We’ve introduced rings last lecture and we said that rings can be commutative, contain a multiplicative identity or can also be a field. We saw the subset that includes elements with multiplicative inverses that’s also a group. This is the Units group or \(R^{\times}\) which contains an element in \(R\) such that it has a multiplication inverse. We also saw an example of a ring which is

$$ \begin{align*} R[i] = \{\text{Formal expressions } a+bi, a, b \in R \}, \quad i^2 = -1 \end{align*} $$
Definition
Given a commutative ring with 1. Define \(R[x]\) to be the set of "Formal Expression" $$ \begin{align*} f = \sum_{k = 0}^n a_kx^k = a_0 + a_1x + ... + a_nx^n \quad a_0,...a_n \in R, \quad x \text{ new symbol} \end{align*} $$


Warning: We want to think of \(3 + 4x + 17x^2 + 0x^3 = 3 + 4x + 17x^2\). So if terms has a zero, then we can take it out.
We really identify an \(f \in R[x]\) with an infinite sequence \((a_k)_{k \geq 0} = (a_0,a_1,a_2,...)\) where each \(a_i \in R\) and because this polynomial is finite and we want an infinite sequence, we’ll say that there exists an \(n\) such that \(a_i = 0\) for all \(i > n\). So after some point \(n\), all the terms will be zero after.

The claim is that \(R[x]\) is a commutative ring with 1. So we need to define multiplication and addition. Therefore, let \(f = \sum_i a_ix^i, g = \sum_j b_jx^j\) where \(a_i,b_j \in R\). Then

$$ \begin{align*} f + g := \sum_k (a_k + b_k)x^k \\ fg := \sum_k (\sum_{i = 0}^k a_ib_{k-i})x^k \end{align*} $$

The identity element is an example of a constant polynomial. A constant polynomial is \(f \in R[i]\) where \(f = \sum_ka_k x^k\) such that \(a_k = 0\) for all \(k \geq 1\) So \(f = a_0\).

Let \(C \subseteq R[i]\) be the subset of constant. Then \(C\) is a subring. Define a bijection from

$$ \begin{align*} \lambda: R &\rightarrow C \\ a &\rightarrow \text{constant polynomial $a$} \end{align*} $$

This bijection is in fact an isomorphism of rings between \(R\) and \(C\).

Convention: We identify an element \(a\) in the original ring \(R\) with the corresponding constant polynomial in \(R[x]\). So we can think of \(R\) as a subring of \(R[x]\).

Remark: In a similar way, we can form \(R[x,y]\) or \(R[x_1,...,x_n]\) (polynomial ring of several variables). In fact, \(R[x,y] = (R[x])[y]\).


The Degree of a Polynomial

We now want to focus on a special case of a polynomial ring where the coefficient ring is actually a field. So let \(R = K\) be a field and let

$$ \begin{align*} f = \sum_k a_kx^k \in K[x], a_k \in K \end{align*} $$

Then, the degree of \(f\), \(\deg(f)\) is the largest integer \(n\) such that \(a_n \neq 0\). For example if

$$ \begin{align*} f = 7x^3 + \frac{1}{2}x - 17 \end{align*} $$

Then, \(\deg(f) = 3\). Warning: if

$$ \begin{align*} f = \sum_{k=0}^{n} a_kx^k \in K[x] \end{align*} $$

Then, we only know that \(\deg(f) \leq n\).


The Degree of the Zero Polynomial

If \(f\) is a constant polynomial, what is the degree of \(f\)? We defined the degree as the largest \(n\) such that \(a_n \neq 0\). In a constant polynomial, all the terms are zero except for \(a_0\). If the constant is non-zero, then the degree is zero. But if the polynomial is zero itself so \(f = 0\), then now all the coefficients \(a_i\)’s are zero so in this case the degree is undefined but we’re going to let the degree in this case be \(-\infty\). So deg\((f)=0\) if and only if \(f = 0\).


The Degree of a Polynomial when the Coefficient Ring is a Field

Next, we’re going to prove a proposition where we will see why we needed the coefficient ring to be a field.

Proposition
If \(K\) is a field and \(f,g \in K[x]\). Then
  1. \(\text{deg}(fg) = \text{deg}(f) + \text{deg}(g)\)
  2. \(\text{deg}(f+g) \leq \max\{\text{deg}(f), \text{deg}(g)\}\)


Convention:

  • \(-\infty + \text{ anything } = -\infty\)
  • \(-\infty \leq \text{ anything }\)

Proof of (1)
Let \(f = a_0 + a_1x + ... + a_mx^m\) and \(g = b_0 + b_1x + ... + b_nx^n\). Then

$$ \begin{align*} fg &= a_0b_0 + (a_1b_0+a_0b_1)x + ... + (a_ma_n)x^{m+n} \\ \end{align*} $$

and if \(a_m \neq 0\) and \(b_n \neq 0\), then \(a_mb_n \neq 0\). This is because \(K\) is a field so if \(a,b \in K\) and \(a \neq 0\), \(b \neq 0\), then \(ab \neq 0\). (because \(K^{\times} = K\{0\}\)).


