Here I will be placing my summaries about the chapters that I have read together with ideas or points of view. Maybe I will solve some exercises from the book or questions I ask myself. I really do not expect someone to be reading this, but anyway if you are reading it I hope you enjoy it.
Part I: Polynomials
I.1: Division Algorithm and Evaluation Homomorphism.
In a certain sense the first part of the book is not really new for me. It talks a lot about polynomials and their properties, but I prefer reading it so that I am not surprised later. I must admit that I did learn something. Let's see.
Let $F$ be a field and $x_1,..., x_n$ a set of formal variables. We will denote by $F[x_1,..., x_n]$ the ring of polynomials in the variables $x_1,..., x_n$ and coefficientes in $F$. This ring has a lot of useful properties that will come in handy for our study many times. I will admit anyway that its properties confuse me from time to time, in certain sense I feel: is this really a proof? or is it a very vaer amazing hand waving? I will show later what I mean.
An element of this ring is a finite sum of terms of the form $Cx_1^{a_1}...x_n^{a_n}$ where $C\in F$ and $a_1,..., a_n \ge 0$ are integers. We will call the degree of the monomial the sum $a_1 + ... + a_n$. The total degree of a polynomial is the highest of the degrees of its monomials. The only polynomial without a well defined degree is 0, sometimes it is customary to define its degree as $-\infty$. We will denote the degree of a polynomial $f$ by $\partial(f)$. It is straightforward that
$\partial(fg) = \partial(f) + \partial(g)$,
hence $F[x_1,...,x_n]$ is an integral domain.
There are some important distinctions between the the polynomial ring in one variable and the polynomial ring in more than one variable. The main difference comes from the fact that in $F[x]$ we have
Division Algorithm: Let $f, g$ be any polynomials in $F[x]$ with $g\neq 0$. Then there exist unique polynomials $q, r$ in $F[x]$ such that $f = qg + r$ and with $\partial(r) < \partial(g)$. ($r$ can be the $0$ polynomial.)
The main consequence of this theorem is on the structure of ideals of $F[x]$. In fact we have the following:
Theorem: The polynomial ring $F[x]$ is a principal ideal domain, that is, all of its ideals are of the form $(g)$ for some $g\in F[x]$.
Proof:
Let $I$ be a nonzero ideal. Pick one $f\in I$ distinct to zero with lowest degree. We claim that $I = (f)$. In order to see this pick any other element of the ideal, say, $g\in I$. Then, by the division algorithm we have
$g = fq + r$, $\partial(r) < \partial(f)$.
But since $I$ is an ideal we get $r = g - fq\in I$. Hence, $r = 0$ for if not it will contradict the minimality of the degree of $f$. We conclude $g = fq \in ( f )$, that is, $I \subset (f)$, and the other inclusion is trivial.
$\blacksquare$
This fact will turn out to be very important later. On the other hand, the polynomial ring $F[x_1, ..., x_n]$ with $n \ge 2$ is not a PID (principal ideal domain).
Example: The ideal $(x, y)\subset F[x, y]$ is not a principal ideal.
A very special feature of this ring is that polynomials can serve as functions. We must make this intuitive idea precise. We start as follows:
Consider a ring $R$ that contains the field $F$ (This means that it conatins an isomorphic copy of $F$, and not precisely $F$ itself. It is a common abuse to say conatins $F$.). Lets pick $n$ elements from R, say, $A_1,..., A_n$.
Definition: We define the evaluation map $F_A : F[x_1,...,x_n] \rightarrow R$ as the ring homomorphism given by
This implies that $b$ shouls satisfy the quadratic equation $b^2 - 10b + 1 = 0$. Using the quadratic formula, which is possible for all primes $p > 2$, we have that the solutions are (in case they exist):
Definition: We define the evaluation map $F_A : F[x_1,...,x_n] \rightarrow R$ as the ring homomorphism given by
$\displaystyle\sum Cx_1^{a_1}...x_n^{a_n} \mapsto \displaystyle\sum CA_1^{a_1}...A_n^{a_n} \in R$.
One particular example that I find awesome (perhaps because it startles me every time) is the following:
Example: Let $F[x_1,..., x_n, x_{n + 1}, ..., x_{n + m}]$ be the polynomial ring in $n + m$ variables. As the tarjet ring we will consider $F[x_1,..., x_n]$, and our elements will be $A_1 = x_1, ..., A_n = x_n$, and $ A_{n + 1}, ..., A_{n + m}$ arbitrary elements in $F$. For example, the monomial $x_1...x_{m + n}$ is mapped to $A_{n + 1}...A_{n +m}x_1...x_n$. This example will be particularly useful when we define the universal polynomial later.
We have the following proposition:
Proposition: The evaluation map is a ring homomorphism.
Proof:
The proof of this proposition is like magic. In a sense it is a triviality but in other it seems like impossible to write it down formaly. I know that is nos true but I am always uncomfortable with this proof. Any way, what we do is to prove that it respects the sum, the product and sends $1$ to $1$. We will denote the evaluation map (for some fixed values $A_1,...,A_n \in R$) by $\phi$.
For the sum it goes like this:
$
\begin{align*}
\phi\left(\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} (C_{i_1,..., i_k} + D_{i_1,..., i_k})x_1^{i_1}...x_{n}^{i_n}\right)
&=
\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} (C_{i_1,..., i_k} + D_{i_1,..., i_k})A_1^{i_1}...A_{n}^{i_n}\\
&=
\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} C_{i_1,..., i_k}A_1^{i_1}...A_{n}^{i_n} \\
&+ \displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} D_{i_1,..., i_k}A_1^{i_1}...A_{n}^{i_n}\\
&=
\phi\left(\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} C_{i_1,..., i_k} x_1^{i_1}...x_{n}^{i_n}\right)\\
&+ \phi\left(\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} D_{i_1,..., i_k}x_1^{i_1}...x_{n}^{i_n}\right)
\end{align*}
$,
as desired.
