Last time we introduced the notion of an ideal in a ring. Stating it again

Definition: Ideal
An ideal in \(R\) is a subset \(I \subseteq R\) such that
  1. \(I\) is a subgroup of the abelian group with addition \((R, +)\). So the ideal must have \(0 \in I\), closed under addition, additive inverses are in \(I\) as well.
  2. If \(a \in I, r \in R\), then \(ra, ar \in I\) so it is also closed under multiplication.

We said that usually we refer to this definition as the “Two sided ideal” since other variants can exists (left and right ideals).

Proposition
If \(\{I_{\alpha}\}\) is a collection of ideals, then \(I = \bigcap_{\alpha} I_{\alpha}\) is an ideal.

As a consequence of this, we have the following defintion

Definition: Ideal Generated by \(S\)
Given a subset \(S \subseteq R\) ring, define $$ \begin{align*} (S) = \bigcap_{\text{all ideas such that }S \subseteq I} I \end{align*} $$ \((S)\) is an Ideal in \(R\) generated by \(S\). It is in fact the smallest Ideal containing the subset \(S\).

It is the smallest, since by definition for any \(I \subseteq R\), \(S\) is contained in \(I\), so \((S)\) is also in \(I\). This is a nice formal definition but it’s hard to use. An explicit description of the Ideal is as follows

Proposition
If \(R\) is a ring with 1 and \(S \subseteq R\) then, $$ \begin{align*} (S) = \{0\} \cup \{a_1s_1b_1 + ... + a_ks_kb_k \ | \ k \geq 1, s_1,...s_k \in S, a_i,b_i \in R\} \end{align*} $$

The elements are of the form \(a_is_ib_i\) since by definition, if an element \(s_i\) is in \(I\), then for any \(a_i \in R\), \(a_is_i \in I\) and for any \(b_i \in R\), \(s_ib_i \in R\). So we have \(a_is_ib_i\) to account for all possible products.
Proof
Let \(T = \{a_1s_2b_1 + ... + a_ks_kb_k \ | \ k \geq 1, s_1,...s_k \in S, a_i,b_i \in R\}\). Then we need to show

  1. Step (1) is showing that \(T\) is an Ideal where \(S \subseteq T\).
  2. Step (2) is if any \(I \subseteq R\) is any idea such that \(S \subseteq I\), then \(T \subseteq I\).

(1) and (2) Together imply that \(T = (S)\).
[TODO: Write full proof]

Principle Ideal

A special case of the above definition is when \(R\) is commutative. If it is, then the definition simplifies to

$$ \begin{align*} (S) = \{0\} \cup \{a_1s_2 + ... + a_ks_k \ | \ k \geq 1, s_1,...s_k \in S, a_i \in R\} \end{align*} $$

We don’t need to multiply on both the left and right since \(a_1s_2 = s_2a_1\) so we get to have all products still with a simpler defintion.
Another special case is the Principle Ideal.

Definition: Principle Ideal
A Principle Ideal \(I\) is an ideal such that \(I = (r)\) for a single \(r = R\) so $$ \begin{align*} (r) = \{a_1rb_1 + ... + a_krb_k \ | \ a_i,b_i \in R\} \end{align*} $$

Note that if \(R\) is a commutative ring with identity then

$$ \begin{align*} (r) &= \{a_1r + ... + a_kr \ | \ a_1,...a_k \in R\} \\ &= \{(a_1 + ... + a_k)r \ | \ a_1,...a_k \in R\} \\ &= \{ ar \ | \ a \in R \} \quad \text{(because $a_1+...+a_k \in R$)} \end{align*} $$

Examples

Example 1: If \(K\) is a field. The only ideals are \(\{0\}\) and \(K\). \(\{0\}\) is generated by \((0)\) and \(K\) is generated by \((1)\). The idea generated by 1 is always the whole ring.
Example 2: Take \(R = \mathbb{Z}\). All ideals are of the form \((d) = \mathbb{Z}d\) for some \(d \geq 0\). In fact, all ideals are principle. Why?
Proof
We know that for any ideal, it is a subgroup (with addition) of \(R=\mathbb{Z}\) so \(I \leq \mathbb{Z}\). If \(I = \{0\}\), then we’re done since \(\{0\}\) is a principle ideal generated by \((0)\). Suppose it is not so \(I \neq \{0\}\), then there exists a non-zero element in \(I\). Choose \(d \in I \cap \mathbb{Z}d\) to be the smallest positive element.
Claim: \(I = \mathbb{Z}d\). We can show this by:
\(\mathbb{Z}d \subseteq I\) (easy): Since \(d \in I\), then any multiple of \(d\) is in \(I\) so \(\mathbb{Z}d \subseteq I\).
\(I \subseteq \mathbb{Z}d\) (hard): Consider any \(a \in I\). Use division with remainder to see that \(a = qd + r\). \(r \in \mathbb{Z}\) where \(0 \leq r < d\). So \(r = a - qd\). \(r\) is in \(I\) because \(a \in I\) and \(qd \in I\). But \(d\) is the smallest element in \(I\). So \(r\) must be zero. Therefore, \(a = qd\) so \(a \in \mathbb{Z}d\). So \(I \in \mathbb{Z}d\). \(\blacksquare\).

More Propositions

We have more propositions about the principle ideal. One is the following

Proposition
Suppose that \(K\) is is a field and \(R = K[x]\). Every ideal in \(R\) is principle. In fact, if \(I \subseteq R\), then there exists a unique \(f\) such that \((f) = I\) and either \(f = 0\) or \(f\) is monic.

