- \( a, b \in I \) implies that \( a - b \in I \)
- \( a \in I \) and \( r \in R \) implies that \( r \otimes a \in I \)

Ideal

An ideal in a crone \( \left( R, \oplus, \otimes \right) \) is a subset \( I \) containing \( 0 _ R \) such that

- \( a, b \in I \) implies that \( a - b \in I \)
- \( a \in I \) and \( r \in R \) implies that \( r \otimes a \in I \)

Trivial Ideal

Suppose \( R \) is a crone, then \( \left\{ 0 _ R \right\} \) is an ideal in \( R \)

Every Crone is an Ideal

For any crone \( R \) it is an ideal in \( R \)

The Kernel of a Crone Homomorphism is a Proper Ideal

If \( \phi : R \to S \) is a crone homomorphism then, then \( \operatorname{ ker } \left( \phi \right) \) is a proper ideal in \( R \)

A Crone Homomorphism is an Injection iff \( \operatorname{ ker } \left( \phi \right) = \left\{ 0 _ R \right\} \)

If \( \phi : R \to S \) is a crone homomorphism then, \( \phi \) is injective iff \( \operatorname{ ker } = \left\{ 0 _ R \right\} \)

Proper Ideal

We say that an ideal \( I \) in a crone \( R \) is proper when \( I \neq R \)

An Ideal is a Normal Subgroup

The Intersection of Ideals is an Ideal

Suppose that \( \mathcal{ I } \) is a family of ideals in a crone \( \mathbb{ R } \) then
\[
\bigcap \mathcal{ I }
\]
is an ideal in \( R \)

Let \( a, b \in \bigcap \mathcal{ I } \), then by definition \( a, b \in I \) for every ideal \( I \in \mathcal{ I } \), therefore \( a - b \in I \) for every \( I \) (because \( I \) is an ideal itself), in otherwords \( a - b \in \bigcap \mathcal{ I } \).

Now suppose that \( a \in \bigcap \mathcal{ I } \) and that \( r \in R \), then we see that \( a \in I \) for every \( I \in \mathcal{ I } \) so that also \( r \otimes a \in I\) since that's for every \( I \) we can conclude that \( r \otimes a \in \bigcap \mathcal{ I } \) as needed.

Ideal Generated By a Set

Suppose that \( R \) is a crone and that \( X \subseteq R \) then we define the **ideal generated by \( X \) ** as the intersection of all ideals in \( R \) that contain \( X \), symbolically let \( \mathcal{ I } _ X \) be the family of all ideals that contain \( X \), then
\[
\left( X \right) _ \diamond := \bigcap \mathcal{ I } _ X
\]

Ideal Lightened Notation

We define the notation \( \left( a _ 1, a _ 2, \ldots , a _ n \right) _ \diamond := \left( \left\{ a _ 1, a _ 2, \ldots , a _ n \right\} \right) _ \diamond \) to lighten the notation.

Ideal Generated by a Set is an Ideal

\( \left( X \right) _ \diamond \) is an ideal in \( R \)

As the intersection of ideals, it is an ideal.

Ideal Generated by a Set is the Smallest Ideal Containing the Set

Suppose \( R \) is a crone and \( X \subseteq R \) , then \( X \subseteq \left( X \right) _ \diamond \) and for any other ideal \( J \) such that \( X \subseteq J \) we have \( \left( X \right) _ \diamond \subseteq J \)

First we show that \( X \subseteq \left( X \right) \), we know that if \( \mathcal{ I } _ X \) is the family of all ideals containing \( X \) then by definition for each \( I \in \mathcal{ I } _ X \) we have \( X \subseteq I \) therefore \( X \subseteq \bigcap \mathcal{ I } _ X = \left( X \right) \) as needed.

Now we show it's the smallest, if \( J \) is a ideal containing \( X \) then by definition \( J \in \mathcal{ I } _ X \), but then \( \left( X \right) = \bigcap \mathcal{ I } _ X \subseteq \bigcap \left\{ J \right\} = J \) as needed.

Left Multiplication Yields an Ideal

Suppose that \( a \in R \) then \( R \otimes \left\{ a \right\} \) is an ideal

Let \( r \otimes a, s \otimes a \in R \otimes \left\{ a \right\} \), then note the following:
\[
\begin{align}
r \otimes a - s \otimes a &= \left( r\otimes a \right) \oplus - \left( s \otimes a \right) \\
&= \left( r \otimes a \right) \oplus \left( \left( -1 \right) \otimes \left( s \otimes a \right) \right) \\
&= \left( r \otimes a \right) \oplus \left( -s \otimes a \right) \\
&= \left( r - s \right) \otimes a
\end{align}
\]
Since \( R \) is a ring, then \( r - s \in R\) which shows that \( r \otimes a - s \otimes a \in R \otimes \left\{ a \right\} \)

Now we show the second property so let \( s \in R \) we must prove \( s \otimes \left( r \otimes a \right) \in R \otimes \left\{ a \right\} \), this follows quickly because of associativity of \( \otimes \), therefore the original expression equals \( \left( s \otimes r \right) \otimes r \) and since \( R \) is closed under \( \otimes \)

Finally \( 0 _ R \in R\) so that \( 0 _ R \otimes a \in R \otimes \left\{ a \right\} \) but we know that \( 0 _ R \otimes a = 0 _ R \) as needed.

