Week 4 - Day 3: Polynomials, Rings, and Units

Understanding Rings and Polynomials

True/False

Consider the following statement:

\[ 1 + 2x = 3. \]

Fact

If \( f \) and \( g \) are polynomials in \( \mathbb{R}[x] \), then:

\[ \deg(fg) = \deg(f) + \deg(g). \]

Polynomials in \( \mathbb{R}[x] \)

A polynomial in \( \mathbb{R}[x] \) has the following form:

\[ \mathbb{R}[x] = \left\{ a_0 + a_1x + \cdots + a_nx^n : n \in \mathbb{Z}, n \geq 0, \text{ each } a_i \in \mathbb{R} \right\}. \]

Key Notes on Proofs

When constructing proofs, avoid these two common pitfalls:

**Never begin a proof with your conclusion.**
**Never end a proof with a fact that is already known to be true.**

Example

Suppose we want to prove that \( 3 \mid p^2 + 2 \). A common but incorrect approach would be to start with the conclusion:

\[ 3 \mid p^2 + 2 \implies p^2 - 2 = 3k \quad \text{for some } k \in \mathbb{Z}. \]

This is incorrect because we are assuming what we are trying to prove. Instead, work step-by-step from known facts and gradually build toward the conclusion.

Another Example

Proving something false by ending with a known true statement doesn’t work:

\[ -1 = 1 \implies (-1)^2 = 1^2 \implies 1 = 1. \]

Just because the last statement is true doesn’t mean the first one is valid! This is an invalid proof technique. \(\square\)

Defining Rings

A **ring** is a set equipped with two operations—typically called addition and multiplication—that satisfy the following properties:

Addition: Closure, associativity, commutativity, identity, inverses.
Multiplication: Closure, associativity.
Both operations: Distributivity between addition and multiplication.

For instance, subtraction in a ring is defined as:

\[ a - b = a + (-b). \]

Proposition: Zero Multiplication in Rings

If \( R \) is a ring, then for all \( a \in R \), we have:

\[ a \cdot 0_R = 0_R. \]

Proof

To prove this, we use the distributive and identity properties:

\[ a \cdot 0 = a \cdot 0 + 0 \]

Subtracting \( a \cdot 0 \) from both sides and applying associativity and distributivity, we ultimately get:

\[ 0 = a \cdot 0. \]

Therefore, the proposition is true. \(\blacksquare\)

Units and Multiplicative Inverses

Definition: Multiplicative Inverse

Let \( R \) be a ring with a multiplicative identity \( 1_R \), and suppose \( a \in R \). An element \( b \in R \) is called a **multiplicative inverse** of \( a \) if:

\[ ab = 1_R = ba. \]

Examples

In \( \mathbb{Z} \) (the integers), does 2 have a multiplicative inverse? The answer is No, since there’s no integer \( b \) such that \( 2b = 1 \).

In \( \mathbb{Q} \) (the rational numbers), does 2 have a multiplicative inverse? Yes! In this case, \( 2 \cdot \frac{1}{2} = 1 \), so the inverse of 2 is \( \frac{1}{2} \).

Fact (HW)

If \( a \in R \) has a multiplicative inverse, then that inverse is unique. We denote it by \( a^{-1} \).

Definition: Unit

An element \( a \) in a ring \( R \) that has a multiplicative inverse is called a **unit**.

Question

In \( \mathbb{Z}[x] \), is \( X \) a unit?

Answer: No. Let’s see why.

Proof by Contradiction (BWOC)

Suppose, by way of contradiction, that \( X \cdot f = 1 \) for some polynomial \( f \in \mathbb{Z}[x] \), where \( \deg(X \cdot f) = 0 \). Then:

\[ \deg(X) + \deg(f) = 0 \]

This implies that \( \deg(f) = -1 \), which is a contradiction because a polynomial can't have a negative degree. Thus, \( X \) is not a unit in \( \mathbb{Z}[x] \).

Polynomials and Units

Let’s expand on this. In \( \mathbb{Z}[x] \), \( \mathbb{Q}[x] \), \( \mathbb{R}[x] \), and \( \mathbb{C}[x] \), no polynomial of positive degree can be a unit.

Why?

Because the inverse of a polynomial like \( X \) would have to be \( \frac{1}{X} \), which is not a polynomial. This makes it impossible for polynomials with positive degrees to have multiplicative inverses.