Multiplication is analogous. Finally the fact the $1$ maps to $1$ is clear.
Example: Consider the polynomial $P = (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) \in F[x_1, x_2, x_3]$. And consider the evaluation map to the ring $F$ given by choosing $A_1 = a, A_2 = b, A_3 = c$. Then we have:
We must admit that it is pretty straight forward to use it.
As a final remark, notice that when $R = F$ the evaluation homomorphism are always surjective.
I.2: Symmetric Group and the Symmetric Polynomials.
As usual we will use $S_n$ to denote the group of permutations of the elements $\{1, ..., n\}$. It is called the Symmetric Groups and its elements are called Permutations. To denote a given permutation we generally use cyclic notation.
Example: $(1 2 3)(4 5)$ it he permutation of $\{1, 2, 3, 4, 5\}$ that takes $1$ to $2$, $2$ to $3$, $3$ to $1$, $4$ to $5$ and $5$ to $4$. It can be considered as a permutation of $S_n$ for $n > 5$ with the understanding that it leaves fixed all the elements $i > 5$.
We have a fundamental action of $S_n$ in $F[x_1, ..., x_n]$ given as follows: If $\sigma\in S_n$ then
The proof that this is indeed an action will be given later (much later...), although it is not hard anyway.
As any action we can ask ourselves two distinct things: which subgroup of $S_n$ fixes certain element and which elements are fixed by certain subgroup.
Definition: A polynomial $p\in F[x_1, ..., x_n]$ is called symmetric if it is fixed by all elements of $S_n$.
Example: In the ring $F[x_1, x_2, x_3]$ the following are symmetric polynomials:
Proposition: The elementary symmetric polynomials are symmetric.
Proof: We only need to see that the permutation induces a bijection in the set of indices, that is, on the sets of $r$ distinct elements of $\{ 1,..., n \}$. Since every permutation is a composition of transpositions it suffices to do this for the transpositions.
Suppose $\tau$ exchanges $i$ and $j$ and leaves everything else fixed. Then the subsets of $r$ elements are divided into these $4$ classes:
This proves that the set of indices remains the same, and son, $\sigma_r$ is unaffected bu $\tau$. This concludes the proof.
Example: The proposition implies that $\sigma_1\sigma_2 + \sigma_2\sigma_3 + \sigma_3\sigma_1$ is a symmetric polynomial. This polynomial is
$x_1^2x_2 + x_1x_2^2 + x_1^2x_3 + x_1x_3^2+ x_2^2x_3 + x_2x_3^2 + x_1^2x_2^2x_3 + x_1^2x_2x_3^2 + x_1x_2^2x_3^2 +\\
x_1^2x_2x_3+ x_1x_2^2x_3 + x_1x_2^2x_3 + 3x_1x_2x_3.$
The next is an outstanding result due to Gauss:
Theorem: Any symmetric polynomial in $F[x_1,..., x_n]$ is a polynomial in the elementary symmetric polynomials.
We will not comment the proof right now, but rather move on to important implications concerning symmetric polynomials. Consider the following map: $\Phi: F[x_1,..., x_n]\rightarrow F[x_1,..., x_n]$ given by $x_i\mapsto\sigma_i$. By a previous proposition we know this map is a ring homomorphism since it is an evaluation map. Its image is generated by $\sigma_1,..., \sigma_n$ and is denoted by $F[\sigma_1,..., \sigma_n]$. By the previous theorem we know that $F[\sigma_1,..., \sigma_n]$ is just the set of symmetric polynomials. Hence a natural question now comes to mind: Is this map injective? Or equivalently, since we know surjectivity, is this map a ring isomorphism? The answer is given by the next
Theorem: The expression of a symmetric polynomial as a polynomial in $\sigma_1, ...,\sigma_n$ is unique.
This theorem implies that $\Phi$ is indeed an isomorphism. Now we have some comments on this fact:
Proposition: Let $F\subset L$ be fields and $f = a_nx^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_0 \in F[x]$ whose roots are $\lambda_1,...,\lambda_n\in L$. Then, $\sigma_i(\lambda_1,...,\lambda_n) = (-1)^ia_i$.
Proof:
An important consequence of this proposition is the following:
Consider the ring $F[x_1,..., x_n, y_1,..., y_m]$ and $\sigma\in S_n$. Then we can define the action:
The use of this proposition will appear later, but it will be mainly used to prove that certain polynomials have coefficientes in some desired ring. The main tool for that is the following corollary of the previous porposition:
Corollary: Let $F\subset L$ be fields and $f\in F[x]$ a polynomial whose roots are $\lambda_1,...,\lambda_n\in L$ and let $p\in F[x_1,..., x_n, y_1,..., y_m]$ be such that $\sigma\cdot p = p$ for all $\sigma\in S_n$. Then $p(\lambda_1,...,\lambda_n, y_1,..., y_m) \in F[y_1,..., y_m]$.
Proof:
A particularly useful case is when $m = 1$. We will return to symmetric polynomials when he have studied more deeply Galois Theory.
I.3: Irreducibility
Irreducible polynomials play the role of prime numbers in the polynomial setting. As we will find out it is very important to understand the behavior of irreducible polynomials. Sadly, to know wether a polynomial is or not irreducible is not an easy task. Here I will describe the basic notions of irreducibility and some methods that can be used to test it.
