A Universal Object and More Polynomial Stuff
Definition: Universal Object
Let \( \mathcal{C} \) be a Category, and let \( F \in \mathcal{C} \) be an object.
- If for every other object, \( G \in \mathcal{C} \) there exists a unique morphism,
\[ \phi : F \rightarrow G \]
then \( F \) is said to be “universally repelling”.
- If for every other object, \( G \in \mathcal{C} \) there exists a unique morphism,
\[ \phi : G \rightarrow F \]
then \( F \) is said to be “universally attracting”.
In either case, \( F \) is said to be “universal.”
Theorem: Z is univeral
Let \( \mathcal{R} \) be the class of rings with units (also called “rings with identity”), then \( \mathbb{Z} \in \mathcal{R} \), the ring of the integers, is universally repelling.
The following proof isn’t terribly enlightening. We provide it for completeness. Feel free to skip it if you think the result is fairly obvious. The gist is that \( 1 \) has to be sent to the multiplicative identity of \(R\) (which is unique) and that determines where everything else is sent, because \( 1 \) generates all of \( \mathbb{Z} \).
Proof
Let \( R \in \mathcal{R} \), then, by definition, \( 1 \in R \), since \( \mathcal{R} \) is the category of rings with units.
For this proof, we will write it as, “\(\mathbb{1}\)” when it stands for the multiplicative identity of \( R \). Similarly, we’ll use “\(\mathbb{0}\)” to represent the additive identity of \( R \).
Define
$$ \phi(n) := \begin{cases} \mathbb{0}, &n = 0 \\ \mathbb{1}, &n = 1 \\ \sum\limits_{k=1}^n \phi(1) = \sum\limits_{k=1}^n \mathbb{1} , &n > 1 \\ -\phi(-n), &n < 0 \\ \end{cases} $$
First, we’ll prove that for any two \( a, b \in \mathbb{Z} \), we have \( \phi(a + b) = \phi(a) + \phi(b) \).
There are seven cases to consider:
-
Either \( a = 0 \) or \( b = 0 \) (or both). In this case, it’s clear that the result holds.
-
\( a > 0, b > 0 \).
$$ \begin{align} \phi(a + b) &= \sum\limits_{k = 1}^{a + b} \mathbb{1} \\ &= \left ( \sum\limits_{k = 1}^{a} \mathbb{1} \right) + \left ( \sum\limits_{k = 1}^{b} \mathbb{1} \right) \\ &= \phi(a) + \phi(b) \\ \end{align} $$
- \( a > 0, b < 0 \), and \( a + b > 0 \) (the case where \( a + b = 0 \) is immediate from the definition).
First, notice that
$$ \begin{align} \mathbb{0} &= \sum\limits_{k=1}^{(-b) - (-b)} \mathbb{1} \\ &= \sum\limits_{k = 1}^{-b} \mathbb{1} - \sum\limits_{k=1}^{-b} \mathbb{1} \\ \end{align} $$
Now, by definition, we have
$$ \begin{align} \phi(a + b) &= \phi(a - (-b)) \\ &= \sum\limits_{k=1}^{a - (-b)} \mathbb{1} \\ &= \sum\limits_{k=1}^{a - (-b)} \mathbb{1} + \left ( \sum\limits_{k = 1}^{-b} \mathbb{1} - \sum\limits_{k=1}^{-b} \mathbb{1} \right) \\ &= \left ( \sum\limits_{k=1}^{a - (-b)} \mathbb{1} + \sum\limits_{k = 1}^{-b} \mathbb{1} \right) - \sum\limits_{k=1}^{-b} \mathbb{1} \\ &= \sum\limits_{k = 1}^{a} \mathbb{1} - \sum\limits_{k = 1}^{-b} \mathbb{1} \\ &= \phi(a) + \phi(b) \\ \end{align} $$
- \( a < 0, b > 0 \), and \( a + b > 0 \).
This follows immediately from the previous case by swapping the variable names (remember that “\( + \)” is commutative, so \( x + y = y + x \) in every ring).
- \( a > 0, b < 0 \), and \( a + b < 0 \).
Then, by definition, we have to use \( -\phi( (-a) + (-b) ) \), but this works by case 4.
- \( a < 0, b > 0 \), and \( a + b < 0 \).
This follows similarly to case 4.
- \( a < 0, b < 0 \).
By definition, this works because
$$ \phi(a + b) = -\phi((-a) + (-b)) $$
which holds by case 1.