Example

Consider \( X(a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n) = 1 \). If you expand this, you get:

\[ a_0 X + a_1 X^2 + \cdots = 1, \]

which leads to contradictions because the left-hand side is a polynomial of degree greater than zero, while the right-hand side is a constant.

Units in \( \mathbb{Z} \) and \( \mathbb{Q} \)

Let’s explore the units in specific rings:

Question

What are the units in \( \mathbb{Z} \)? The units are simply \( \{ \pm 1 \} \), because:

\[ 1 \cdot 1 = 1 \quad \text{and} \quad (-1) \cdot (-1) = 1. \]

Question

What are the units in \( \mathbb{Q} \)? The units are all nonzero rationals:

\[ \mathbb{Q} \setminus \{0\}. \]

The Most Important Type of Ring: Fields

Definition: Field

A nonzero **commutative ring** with identity is called a **field** if every nonzero element is a unit (i.e., every nonzero element has a multiplicative inverse).

Application to Linear Algebra

In introductory linear algebra, scalars typically come from the field \( \mathbb{R} \) (the real numbers). However, in advanced linear algebra, scalars can come from any field \( \mathbb{F} \), making fields central to more abstract linear algebra.

Field Examples

Some rings are fields, but others are not. Consider the following:

In \( 2 \mathbb{Z} \) (the even integers), \( 2 \neq 0 \), but 2 is not a unit because there’s no integer that multiplies with 2 to get 1.
\( 2 \mathbb{Z} \) doesn’t have an identity either, because there is no element \( e \) such that \( 2e = e2 = e \) for all elements in \( 2 \mathbb{Z} \).
In \( \mathbb{R}[x] \), \( X \neq 0 \), but \( X \) is not a unit because, as discussed, no polynomial with a positive degree can have an inverse.

Note: \( \mathbb{C} \) is a Field

It might not be immediately obvious, but \( \mathbb{C} \), the set of complex numbers, is a field. Let’s explore why.

The Complex Numbers \( \mathbb{C} \)

The complex numbers are defined as:

\[ \mathbb{C} = \{ a + b i : a, b \in \mathbb{R}, i^2 = -1 \}. \]

Example: Inverse of a Complex Number

What are the real numbers \( a \) and \( b \) for \( \left(1 + 2i \right)^{-1} = \frac{1}{1 + 2i} \)? To find the inverse, we can use the trick of multiplying by the conjugate of the denominator:

\[ \frac{1}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i} = \frac{1 - 2i}{1 + 4} = \frac{1 - 2i}{5} = \frac{1}{5} + \frac{-2}{5}i. \]

Thus, the inverse of \( 1 + 2i \) is \( \frac{1}{5} + \frac{-2}{5}i \).

Rings with New Operations

Let’s look at a new way of defining a ring with different operations.

Example: A Modified Ring

Consider \( R = \mathbb{Z} \) with new operations defined as follows:

For \( r, s \in \mathbb{Z} \), define \( r \odot s = r \times s - 1 \).
Under this operation, \( r \odot s \) becomes \( rs - r + s \).

Proving \( R \) is a Ring

Let’s explore whether this modified structure satisfies the properties of a ring:

Closure: We need to check whether \( r \odot s \in R \) for all \( r, s \in R \).
Associativity: We check whether \( r \odot (s \odot t) = (r \odot s) \odot t \).
Identity: What’s the additive identity? Let’s explore that next.

Finding the Identity \( 0_R \)

What’s \( 0_R \), the additive identity? We want \( r \odot 0_R = r \) for all \( r \in R \). Let’s do some scratch work:

\[ r + 0_R - 1 = r. \]

This implies that \( 0_R = 1 \).

Finding \( 1_R \), the Multiplicative Identity

Next, we need to determine whether \( R \) has a multiplicative identity, \( 1_R \). From the operation:

\[ r \odot 1 = r + 1 - 1 = r \quad \text{and} \quad 1 \odot r = 1r - 1 = r. \]

Thus, the number 1 serves as the identity in this ring.

Does \( R \) have a \( 1_R \)?

Let’s check if the ring has a multiplicative identity by setting \( a \odot b = ab + a + b \). If we choose \( b = 0 \), then:

\[ 1_R = 0. \]

So, \( R \) does not have a traditional multiplicative identity under this operation.

Proving Left Distributivity

Let’s end with a final proof of distributivity in this modified ring:

\[ a \odot (b \oplus c) = \dots \quad \text{and} \quad (a \odot b) \oplus (a \odot c) = \dots \]