Definition: A nonconstant polynomial $f\in F[x]$ is irreducible if $f = gh$ where $g$ and $h$ have strictly smaller degree implies that $g$ or $h$ is a constant.
There are several equivalences for this notion of irreducibility:
Proposition: Let $f\in F[x]$ be a nonconstant polynomial. The following are equivalent:
Suppose that $f$ is an irreducible polynomial and that $(f)\subset I$, where $I$ is an ideal. Since in $F[x]$ all ideals are principal we must have $I = (g)$ for some polynomial $g\in F[x]$. This means that $g$ divides $f$, that is, $f = gh$. Since $f$ is irreducible then either $g$ or $h$ is a constant. If $g$ is a constant then $I = F[x]$, if $h$ is a constant then $I = (f)$. This means that $(f)$ is a maximal ideal as desired.
The other implication is similar.
Although the theory we will develop is quite general we will consider several examples that take place over $\mathbb{Q}$ and so we will give criteria that are suited for this case. The first important theorem(which we do not prove here) is the very well known Gauss' Lemma:
Theorem[Gauss' Lemma]: Let $f\in\mathbb{Q}[x]$ have integer coefficients. If $f = gh$ for some polynomials $g, h \in \mathbb{Q}[x]$ then there exist a rational $\delta\neq 0$ such that $g' = \delta g$ and $h' = \delta^{-1}h$ have integer coefficientes and $ f = g'h'$.
This theorem implies that if a polynomial with integer coefficients is product of polynomials with rational coefficients then it is actually a product of polynomials with integer coefficents. This is very useful since it allows us to study the polynomial with congruences, an approach not available if we did not know the coefficientes of the product could be taken in $\mathbb{Z}$.
We will study three methods for irreducibility over $\mathbb{Z}$ (and hence over $\mathbb{Q}$). The first two are very well known, but the last is quite amazing and I have never seen it before studying this book. I will not prove all details in the first two criteria.
Method I: Schönemann-Eisenstein Criterion.
Proposition: Let $f = a_nx^n + ... a_1x + a_0 \in \mathbb{Z}[x]$ be a polynomial. If there exists a prime number $p\in \mathbb{Z}$ such that
This is the Schönemann-Eisenstein Criterion. Although it can be used to create a lot of polynomials that satisfy the hypothesis, a general polynomial will not satisfy the criterion. For example, the following polynomials satisfy the criterion:
Step 2: Use a suitable linear polynomial.
We will use the polynomial $f = x + 1$. Realize that
When we substitute we obtain:
Step 3: Use the criterion with the prime $p$.
It is a very well known fact that for $p$ prime the numbers $\dbinom{p}{i}$ for $1\le i \le p -1$ are divisible by $p$. Hence, we can use the criterion with the prime $p$.
Thanks to the previous steps we have proven that $\Phi_p$ is an irreducible polynomial for any prime $p$.
These polynomials are related to the roots of unity and will be one of our main attractions when we study Cycclotomic Extensions. Also, if we understand this polynomials for certain numbers it can help us to understand ruler and compass constructions. We will come back to these topics later.
Method II: Using congruences.
Given a polynomial with coefficients in $\mathbb{Z}$ and a prime $p$ we can consider the polynomial $f_p$ obtained by reducing its coefficients mod $p$ and considering as a polynomial in $\mathbb{F}_p[x]$.
Proposition: Let $f = a_nx^n + ... + a_1x + a_0\in\mathbb{Z}[x]$ be a polynomial. If there exists a prime $p$ such that:
which can be easily verified bu expanding the right hand side.
Case II: b = d = 1.
In this case the congruence that we need to solve is $a^2 = -8$. By quadratic reciprocity we know that means:
$\begin{align*}
\left(\dfrac{-8}{p}\right) &= \left(\dfrac{-1}{p}\right) \left(\dfrac{2}{p}\right)^3\\
&= (-1)^{\frac{p-1}{2}}\left(\dfrac{2}{p}\right)\\
&= (-1)^{\frac{p-1}{2}}(-1)^{\frac{p^2 -1}{8}}\\
&= (-1)^{\frac{p^2 + 4p - 5}{8}}
\end{align*}$
We want the exponent to be even, so we need $p^2 + 4p - 5 \equiv 0 \pmod{16}$. Checking case by case each one of $\pm 1, \pm 3, \pm 5, \pm 7$ we find out that only $1, 3, 9, 11$ satisfy this. Hence for these primes $-8$ is an square and we can solve the system of equations.
The remaining possibility is:
Case III: c = 0.
In this case we have $a = c = 0$, which reduces the system of equations to
Example: Let $F[x_1,..., x_n, x_{n + 1}, ..., x_{n + m}]$ be the polynomial ring in $n + m$ variables. As the tarjet ring we will consider $F[x_1,..., x_n]$, and our elements will be $A_1 = x_1, ..., A_n = x_n$, and $ A_{n + 1}, ..., A_{n + m}$ arbitrary elements in $F$. For example, the monomial $x_1...x_{m + n}$ is mapped to $A_{n + 1}...A_{n +m}x_1...x_n$. This example will be particularly useful when we define the universal polynomial later.
We have the following proposition:
Proposition: The evaluation map is a ring homomorphism.
Proof:
The proof of this proposition is like magic. In a sense it is a triviality but in other it seems like impossible to write it down formaly. I know that is nos true but I am always uncomfortable with this proof. Any way, what we do is to prove that it respects the sum, the product and sends $1$ to $1$. We will denote the evaluation map (for some fixed values $A_1,...,A_n \in R$) by $\phi$.