Principal Ideal Generated by An Element

Suppose that \( R \) is a crone and that \( a \in R \), then \( R \otimes a \) is the **principal ideal generated by \( a \) **

Principal Ideal Equals Generated Ideal

\( R \otimes \left\{ a \right\} = \left( a \right) _ \diamond \)

Firstly we know that \( \left( a \right) := R \otimes \left\{ a \right\} \) and that \( R \otimes \left\{ a \right\} \) is an ideal, clearly it contains \( a \) because \( a - 0 _ R \) must be in it since it's an ideal.

\( \left( \left\{ a \right\} \right) \) is defined as \( \bigcap \mathcal{ I } _ \left\{ a \right\} \) where \( \mathcal{ I } _ \left\{ a \right\} \) is the family of all ideals in \( R \) that contain \( \left\{ a \right\} \). From this we can straight away see that \( \bigcap \mathcal{ I } _ \left\{ a \right\} \subseteq R \otimes \left\{ a \right\} \) since \( R \otimes \left\{ a \right\} \in \mathcal{ I } _ \left\{ a \right\} \) and the intersection can only get smaller. Thus we've just shown that \( \left( \left\{ a \right\} \right) \subseteq \left( a \right) \)

We'll now prove that \( \left( a \right) \subseteq \left( \left\{ a \right\} \right) \), let \( x \in \left( a \right) := R \otimes \left\{ a \right\} \) thus \( x = r \otimes a \) for some \( r \in R \), our goal is to show that \( r \otimes a \in \bigcap \mathcal{ I } _ \left\{ a \right\} \) , that is we must show that \( r \otimes a \in I \) for any \( I \in \bigcap \mathcal{ I } _ \left\{ a \right\} \).

Well if the above is true then \( I \) is an ideal that contains \( a \), therefore by the second property of an ideal we can see that \( r \otimes a \in I \), which is what we needed to show, so we can see that \( x \in \bigcap \mathcal{ I } _ \left\{ a \right\} := \left( \left\{ a \right\} \right) \). Thus we conclude that \( \left( a \right) = \left( \left\{ a \right\} \right) \) as needed.

Ideal Generated by a Finite Set is their Linear Combinations

Let \( A = \left\{ a _ 1, a _ 2, \ldots a _ n \right\}\) For any \( n \in \mathbb{ N } _ 1 \) we have
\[
\left( A \right) _ \diamond = \left\{ \sum _ { i = 1 } ^ n r _ i a _ i : r _ i \in R, i \in [ 1 ... n ] \right\}
\]

Note \( I \) is an ideal that contains \( A \), this is because we can systematically set a specific \( r _ i = 1 \) and for each \( i \neq j \) have \( r _ j = 0 \) which turns the sum into \( a _ i \) , then we know that \( \left( A \right) \subseteq I \), this is because \( \left( A \right) \) is the smallest ideal containing \( A \)

Now we'll prove that \( I \subseteq \left( A \right) \) let \( s = \sum _ { i = 1 } ^ n r _ i a _ i \in I \), to do so we recognize that \( \left( A \right) := \bigcap \mathcal{ I } _ A \) so that we must show that \( s \in I \) for every \( I \in \mathcal{ I } _ A \).

This \( \left( A \right) \) is an ideal that contains \( A \) therefore note that \( r _ i a _ i \) is part of \( \left( A \right) \), therefore we can see that each \( a _ i r _ i \in \left( A \right) \),

Since \( \left( A \right) \) is an ideal then we know it's closed under finite addition meaning that \( \sum _ { i = 1 } ^ n x _ i \in \left( A \right) \) for any \( x _ i \in \left( A \right) \), since we know each \( a _ i r _ i \in \left( A \right) \) then by setting \( x _ i = a _ i r _ i \) we have shown that \( \sum _ { i = 1 } ^ n a _ i r _ i \in \left( A \right) \) as needed and thus \( I = \left( A \right) \)

Product of Ideals is Contained in their Intersection

Let \( I, J \) be two ideals and define their product
\[
I J := \left\{ \sum _ { i = 1 } ^ n r _ i a _ i b _ i : a _ i \in I, b _ i \in J, r _ i \in R, n \in \mathbb{ N } _ 1 \right\}
\]
Prove that \( IJ \subseteq I \cap J \)