Now we’ll see that for all \( a, b \in \mathbb{Z} \), we have
$$ \phi(ab) = \phi(a) \cdot \phi(b) $$
Again, there are a few cases to consider. We begin with case “0”.
- \( a = 0, b = 0 \) (or both).
This is immediate from the definition.
- \( a, b > 0 \).
$$ \begin{align} \phi(ab) &= \sum\limits_{k=1}^{ab} \mathbb{1} \\ &= \underbrace{ \sum\limits_{k=1}^b \mathbb{1} + … + \sum\limits_{k=1}^b \mathbb{1}}_{a - times} \\ &= \underbrace{ \phi(b) + … + \phi(b) }_{a - times} \\ &= \underbrace{ \mathbb{1} \cdot \phi(b) + … + \mathbb{1} \cdot \phi(b) }_{a - times} \\ &= ( \underbrace{ \mathbb{1} + … + \mathbb{1} }_{a - times} ) \cdot \phi(b) \\ &= \phi(a) \cdot \phi(b) \\ \end{align} $$
- \( a < 0, b > 0 \).
Since \( ab < 0 \), we have, by definition,
$$ \begin{align} \phi(ab) &= -\phi((-a)b) \\ &= -(\phi( (-a) ) \cdot \phi(b)) \\ &= (-\phi(-a)) \cdot \phi(b) \\ &= \phi(a) \cdot \phi(b) \\ \end{align} $$
where the second line follows from the first by case 1.
- \( a > 0, b < 0 \).
This is similar to the previous case.
- \( a < 0, b < 0 \).
Since \( ab = (-a)(-b) > 0 \), we have
$$ \begin{align} \phi(ab) &= \sum\limits_{k = 1}^{ab} \mathbb{1} \\ &= \sum\limits_{k=1}^{(-a)(-b)} \mathbb{1} \\ &= \phi(-a) \cdot \phi(-b) \\ &= -\phi(a) \cdot -\phi(b) \\ &= \phi(a) \cdot \phi(b) \\ \end{align} $$
where the last line follows from the times negative mini-lemma, since
$$ \begin{align} -x &= \mathbb{1} \cdot -x \\ &= -\mathbb{1} \cdot x \\ \end{align} $$
and \( (-\mathbb{1})^2 = \mathbb{1} \) in any ring.
Thus, \( \phi \) is a ring homomorphism.
Uniqueness
Notice that for any homomorphism
$$ \psi : \mathbb{Z} \rightarrow R $$
we must have
$$ \begin{align} \psi(1) &= \mathbb{1} \in R \\ \psi(0) &= \mathbb{0} \in R \\ \end{align} $$
Therefore, for all \( a \in \mathbb{Z} \), we have \( \psi(a) = \phi(a) \). Thus \( \phi \) is unique.
QED
That proof was cumbersome and the result may not have seemed worth it, but it does allow us to easily show that the following definition makes sense.
Definition: Characteristic of a Field
Let \( F \) be a field and let \( \phi : \mathbb{Z} \rightarrow F \) be a homomorphism. Then
$$ \text{Ker}(\phi) = n\mathbb{Z} $$
for a unique \( n \geq 0 \). \( n \) is said to be the characteristic of \( F \).
Note: \( n \neq 1 \), because we don’t allow \( 0 = 1 \in F \), so the kernel cannot be all of \( \mathbb{Z} \).
This makes sense because, \( F \) (since it is a field) is a ring with units. Therefore, there is a unique homomorphism, \( \phi : \mathbb{Z} \rightarrow F \), by the previous theorem. The kernel of a ring homomorphism is an ideal of the domain of \( \phi \).
We saw in an earlier entry that \( \mathbb{Z} \) is a Euclidean Domain and that a Euclidean Domain is a PID (Principal Ideal Domain), which means that \( \text{Ker}(\phi) = \langle n \rangle \) (i.e. \(\{ n \cdot x | x \in \mathbb{Z} \}\)), for some \( n \in \mathbb{Z} \).
\( n \) is unique because no two non-negative integers have exactly the same set of multiples.
Lemma: Positive Characteristic is Always a Prime
Let \( F \) be a field with characteristic \( p > 0 \), then \( p \) is a prime number.
You don’t think we’d use “\( p \)” for a composite number, do you?