For the sum it goes like this:
$
\begin{align*}
\phi\left(\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} (C_{i_1,..., i_k} + D_{i_1,..., i_k})x_1^{i_1}...x_{n}^{i_n}\right)
&=
\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} (C_{i_1,..., i_k} + D_{i_1,..., i_k})A_1^{i_1}...A_{n}^{i_n}\\
&=
\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} C_{i_1,..., i_k}A_1^{i_1}...A_{n}^{i_n} \\
&+ \displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} D_{i_1,..., i_k}A_1^{i_1}...A_{n}^{i_n}\\
&=
\phi\left(\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} C_{i_1,..., i_k} x_1^{i_1}...x_{n}^{i_n}\right)\\
&+ \phi\left(\displaystyle\sum_{i_1,..., i_k \in \mathbb{N}\cup\{0\}} D_{i_1,..., i_k}x_1^{i_1}...x_{n}^{i_n}\right)
\end{align*}
$,
as desired.
Multiplication is analogous. Finally the fact the $1$ maps to $1$ is clear.
$\blacksquare$
The idea of the proof is that multiplication and addition behave in the same way in very similar ways in both rings. Anyway, it is important to understand how to apply this fact (Specially when it is not mentioned that it is being used).
$\phi(P) = \phi (x_1 - x_2)\phi(x_2 - x_3)\phi(x_3 - x_1) = (a - b)(b - c)(c - a)$.
We must admit that it is pretty straight forward to use it.
As a final remark, notice that when $R = F$ the evaluation homomorphism are always surjective.
I.2: Symmetric Group and the Symmetric Polynomials.
As usual we will use $S_n$ to denote the group of permutations of the elements $\{1, ..., n\}$. It is called the Symmetric Groups and its elements are called Permutations. To denote a given permutation we generally use cyclic notation.
Example: $(1 2 3)(4 5)$ it he permutation of $\{1, 2, 3, 4, 5\}$ that takes $1$ to $2$, $2$ to $3$, $3$ to $1$, $4$ to $5$ and $5$ to $4$. It can be considered as a permutation of $S_n$ for $n > 5$ with the understanding that it leaves fixed all the elements $i > 5$.
We have a fundamental action of $S_n$ in $F[x_1, ..., x_n]$ given as follows: If $\sigma\in S_n$ then
$\sigma\cdot p(x_1,..., x_n) = p(x_{\sigma(1)},...,x_{\sigma(n)})$.
That is, $\sigma\cdot p$ is the image of $p$ under the evaluation map with $R = F[x_1,..., x_n]$ that takes $x_i$ to $x_{\sigma(i)}$.
Example: The permutation $(1 2 3)$ takes $x_1 + x_2x_3$ to $x_2 + x_3x_1$.
The proof that this is indeed an action will be given later (much later...), although it is not hard anyway.
As any action we can ask ourselves two distinct things: which subgroup of $S_n$ fixes certain element and which elements are fixed by certain subgroup.
Definition: A polynomial $p\in F[x_1, ..., x_n]$ is called symmetric if it is fixed by all elements of $S_n$.
Example: In the ring $F[x_1, x_2, x_3]$ the following are symmetric polynomials:
$\sigma_1 = x_1 + x_2 + x_3$,
$\sigma_2 = x_1x_2 + x_2x_3 + x_3x_2$
$\sigma_3 = x_1x_2x_3$.
In general, the polynomial formed by adding all possible terms formed by multiplying $r$ variables together is a symmetric polynomial. We will study now these polynomials.
Definition: The $k-$th elementary symmetric polynomial in $F[x_1, ..., x_n]$ is the polynomial
$\sigma_r = \sigma_r(x_1, ..., x_n) = \displaystyle\sum_{1 \le i_1,..., i_r \le n, \\ \mbox{all distinct}}x_{i_1}...x_{i_r}$.
Proposition: The elementary symmetric polynomials are symmetric.
Proof: We only need to see that the permutation induces a bijection in the set of indices, that is, on the sets of $r$ distinct elements of $\{ 1,..., n \}$. Since every permutation is a composition of transpositions it suffices to do this for the transpositions.
Suppose $\tau$ exchanges $i$ and $j$ and leaves everything else fixed. Then the subsets of $r$ elements are divided into these $4$ classes:
- Sets that contain $i$ and $j$.
- Sets that contain $i$ but not $j$.
- Sets that contain $j$ but not $i$.
- Sets that contain none of $i$ or $j$.
This proves that the set of indices remains the same, and son, $\sigma_r$ is unaffected bu $\tau$. This concludes the proof.
$\blacksquare$
Another very useful property of the action of $S_n$ on the polynomials is the following: for any polynomials $f, g \in F[x_1,..., x_n]$ and any $\sigma\in S_n$ we have
- $\sigma\cdot(f + g) = \sigma\cdot f + \sigma\cdot g$,
- $\sigma(fg) = (\sigma\cdot f)(\sigma\cdot g)$.
When we prove that the action is indeed an action we will also prove this. Now, thanks to these facts we know the following
Proposition: Any polynomial which is obtained as sum or products of the elementary symmetric polynomials is symmetric.
$x_1^2x_2 + x_1x_2^2 + x_1^2x_3 + x_1x_3^2+ x_2^2x_3 + x_2x_3^2 + x_1^2x_2^2x_3 + x_1^2x_2x_3^2 + x_1x_2^2x_3^2 +\\
x_1^2x_2x_3+ x_1x_2^2x_3 + x_1x_2^2x_3 + 3x_1x_2x_3.$
The next is an outstanding result due to Gauss:
Theorem: Any symmetric polynomial in $F[x_1,..., x_n]$ is a polynomial in the elementary symmetric polynomials.