Non-example: Take \(R = \mathbf{Z}_4 = \{0,1,2,3\}\). \(\mathbf{Z}_4\) is not a field since element 2 doesn’t have an inverse in \(\mathbf{Z}_4\). Note that only elements 1 and 3 have inverses. (\(3*3 = 1\) and \(1*1 = 1\) in \(\mathbf{Z}_4\). Now consider \(\mathbf{Z}_4[x]\) and let \(f = 1 + 2x \in \mathbf{Z}_4[x]\). Observe that

$$ \begin{align*} ff = f^2 = (1 + 2x)^2 = 1 + 2x + 2x + (2x)(2x) = 1 + 4x + 4x^2 = 1 \end{align*} $$

The proposition says the degree of \(fg\) should be \(1+1=2\) but here so the degree of \(f^2\) is zero. So we’re failing here because we’re not working in a field. Note that this shows that \(f \in \mathbf{Z_4}[x]^{\times}\) is a unit and it’s multiplicative inverse is itself. Here is a consequence of the proposition:

Corollary
If \(K\) is a field, then the units in the polynomial ring \(K[x]\) are exactly the elements of \(K^{\times}\). In other words, \(K[x]^{\times} = K^{\times}\).


So if \(K\) is a field, then all the units in the ring \(K[x]\) will be the nonzero constant polynomials. Those units are called \(K[x]^{\times}\). They form the multiplicative group of the ring \(K[x]\). Reminder: \(K\) is a field so every element except zero has a multiplicative inverse. \(K^{\times}\) is not a field, it is a group \((K^{\times},\cdot)\) that excludes zero so every single element has a multiplicative inverse.

Proof
One direction: If we have a constant non-zero polynomial and we’re working in a field, then it has an inverse that’s also a constant polynomial.
Other direction: If we have two polynomials \(f,g\) such that \(fg = 1\), then computing the degree of both sides we see that

$$ \begin{align*} \deg(f,g) &= \deg(1) \\ \deg(f) + \deg(g) &= 0 \\ \deg(f) &= -\deg(g). \end{align*} $$

But the only solution is that \(\deg(f) = \deg(g) = 0\). So the units always have to be constant.


Division Algorithm for \(K[x]\)

Proposition
Let \(K\) be a field. Let \(p, d \in K[x]\) where \(\deg(d) \geq 0\) so \(d\) is not a constant polynomial. Then there exists unique \(q,r \in K[x]\) such that
  1. \(p = qd + r\)
  2. \(\deg(r) < \deg(d)\)


Proof
Let

$$ \begin{align*} p &= \sum_i a_ix^i \text{ with } \deg(p)=m, \ \text{ so } \ a_m \neq 0, a_k > 0 \text{ if } k > m\\ d &= \sum_j b_ix^j, \text{ with } \deg(d)=n \geq 0, \ \text{ so } \ b_n \neq 0, b_k > 0 \text{ if } k > n \end{align*} $$

First, show that if \(m > n\) so the degree of \(p\) is greater than the degree of \(d\), then there exists a monomial \(cx^k\) such that

  1. \(p = (cx^k)d + p'\)
  2. \(\deg(p') < m = \deg(p)\)

So we can in a sense subtract that extra monomial where its degree is less than \(m\). Now, re-write both polynomials such that highest term is in the front

$$ \begin{align*} p &= a_mx^m + \text{ lower polynomial } \\ d &= b_nx^n + \text{ lower polynomial } \end{align*} $$

Let \(c = a_mb_n^{-1} = \frac{a_m}{b_n}\). This is fine becasuse \(b_n \neq 0\). Let \(k = m - n\). Then

$$ \begin{align*} p' &= p - (cx^k)d = (a_mx^m + \text{ lower polynomial }) - (a_nb_n^{-1}x^{m-n})(b_nx^n + \text{ lower polynomial }) \\ &= (a_mx^m - a_nx^m) \text{ lower polynomial }) \\ &= \text{ lower polynomial } \end{align*} $$

Therefore, \(\deg(p') < m\).

To prove the proposition, we use induction on \(m = \deg(p)\).
If \(m < n = \deg(d)\), let \(q = 0, r = p\) so that \(p = 0d + r\). Therefore, \(\deg(r) = \deg(p) < n\).

If \(m \geq n\), use induction. I can write \(p = (cx^k)d + p'\) where \(\deg(p') < m\). By induction, there exists a \(q'\) such that \(q'd + r\) where \(\deg(r) < n\) so

$$ \begin{align*} p &= (cx^k + p')d + r. \end{align*} $$

[TODO: This proof is unclear and a mess … ]


References