

\textbf{(c)} Recall the ring $\mathbb{S}=\mathbb{Q}\left[  \sqrt{5}\right]  $
from \S 1.1.2 (Lecture 2). Then, the map%
\begin{align*}
\mathbb{Q}\left[  x\right]  /\left(  x^{2}-5\right)   &  \rightarrow
\mathbb{S},\\
\overline{p}  &  \mapsto p\left[  \sqrt{5}\right]
\end{align*}
is a $\mathbb{Q}$-algebra isomorphism.
\end{proposition}

\begin{proposition}
\label{prop.fieldext.xx-1}We have%
\begin{align*}
\mathbb{Q}\left[  x\right]  /\left(  x^{2}-1\right)   &  \cong\mathbb{Q}%
\left[  C_{2}\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{the group algebra of
the cyclic group }C_{2}\right)  \\
&  \cong\left\{  \left(
\begin{array}
[c]{cc}%
a & b\\
b & a
\end{array}
\right)  \ \mid\ a,b\in\mathbb{Q}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{a
subring of }\mathbb{Q}^{2\times2}\right)  \\
&  \cong\mathbb{Q}\times\mathbb{Q}\ \ \ \ \ \ \ \ \ \ \left(  \text{a direct
product of two }\mathbb{Q}\text{s}\right)
\end{align*}
as $\mathbb{Q}$-algebras.
\end{proposition}

In the above examples, the quotient ring has always had the form $R\left[
x\right]  /b$, where $b$ is a non-constant polynomial whose leading
coefficient is a unit (actually $1$). This has the nice consequence that the
resulting ring contains a copy of the original ring $R$ as a subring (although
we have yet to prove this in full generality). Let us see what happens if the
leading coefficient is not a unit or $b$ is constant:

\begin{proposition}
\label{prop.fieldext.Zx/2}\textbf{(a)} For any integer $m$, we have
$\mathbb{Z}\left[  x\right]  /m\cong\left(  \mathbb{Z}/m\right)  \left[
x\right]  $ as $\mathbb{Z}$-algebras.

\textbf{(b)} The ring $\mathbb{Z}\left[  x\right]  /1$ is trivial.
\end{proposition}

\begin{proposition}
Fix a nonzero integer $m$. Then,%
\[
\mathbb{Z}\left[  x\right]  /\left(  mx-1\right)  \cong R_{m},
\]
where $R_{m}$ is the ring of $m$-integers as defined in Exercise 5 of homework
set \#1. Specifically, there is a $\mathbb{Z}$-algebra isomorphism%
\begin{align*}
\mathbb{Z}\left[  x\right]  /\left(  mx-1\right)   &  \rightarrow R_{m},\\
\overline{p}  &  \mapsto p\left[  \dfrac{1}{m}\right]  .
\end{align*}
(This is not quite trivial.)
\end{proposition}

\begin{proof}
See Proposition 4.5.6 in the text.
\end{proof}

As we already announced, the common thread here is that if you have a
commutative ring $R$ and a univariate polynomial $b\in R\left[  x\right]  $,
then the quotient ring $R\left[  x\right]  /b$ is like \textquotedblleft$R$
with a root of $b$ thrown in\textquotedblright. In fact, the residue class
$\overline{x}\in R\left[  x\right]  /b$ will be this new root of $b$. This
ring $R\left[  x\right]  /b$ is not always as nice as one might hope. In
particular, it can happen that distinct elements of $R$ become equal in
$R\left[  x\right]  /b$ (for example, $1\neq3$ in $\mathbb{Z}$, but
$\overline{1}=\overline{3}$ in $\mathbb{Z}\left[  x\right]  /2$), but often
enough this is not the case. When this does not happen, the subset $\left\{
\overline{r}\ \mid\ r\in R\right\}  $ of $R\left[  x\right]  /b$ is a subring
isomorphic to the original ring $R$, and thus can be viewed as a
\textquotedblleft copy of $R$\textquotedblright\ inside $R\left[  x\right]
/b$; thus, in this case, we can pretend that $R\left[  x\right]  /b$ contains
$R$ as a subring (i.e., we can pretend that $R\left[  x\right]  /b$ is an
extension of $R$).

Cardano was lucky in this sense: The complex numbers he introduced as
$\mathbb{R}\left[  x\right]  /\left(  x^{2}+1\right)  $ (even if he did not
write it this way) were really an extension of the real numbers $\mathbb{R}$;
distinct real numbers do not become equal in $\mathbb{R}\left[  x\right]
/\left(  x^{2}+1\right)  $. We are not that lucky in Proposition
\ref{prop.fieldext.Zx/2} \textbf{(a)} (at least not for $m\neq0$), and
certainly not in Proposition \ref{prop.fieldext.Zx/2} \textbf{(b)}.

Cardano was also lucky that the complex numbers form a field. We are not that
lucky in Proposition \ref{prop.fieldext.xx-1}.

When are we lucky and when are we not? Is there a general criterion for when a
quotient ring $R\left[  x\right]  /b$ contains a copy of $R$ (as opposed to
making distinct elements of $R$ equal) ? Is there a general criterion for when
$R\left[  x\right]  /b$ is a field?\footnote{This latter question is best
asked if $R$ itself is a field.}

Next time, we will see such criteria, and then we will apply them to a natural
question: What finite fields are there? We know the finite fields
$\mathbb{Z}/p$ for all primes $p$. The best way to get other finite fields is
to start with $\mathbb{Z}/p$ and to \textquotedblleft throw\textquotedblright%
\ new \textquotedblleft numbers\textquotedblright\ into them -- using the very
same $R\left[  x\right]  /b$ construction that Cardano used to define
$\mathbb{C}$. The technical term for \textquotedblleft throwing
in\textquotedblright\ is \textquotedblleft\textbf{adjoining}\textquotedblright%
, and so this method is called \textbf{root adjunction} (since the new
\textquotedblleft numbers\textquotedblright\ we adjoin are roots of given polynomials).
\end{document}