We will not comment the proof right now, but rather move on to important implications concerning symmetric polynomials. Consider the following map: $\Phi: F[x_1,..., x_n]\rightarrow F[x_1,..., x_n]$ given by $x_i\mapsto\sigma_i$. By a previous proposition we know this map is a ring homomorphism since it is an evaluation map. Its image is generated by $\sigma_1,..., \sigma_n$ and is denoted by $F[\sigma_1,..., \sigma_n]$. By the previous theorem we know that $F[\sigma_1,..., \sigma_n]$ is just the set of symmetric polynomials. Hence a natural question now comes to mind: Is this map injective? Or equivalently, since we know surjectivity, is this map a ring isomorphism? The answer is given by the next
Theorem: The expression of a symmetric polynomial as a polynomial in $\sigma_1, ...,\sigma_n$ is unique.
This theorem implies that $\Phi$ is indeed an isomorphism. Now we have some comments on this fact:
- First realize that $F[\sigma_1,..., \sigma_n]$ is isomoprhic to one of its proper subrings. in this case to: $F[\sigma_1, ..., \sigma_n]$. I find this fact rather impressive, not because it is uncommon but because it is beautiful.
- In this situation we call $\sigma_1,...,\sigma_n$ algebraically independent.
- It will be of central importance to consider the image of this subring via the evaluation maps. We will call this ring the symmetric ring of $F[x_1,..., x_n]$. This name is not standard.
Now we will start proving several facts of the symmetric subring that at first seem quite simple but that turn out to be very powerful.
Proof:
An important consequence of this proposition is the following:
Corollary: Let $F\subset L$ be fields and $f\in F[x]$ a polynomial whose roots are $\lambda_1,...,\lambda_n\in L$. Then $p(\lambda_1,...,\lambda_n) \in F$ for any $p\in F[\sigma_1,...,\sigma_n]$.
Proof:
Now we will consider a slight generalization of the symmetric polynomials that will turn out to be fundamental.Proof:
Consider the ring $F[x_1,..., x_n, y_1,..., y_m]$ and $\sigma\in S_n$. Then we can define the action:
$\sigma\cdot p(x_1,..., x_n, y_1,..., y_m) = p(x_{\sigma(1)},..., x_{\sigma(n)}, y_1,..., y_m)$.
We can ask ourselves again what can be said about those polynomials such that $\sigma\cdot p = p$. We have the following proposition explaining this:
Proposition: Let $p\in F[x_1,..., x_n, y_1,..., y_m]$ be written as $p = \displaystyle\sum_{\alpha \in A}p_{\alpha}(x_1,..., x_n)y^{\alpha}$. Then $\sigma\cdot p = p$ for every $\sigma\in S_n$ if and only if $p^{\alpha}$ is a symmetric polynomial for all $\alpha\in A$.
Proof:
The use of this proposition will appear later, but it will be mainly used to prove that certain polynomials have coefficientes in some desired ring. The main tool for that is the following corollary of the previous porposition:
Corollary: Let $F\subset L$ be fields and $f\in F[x]$ a polynomial whose roots are $\lambda_1,...,\lambda_n\in L$ and let $p\in F[x_1,..., x_n, y_1,..., y_m]$ be such that $\sigma\cdot p = p$ for all $\sigma\in S_n$. Then $p(\lambda_1,...,\lambda_n, y_1,..., y_m) \in F[y_1,..., y_m]$.
Proof:
A particularly useful case is when $m = 1$. We will return to symmetric polynomials when he have studied more deeply Galois Theory.
I.3: Irreducibility
Irreducible polynomials play the role of prime numbers in the polynomial setting. As we will find out it is very important to understand the behavior of irreducible polynomials. Sadly, to know wether a polynomial is or not irreducible is not an easy task. Here I will describe the basic notions of irreducibility and some methods that can be used to test it.
Definition: A nonconstant polynomial $f\in F[x]$ is irreducible if $f = gh$ where $g$ and $h$ have strictly smaller degree implies that $g$ or $h$ is a constant.
There are several equivalences for this notion of irreducibility:
Proposition: Let $f\in F[x]$ be a nonconstant polynomial. The following are equivalent:
- $f$ is irreducible.
- $(f)$ is a maximal ideal in $F[x]$.
- $\frac{F[x]}{(f)}$ is a field.
Suppose that $f$ is an irreducible polynomial and that $(f)\subset I$, where $I$ is an ideal. Since in $F[x]$ all ideals are principal we must have $I = (g)$ for some polynomial $g\in F[x]$. This means that $g$ divides $f$, that is, $f = gh$. Since $f$ is irreducible then either $g$ or $h$ is a constant. If $g$ is a constant then $I = F[x]$, if $h$ is a constant then $I = (f)$. This means that $(f)$ is a maximal ideal as desired.
The other implication is similar.
$\blacksquare$
Theorem[Gauss' Lemma]: Let $f\in\mathbb{Q}[x]$ have integer coefficients. If $f = gh$ for some polynomials $g, h \in \mathbb{Q}[x]$ then there exist a rational $\delta\neq 0$ such that $g' = \delta g$ and $h' = \delta^{-1}h$ have integer coefficientes and $ f = g'h'$.
This theorem implies that if a polynomial with integer coefficients is product of polynomials with rational coefficients then it is actually a product of polynomials with integer coefficents. This is very useful since it allows us to study the polynomial with congruences, an approach not available if we did not know the coefficientes of the product could be taken in $\mathbb{Z}$.