Proof
Suppose, for a contradiction, that the characteristic is equal to a composite, \( a b \) (where both are greater than \( 1 \)). Then we have
$$ \text{Ker}(\phi) = \{ ab \cdot x | x \in \mathbb{Z} \} $$
which means that \( a \not\in \text{Ker}(\phi) \) and \( b \not\in \text{Ker}(\phi) \), and so \( \phi(a) \neq 0 \) and \( \phi(b) \neq 0 \). However, we also have
$$ \begin{align} 0 &= \phi(ab) \\ &= \phi(a) \cdot \phi(b) \\ \end{align} $$
which is impossible, because a field cannot have zero divisors.
QED
Definitions: Maximal and Prime Ideals
Let \( R \) be a ring with units (what is also called a “ring with identity”), and \( \mathfrak{a} \subseteq R \) be an ideal.
- If for every ideal, \( \mathfrak{b} \subseteq R \), we have
$$ \begin{align} &\mathfrak{a} \subseteq \mathfrak{b} \implies \\ &\mathfrak{a} = \mathfrak{b} \text{ or } \mathfrak{b} = R \\ \end{align} $$
then \( \mathfrak{a} \) is said to be a “maximal ideal”.
- If for every \( x, y \in R \), we have
$$ \begin{align} &xy \in \mathfrak{a} \implies \\ &x \in \mathfrak{a} \text{ or } y \in \mathfrak{a} \\ \end{align} $$
then \( \mathfrak{a} \) is said to be a “prime ideal”.
You can easily extend the definition to left, right and two-sided ideals. We’ll be using them in the context of commutative rings, at least for a while, so the distinction won’t matter.
Lemma: Maximal Ideal is Prime
Let \( R \) be a commutative ring with units, and \( \mathfrak{a} \subseteq R \) be maximal. Then \( \mathfrak{a} \) is a prime ideal.
Proof
Suppose \( xy \in \mathfrak{a} \), but \( x \not\in \mathfrak{a} \). We want to show that \( y \in \mathfrak{a} \).
Since \( x \not\in \mathfrak{a} \), we have \( \langle x \rangle + \mathfrak{a} = R \), where \( \langle x \rangle \) is the ideal generated by \( x \) and the sum of two ideals is as we described it earlier (we explained it here: operations on ideals).
This means that \( 1 \in \langle x \rangle + \mathfrak{a} \), which implies that there is an \( s \in R \) and a \( t \in \mathfrak{a} \), such that
$$ s \cdot x + t = 1 $$
When we multiply through by \( y \), we get
$$ s \cdot xy + ty = y $$
Since \( t \in \mathfrak{a} \), we have \( ty \in \mathfrak{a} \). We also assumed that \( xy \in \mathfrak{a} \), so we must have \( s \cdot xy \in \mathfrak{a} \). This proves that \( y \in \mathfrak{a} \), because ideals are closed under addition.
QED
Lemma: R mod a Prime Ideal is an Integral Domain
Let \( R \) be a commutative integral domain with units. Let \( \mathfrak{p} \subseteq R \) be a prime ideal, then \( R / \mathfrak{p} \) is an integral domain.
proof
Let \( a + \mathfrak{p} \in R / \mathfrak{p} \) and \( b + \mathfrak{p} \in R / \mathfrak{p} \) be such that \( ab = 0 + \mathfrak{p} \). This means that \( ab \in \mathfrak{p} \). Since \( \mathfrak{p} \) is a prime ideal, we have either \( a \in \mathfrak{p} \) or \( b \in \mathfrak{p} \) (or both).
Without loss of generality, suppose \( a \in \mathfrak{p} \). Then
$$ a + \mathfrak{p} = 0 + \mathfrak{p} $$
i.e. \( a = 0 \in R / \mathfrak{p} \).
QED
lemma: R mod a Maximal Ideal is a Field
Let \( R \) be a commutative integral domain with units. Let \( \mathfrak{m} \subsetneq R \) be a maximal ideal (so \( \mathfrak{m} \) is a proper subset of \( R \)), then \( R / \mathfrak{m} \) is a field.
Proof
By the previous couple of lemmas, we know that \( R / \mathfrak{m} \) is an integral domain. Now we must show that every non-zero element is a unit.
Suppose \( 0 \neq a \in R / \mathfrak{m} \). We know this is possible since we’re assuming that \( \mathfrak{m} \subsetneq R \).