A monic polynomial is \(f = a_nx^n + a_{n-1}x^{n-1} + ... + a_0\) where \(a_n = 1\).
Proof
If \(I \subseteq K[x]\) and \(I\) is an ideal. Then if \(I = \{0\}\), then \(I\) is generated by \(0\) so \(I = (0)\). Now, suppose that \(I \neq \{0\}\). Then, pick an element \(f \in I/\{0\}\) of minimal degree. We can choose

$$ \begin{align*} f = a_nx^n +... +a_0 \end{align*} $$

to be monic. If it’s not monic, then use \(f' = a_n^{-1}f\). This is fine because \(a_n\) is a unit in \(K\) and has an inverse. Both \(f\) and \(f'\) have the same degree.
Note: This polynomial of minimal degree is unique. Why? Suppose \(f, g \in I/\{0\}\) and \(f\) and \(g\) are monic of the same minimal degree. Then their difference is in the ideal so \(f-g \in I\). (side note: reminder for any two polynomials in \(I\), their difference is in \(I\) since \(I\) is closed under addition/multiplication). In fact, since \(f\) and \(g\) are monic, then their leading terms are 1 so when we take the difference, this new polynomial \(f-g\) must have degree lower than \(f\) and \(g\). But this is a contradiction since we said that \(f\) and \(g\) are both of minimal degree. Thus, \(f - g = 0\) and \(f = g\). So the monic polynomial of minimal degree in \(I\) is unique. (Another side note: This is not true inside all of \(K[x]\).)
So now the claim is that \(I = (f)\) so \(I\) is generated by this minimal monic polynomial.
\((f) \subseteq I\) (easy): This is immediate since \(f\) is in \(I\) so all multiplies of \(f\) are in \(I\).

\(I \subseteq (f)\) (hard): We know \((f) = \{fg \ | \ g \in K[x]\}\). So suppose that \(p \in I\). Then use division with remainder \((p \div f)\) to get \(p = qf + r\) for some polynomials \(q, r \in K[x]\) with \(\deg(r) < \deg(f)\). So now \(r = p - qf\). We know \(p \in I\) and \(qf \in I\) since \(f \in I\). So \(r\) must be in \(I\). But that’s impossible since we said that \(f\) has the smallest degree in \(I\) so \(r = 0\). Therefore, \(p = qf\) is in \((f)\) as we wanted to show. \(\ \blacksquare\)

Example

Suppose that \(K\) is a field with \(c \in K\). Consider the evaluation homomorphism of rings where when \(r \in K \subseteq K[x]\), then \(ev_c(r) = r\)

$$ \begin{align*} ev_c: K[x] &\rightarrow K \\ ev_c(f) &= f(c) \end{align*} $$

The kernel is an ideal. \(\ker(ev_c) = I\). By definition

$$ \begin{align*} \ker(ev_c) = \{f(x) \in K[x] \ | \ f(c) = 0\}. \end{align*} $$

So it’s the set of polynomials that vanish at \(c\) or have \(c\) as a root. But since \(K\) is a field, then \(I\) is a principle ideal so it’s generated by some unique monic polynomial. One monic polynomial that satisfies this is \(x - c\). So \(\ker(ev_c) = (x-c)\).

Consequences

Proposition
Suppose that \(K\) is is a field, then every non-zero polynomial \(f \in K[x]\) has only finitely many roots in \(K\) in \(R\).

Proof
If \(c \in K\) and \(f(c) = 0\), then \(f\) is generated by \((x - c)\). The ideal generated by \((x-c)\) is the set of multiples of this generator so

$$ \begin{align*} I = \{ (x-c)g \ | \ g \in K[x]\}. \end{align*} $$

So \(f\) must be of the from \((x-c)g\) for some polynomial \(g \in K[x]\) and if \(c\) is a root, then we can factor off the linear term \((x - c)\). Inductively, we can iterate this process on \(g\) and in each iteration notice that the degree of the polynomial goes down by one until we reach the form

$$ \begin{align*} f = (x - c_1) (x - c_2)...(x - c_k)g', \quad c_1,...,c_k \in K \end{align*} $$

where \(g' \in K[x]\) is a polynomial with no roots in \(K\). Now, we can see that we found the roots \(c_1,c_2,...,c_k \in K\) but \(g\) has no roots in \(K\) (otherwise, we would’ve kept factoring). The degree of \(f\) is

$$ \begin{align*} \deg(f) &= \deg((x - c_1) (x - c_2)...(x - c_k)g') \\ n &= k + \deg(g') \end{align*} $$

So \(k\) at most \(n\).

Example

Let \(K = \mathbb{R} \subseteq \mathbb{C}\) be a subring. Define the following function using the substitution principle

$$ \begin{align*} ev_i : \mathbb{R}[x] &\rightarrow \mathbb{C} \\ ev_i(r) &= r \\ rv_i(x) &= i. \end{align*} $$

In other words, \(ev_i(f) = f(i) \in \mathbb{C}\). So it’s like plugging in \(i\) in the function. What is the kernel? We know the kernel is an ideal that is generated by a single monic polynomial. In the previous example, we were evaluating at \(c\) where \(c \in K\). But now, \(i\) is not in the field \(K\). The question is now: Do we know a monic polynomial that has \(i\) as a root? It is \(x^2 + 1\) so \(I = \ker(ev_i) = (x^2 + 1)\).

References