We will study three methods for irreducibility over $\mathbb{Z}$ (and hence over $\mathbb{Q}$). The first two are very well known, but the last is quite amazing and I have never seen it before studying this book. I will not prove all details in the first two criteria.
Method I: Schönemann-Eisenstein Criterion.
Proposition: Let $f = a_nx^n + ... a_1x + a_0 \in \mathbb{Z}[x]$ be a polynomial. If there exists a prime number $p\in \mathbb{Z}$ such that
- $p|a_i$ for all $0\le i\le n -1$,
- $p$ does not divide a_n$,
- $p^2$ does not divide $a_0$,
This is the Schönemann-Eisenstein Criterion. Although it can be used to create a lot of polynomials that satisfy the hypothesis, a general polynomial will not satisfy the criterion. For example, the following polynomials satisfy the criterion:
- $x^n - p$ with the prime $p$.
- $x^3 + 9x + 3$ with the prime $p = 3$.
An important example that we will study recurrently is the following:
Example: Let $p$ be a prime number. Then $x^{p - 1} + ... + x + 1 = \Phi_p(x)$ is an irreducible polynomial.
Directly we cannont use the criterion with any prime but if we use some tricks we will succed. We go by steps:
Step 1: Supose that $g = x - n$ with $n\in\mathbb{Z}$ and consider the polynomial $h = f\circ g$. Then we claim that $h$ is irredcible if and only if $f$ is.
It is easier to prove the counterpart: we will prove that $h$ is reducible if and only if $f$ is.
If $f$ is reducible, say $ f = ab$ with $a$ and $b$ non constant then:
$h = f\circ g = (ab)\circ g = (a \circ g)(b\circ g)$,
and so $h$ is reducible. I want to mention now how this is perfectly rigurous. Consider the evaluation map $\Phi_g: F[x]\rightarrow F[x]$ given by evaluating $x$ at $g$. We obtain then that $h = f\circ g$ is really $\Phi_g(f)$, and so, the previous lines can be written as:
$h = \Phi_g(f) = \Phi_g(ab) = \Phi_g(a)\Phi_g(b)$,
where we have used that $ \Phi_g$ is an homomorphism. It is very easy to see that this homomorphism preserves the degree (since we are evaluating at a polynomial of degree $1$). Hence, $h$ is reducible.
Now, suppose $h$ is reducible. Then, precomposing with $k =x + n$ we get
$f = f\circ(g \circ k) = h\circ k$,
and so by the previous argument $f$ is reducible.
Step 2: Use a suitable linear polynomial.
We will use the polynomial $f = x + 1$. Realize that
$ x^{p -1} + ... + x + 1 = \dfrac{x^p - 1}{x - 1}$
$\dfrac{(x + 1)^p - 1}{x + 1 - 1} = \dfrac{\displaystyle\sum_{i = 0}^p\binom{p}{i}x^i - 1}{x} = \displaystyle\sum_{i = 1}^p\binom{p}{i}x^{i - 1}$.
Step 3: Use the criterion with the prime $p$.
It is a very well known fact that for $p$ prime the numbers $\dbinom{p}{i}$ for $1\le i \le p -1$ are divisible by $p$. Hence, we can use the criterion with the prime $p$.
Thanks to the previous steps we have proven that $\Phi_p$ is an irreducible polynomial for any prime $p$.
Method II: Using congruences.
Given a polynomial with coefficients in $\mathbb{Z}$ and a prime $p$ we can consider the polynomial $f_p$ obtained by reducing its coefficients mod $p$ and considering as a polynomial in $\mathbb{F}_p[x]$.
Proposition: Let $f = a_nx^n + ... + a_1x + a_0\in\mathbb{Z}[x]$ be a polynomial. If there exists a prime $p$ such that:
- $p$ does not divide $a_n$,
- $f_p$ is irreducible in $\mathbb{Z}_p$,
then $f$ is irreducible.
Proof: Suppose that $f = gh$ is reducible over $\mathbb{Q}$. Then, by Gauss' Lemma, it is redcible over $\mathbb{Z}$ and we may assume that $g, h$ have integer coefficients. Then we have:
$f_p =g_ph_p$,
and since $f_p$ is irreducible it follows that one of $g_p$ or $h_p$ is a constant. Suppose that $g_p$ is a constant, then if the leading term of $g$ is $b_m$ then we have that $b_m\equiv 0$ mod $p$. Since the product of the leading terms of $g$ and $h$ is $a_n$ it follows that $b_m | a_n$. Hence, $a_n \equiv 0$ mod $p$. This is a contradiction, hence $f$ is irreducible.
$\blacksquare$
Example: The condition that $p$ does noot divide $a_n$ is fundamental. Consider the polynomial $f = 2x^2 + 5x + 2$ and use $p = 2$. Then $f_2 = x$ which is irreducible, but $f$ is not irreducible since $2x^2 + 5x + 2 = (2x + 1)(x + 2)$.
Again this method may seem ad hoc and it may be not very useful when we have a given particular polynomial.
I also want to comment that although congruences are very useful they can play tricks on us some time. To see an interesting phenomenon consider the following example:
Example: Let $f = x^4 + 10x^2 + 1$. We will prove that for any prime number $p$ the polynomial $f_p$ is reducible. In order to do that we will use quadratic reciprocity. We want to find $a, b, c, d\in\mathbb{F}_p$ such that:
$
\begin{align*}
\left(\dfrac{-12}{p} \right) &= \left(\dfrac{-1}{p}\right) \left(\dfrac{4}{p}\right) \left(\dfrac{3}{p}\right)\\
&= (-1)^{\frac{p-1}{2}} (-1)^{\frac{p -1}{2}}\left(\dfrac{p}{3}\right) \\
&= \left( \dfrac{p}{3} \right)
\end{align*}
$
So if $\left(\dfrac{p}{3}\right) = 1$ we can solve the system. This condition is satisfied if $p \equiv 1 \pmod{3}$.