We’ll now make an argument that is fairly similar to what we did before when we proved that maximal ideals are prime. We know that the ideal
$$ \langle a \rangle + \mathfrak{m} = R $$
because it is an ideal that properly contains \( \mathfrak{m} \) (a maximal ideal). Thus, we have an \( s \in R \) and a \( t \in \mathfrak{m} \), such that
$$ s \cdot a + t = 1 $$
which implies that
$$ \begin{align} (s \cdot a + t) + \mathfrak{m} &= 1 + \mathfrak{m} \\ \implies s \cdot a + (t + \mathfrak{m}) &= 1 + \mathfrak{m} \\ \implies (s \cdot a) + \mathfrak{m} &= 1 + \mathfrak{m} \\ \end{align} $$
i.e. \( s \cdot a = 1 \in R / \mathfrak{m} \), so \( a \) is a unit.
QED
We’ll use this to show that fields with positive characteristic actually exist.
Corollary: For Every Prime Number, there is a Field with that Characteristic
Let \( p > 0 \) be a prime number, then there exists a field, \( F \), such that the characteristic of \( F \) is \( p \).
We’ll see later that there are infinitely many fields of every possible characteristic.
proof
In fact, there is a field with exactly \( p \) elements.
The only thing left to show is that \( p\mathbb{Z} \subset \mathbb{Z} \) is a maximal ideal, whenever \( p \) is a prime.
Suppose, for a contradiction, that there exists and ideal, \( \mathfrak{a} \), such that
$$ p\mathbb{Z} \subsetneq \mathfrak{a} \subsetneq \mathbb{Z} $$
then there exists an \( a \in \mathfrak{a} \setminus p\mathbb{Z} \). Since
$$ p\mathbb{Z} = \{ pn | n \in \mathbb{Z} \} $$
we know that \( p \nmid a \), and thus, \( \gcd(p, a) = 1 \). We know from our work on the Euclidean Algorithm that there are \( s, t \in \mathbb{Z} \), such that
$$ s \cdot p + t \cdot a = \gcd(p, a) = 1 $$
Now, since \( p \in \mathfrak{a} \) (because \( p\mathbb{Z} \) is a subset of \( \mathfrak{a} \)), we also have \( s \cdot p \in \mathfrak{a} \). Similarly, \( t \cdot a \in \mathfrak{a} \). This means that we must also have \( s \cdot p + t \cdot a = 1 \in \mathfrak{a} \), because ideals are closed under addition. Therefore, \( \mathfrak{a} = \mathbb{Z} \), a contradiction.
Therefore \( \mathbb{Z} / p\mathbb{Z} \) is a field. It has exactly \( p \) elements (one for each coset of \( p\mathbb{Z} \)).
QED
When \( \mathbb{Z} / p\mathbb{Z} \) is viewed as a field, it is often written, “\( \mathbb{F}_{p} \)”, or more simply as “\( F_p \)”. We’ll mostly use “\( \mathbb{F}_p \)”.
Corollary: All Fields with p Elements are Isomorphic
Let \( F_1 \) and \( F_2 \) be two fields with exactly \( p \) elements, where \( p \in \mathbb{Z} \) is prime. Then \( F_1 \) and \( F_2 \) are isomorphic.
At this point it might be tempting to think that there’s nothing left to show, but we haven’t actually proven that much about the structure of finite fields yet. For example, we haven’t shown that the characteristic divides the size of a finite group.
proof
We’ll start by proving that the characteristic of a finite field divides the size of the field. We do that by noticing that when we restrict our focus to just the “\( + \)” operator, a field, \( F \) is an Abelian group. What’s more, the unique homomorphism,
$$ \phi : \mathbb{Z} \rightarrow F $$
is a group homomorphism, when we consider only the action of the “\( + \)” operator. Recall from an earlier entry that the image of a homomorphism is a subgroup. Also, by Lagrange’s Theorem, the order of a subgroup must divide the order the parent group (if the parent group is finite, so that divisibility makes sense). Thus, the characteristic of a finite field must divide the size of the field.
This means that if a field has size \( p \), where \( p \) is a prime, then it also must have characteristic \( p \). Which means that the field must be exactly the image of the homomorphism, \( \phi \).
Thus,
$$ \begin{align} F_1 &= \mathbb{Z} / p\mathbb{Z} &&(= \mathbb{F}_p) \\ &= F_2 \\ \end{align} $$
QED
That’s it for this entry. Thanks for reading!!