To give an explicit example of this case consider $p = 7$. Then $a^2 = -12 = 2$ and a solution is $a = 3$. Hence we can take $a= 3, c = -3$. We have then:
Example: Let $f = x^4 + 10x^2 + 1$. We will prove that for any prime number $p$ the polynomial $f_p$ is reducible. In order to do that we will use quadratic reciprocity. We want to find $a, b, c, d\in\mathbb{F}_p$ such that:
$x^4 + 10x^2 + 1 = (x^2 + ax + b)(x^2 + cx + d)$.
Multiplying the right hand side we get the system of equations:
$a + c = 0, b + ac + d = 10, bc + ad = 0, bd = 1$.
We have that $a = -c$, and so $bc + ad = 0 \Rightarrow c(b - d) = 0$. If $b = d$, then last equation tells us that $b = d = 1$ or $b = d = -1$.
Case I: b = d = -1.
In that case $a$ and $c$ must satisfy the system:
$ac = 12, a + c = 0$.
This means that $a^2 = - 12$, and so our question reduces to the question: In which cases is $-12$ a quadratic residue modulo $p$? Quadratic reciprocity tells us:
$
\begin{align*}
\left(\dfrac{-12}{p} \right) &= \left(\dfrac{-1}{p}\right) \left(\dfrac{4}{p}\right) \left(\dfrac{3}{p}\right)\\
&= (-1)^{\frac{p-1}{2}} (-1)^{\frac{p -1}{2}}\left(\dfrac{p}{3}\right) \\
&= \left( \dfrac{p}{3} \right)
\end{align*}
$
To give an explicit example of this case consider $p = 7$. Then $a^2 = -12 = 2$ and a solution is $a = 3$. Hence we can take $a= 3, c = -3$. We have then:
$x^4 + 10x^2 + 1 = (x^2 + 3x - 1)(x^2 -3x - 1)$
which can be easily verified bu expanding the right hand side.
Case II: b = d = 1.
In this case the congruence that we need to solve is $a^2 = -8$. By quadratic reciprocity we know that means:
$\begin{align*}
\left(\dfrac{-8}{p}\right) &= \left(\dfrac{-1}{p}\right) \left(\dfrac{2}{p}\right)^3\\
&= (-1)^{\frac{p-1}{2}}\left(\dfrac{2}{p}\right)\\
&= (-1)^{\frac{p-1}{2}}(-1)^{\frac{p^2 -1}{8}}\\
&= (-1)^{\frac{p^2 + 4p - 5}{8}}
\end{align*}$
We want the exponent to be even, so we need $p^2 + 4p - 5 \equiv 0 \pmod{16}$. Checking case by case each one of $\pm 1, \pm 3, \pm 5, \pm 7$ we find out that only $1, 3, 9, 11$ satisfy this. Hence for these primes $-8$ is an square and we can solve the system of equations.
The remaining possibility is:
Case III: c = 0.
In this case we have $a = c = 0$, which reduces the system of equations to
$b + d = 10, bd = 1$.
This implies that $b$ shouls satisfy the quadratic equation $b^2 - 10b + 1 = 0$. Using the quadratic formula, which is possible for all primes $p > 2$, we have that the solutions are (in case they exist):
$\dfrac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6}$.
So everything reduces to the question of whether there is or not a squarte root of $6$. Quadratic reciprocity implies
$\begin{align*}
\left(\dfrac{6}{p}\right) &= \left(\dfrac{2}{p}\right) \left(\dfrac{3}{p} \right)\\
&= (-1)^{\frac{p^2-1}{8}}(-1)^{\frac{p-1}{2}}\left(\dfrac{p}{3}\right)\\
&= (-1)^{\frac{p^2 + 4p - 5}{8}}\left(\dfrac{p}{3}\right)
\end{align*}$
Now, the remaining cases from case II were the congruences $5, 7, 13, 15 \pmod{16}$. If we substitute any of these numbers in $p^2 + 4p - 5$ we obtain $8$, which means that the quotient in the exponent is an odd number. So for these cases:
$\left(\dfrac{6}{p}\right) = -\left(\dfrac{p}{3}\right)$,
and since the Case I took care of the cases that were squares mod $3$ then we can assume this Jacobi symbol is $-1$. Hence,
$\left(\dfrac{6}{p}\right) = 1$.
So, since any odd prime was considered in one of cases I, II or III then we only need to check what happens when $p = 2$. For this case it is easy to see that
$x^4 + 1 = (x^2 + 1)^2$.
We have proven that for any prime $p$ the polynomial $f_p$ is reducible. Hence it is a very natural question to ask: Must $f$ itself be reducible? The answer is no. This polynomial is not reducible over the integers (and hence over $\mathbb{Q}$). This is not hard to prove and we will not do it now (we will do it later, since this polynomial will come out many times.)
Both of the methods discussed so far can be used when the polynomials allow the existence of a prime that satisfies certain conditions. Now we will work with a third method that can be always applied. It is an algorithm, so even do it is always appicable it may take some time to use it.
Method 3: Lagrange's Interpolation Formula
The idea of Lagrange Interpolation formula is to find polynomials that at certain chosen points take selected values. We now make this precise. Let $F$ be any field. We need the following set of data:
- The selected points where we want the polynomial to take certain values. This must be distinct points $s_0, ..., s_d \in F$.
- The selected values. This must be $v_0,..., v_d\in F$.
Then Lagrange Inerpolation Formula refers to the following formula:
$G(x) = \displaystyle\sum_k^d v_k\displaystyle\prod_{j\neq k}\dfrac{x - s_j}{s_k - s_j}$.
The following proposition is obvious but expresses the fact that this polynomial takes the values we wanted in the places we wanted.
Proposition: $G\in F[x]$ is a polynomial of degree smaller or equal to$d$ such that $G(s_k) = v_k$ for $k= 0, ..., d$.
Another nice proposition that is useful is:
Proposition: If $h\in F[x]$ is such that $\deg h \le d$ and $h(s_k) = v_k$ for $k = 0,..., d$, then $h = G$.
Proof: The polynomial $G - h$ has degree at most $d$ and has $d + 1$ roots: $s_0,..., s_d$. Hence, it must be the zero polynomial.
$\blacksquare$
Example: Suppose we are looking for a polynomial $P\in\mathbb{Q}[x]$ such that $P(1) = 2, P(2) = 4, P(3) = 6$. The polynomials that go in the sum are the following:
- $\dfrac{x - 2}{1 - 2}\dfrac{x - 3}{1 - 3} = \dfrac{x^2 - 5x + 6}{2}$
- $\dfrac{x - 1}{1}\dfrac{x - 3}{-1} = \dfrac{x^2 - 4x + 3}{-1}$
- $\dfrac{x - 1}{2}\dfrac{x - 2}{1} = \dfrac{x^2 - 3x + 2}{2}$.
Hence, Lagrange Interpolation Formula tells us that the polynomial is:
$G(x) = 2\left(\dfrac{x^2 - 5x + 6}{2}\right) + 4\left(\dfrac{x^2 - 4x + 3}{-1}\right) + 6\left(\dfrac{x^2 - 3x + 2}{2}\right) = 2x$.
This example shows three things that are important to realize:
- The polynomial obtained this way not necessarily has degree $d$. For example, in this case we obtain a linear polynomial even do $d = 2$.
- If the set of values are known to coincide with the values a polynomial of degree $\le d$ at the same points then that polynomial is the one we will obtain. For example, just by looking we knew that $2x$ was a polinomial that satisfied the $3$ equations we wanted. Hence, since its degree was less than $2$ we knew by the previous proposition that we were ging to obtain exactly $2x$.
- Even if the answer is simple the computations may be tedious.
We will describe an algorithm to test reducibility over $\mathbb{Q}$ of a polynomial in $\mathbb{Z}[x]$. Suppose we have a polynomial $f\in\mathbb{Z}[x]$ of degree $n$ such that $f(i)\neq 0$ for $0\le i \le n - 1$. This is not restrictive for if we knew $f(i) = 0$ then the polynomial is certainly not irreducible.
We can create a polynomial following the next steps:
- Fic an integer $0 < d < n$.
- Fix divisors $a_0,..., a_d\in\mathbb{Z}$ of $f(0),..., f(d) \in\mathbb{Z}$.
- Construct a polynomial $g$ of degree at most $d$ such that $g(i) = a_i$ for $i = 0, ..., d$.
- Call $g$ acceptable polynomial if $g\in\mathbb{Z}[x]$ and its degree is $d$.
If we do this for all $d$ and all possible sets of divisors $a_0,..., a_d$ we obtain a finite set $A$ of acceptable polynomials.
Proposition: The polynomial $f$ is irreducible over $Q$ if and only if it is not divisible by any of the polynomials of the set $A$.
Remark: Before going into the proof realize that if you think about it, this sounds a lot to Gauss' Lemma.
Proof: One of the implications is obvious. We need to prove that if it is not divisible by any of the polynomials of the set $A$ then it is irreducible.
We prove the contrapositive. Suppose that $f$ is reducible and that $f = gh$ for some nonconstant polynomials. Gauss' Lemma implies that we can consider $g, h\in\mathbb{Z}[x]$. We will prove that $g\in A$. Let $d$ be the degree of $g$. Then, for any $i = 0,..., d$ we have that $f(i) = g(i)h(i)$ and so $g(i) | f(i)$. Hence, when we choose $d$ in the first step and $a_i = g(i)$ we construct a polynomial $\tilde{g}$ of degree at most $d$ such that $\tilde{g}(i) = a_i = g(i)$. Hence, by the previous proposition it must happen that $g = \tilde{g}$. Hence, $\tilde{g}\in A$.
The last aspect we will explore of polynomials in general will be the problem of the existence of its roots.
We prove the contrapositive. Suppose that $f$ is reducible and that $f = gh$ for some nonconstant polynomials. Gauss' Lemma implies that we can consider $g, h\in\mathbb{Z}[x]$. We will prove that $g\in A$. Let $d$ be the degree of $g$. Then, for any $i = 0,..., d$ we have that $f(i) = g(i)h(i)$ and so $g(i) | f(i)$. Hence, when we choose $d$ in the first step and $a_i = g(i)$ we construct a polynomial $\tilde{g}$ of degree at most $d$ such that $\tilde{g}(i) = a_i = g(i)$. Hence, by the previous proposition it must happen that $g = \tilde{g}$. Hence, $\tilde{g}\in A$.
$\blacksquare$
This is a nice method from the point of view of order: what you have done with this algorithm is to find a method of ordering all the polynomials in $\mathbb{Z}[x]$ that can divide $f$. The bad thing is that the set $A$ can be very big. Just try to compute $A$ for $x^4 + 10x^2 + 1$ and you will see.
I.4: Roots of Polynomials.
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