Let \( \sum _ { i = 1 } ^ n r _ i a _ i b _ i \in I J \), we'll first show that this is an element of \( I \), we first recall that both \( I, J \subseteq R \), therefore since \( b _ i \in J \) then \( \left( b _ i \in R \right) \) , since we are working with a crone, then we know that \( r _ i a _ i b _ i = r _ i b _ i a _ i \) since we have commutativity, by associativity and closure of multiplication in \( R \) then \( r _ i b _ i \in R \) therefore \( r _ i b _ i a _ i \in I \) by the second property of an ideal. We can also see by the first property that each since each of the summands are elements of \( I \) so will be their finite summation. Thus \( \sum _ { i = 1 } ^ n r _ i a _ i b _ i \in I \). By symmetry we can do the same thing to show that the sum is also in \( J \), this shows that \( I J \subseteq I \cap J \)

Can't get to All Polynomials From a Generated Ideal

Let \( I \) be the ideal in \( \mathbb{ Z } [ x ] \) generated by \( \left\{ 2, x \right\} \), prove that \( I ^ 2 := I I \) contains elements not of the form \( ab \) for \( a, b \in I \)

An Ideal of Continuous Functions

Let \( R := C \left( \left[ 0, 1 \right] \right) \) be the set of continuous functions \( f: \left[ 0, 1 \right] \to \mathbb{ R } \). For any \( c \in \mathbb{ R } \) we define \( I _ c := \left\{ f \in R : f \left( c \right) = 0 \right\} \)

- Show that the set \( R \) is a crone, and the set \( I _ c \) is an ideal of \( R \)
- Is \( I _ { c _ 1 } \cup I _ { c _ 2 } \) an ideal? What about \( I _ { c _ 1 } \cap I _ { c _ 2 } \)?
- Show that \( R / I _ c \cong \mathbb{ R } \) Hint: consider the map \( \phi \left( f + I _ c \right) = f \left( c \right) \)

We start by showing that \( R \) is a crone, first recall that by first year calculus we know that the product and sum of two continous functions on a certain domain is still continuous on that domain, additionally we have defined \( \left( f + g \right) \left( x \right) = f \left( x \right) + g \left(x \right) \) but addition and multiplication commute in \( \mathbb{ R } \) so it's easy to prove that \( f + g = g + f \) and \( f \cdot g = g \cdot f \) moreover we can also get associativity this way, all these facts together show that \( R \) is closed with respect to \( +, \cdot \) and \( \cdot, + \) are both associative and commutative, we also have inverses with respect to \( + \) this is because for any function \( f \), the function, \( - f \) which is defined as \( - f \left( x \right) = \left( -1 \right) f \left( x \right) \) has the property that \( f + \left( - f \right) = 0 \) where on the right we have the constant function that maps everything to zero (which is the additive inverse). So it is a crone.

We need to prove that \( I _ c \) is an ideal, therefore suppose that \( f, g \in I _ c \), we want to show that \( f - g \in I _ c \), but we can see that \( \left( f - g \right) \left( c \right) := f \left( c \right) - g \left( c \right) = 0 \), so we know that \( f - g \in I _ c \). Now let \( h : \left[ 0, 1 \right] \to \mathbb{ R } \) be any continuous function, and suppose that \( k \in I _ c \) then \( \left( h \cdot k \right) \left( c \right) := h \left( c \right) \cdot k \left( c \right) = h \left( c \right) \cdot 0 = 0 \), therefore \( h \cdot c \in I _ c \) as needed, so that \( I _ c \) is an ideal.

We see that \( I _ { c _ 1 } \cup I _ { c _ 2 } \) is not an ideal, this is because they may vanish at different places, consider the following counter example: \( f, g \in I _ { c _ 1 } \cup I _ { c _ 2 } \) then perhaps \( f \left( c _ 1 \right) = 0 \) but \( f \left( c _ 2 \right) = 1 \) and \( g \left( c _ 1 \right) = 1 \) and \( g \left( c _ 2 \right) = 0 \), then if we consider \( \left( f - g \right) \left( c _ 1 \right) := f \left( c _ 1 \right) - g \left( c _ 1 \right) = 0 + 1 \neq 0 \) and then also \( \left( g - f \right) \left( c _ 1 \right) := g \left( c _ 1 \right) - f \left( c _ 1 \right) = 1 + 0 \neq 0 \) so therefore \( f - g \notin I _ { c _ 1 } \cup I _ { c _ 2 } \) so it cannot be an ideal.

We should have that \( I _ { c _ 1 } \cap I _ { c _ 2 } \) is an ideal, this is because now an element would have two vanishing points and things would fallout like in our original verification that \( I _ c \) was an ideal, well anyway suppose that \( f, g \in I _ { c _ 1 } \cap I _ { c _ 2 } \) now note that \( f - g \left( c _ 1 \right) = 0 = f - g \left( c _ 2 \right) \) because they both vanish at these points so \( f - g \in I _ { c _ 1 } \cap I _ { c _ 2 } \). Now suppose that \( h \in R \) and \( k \in I _ { c _ 1 } \cap I _ { c _ 2 } \) then \( h k \left( c _ 1 \right) = 0 = h k \left( c _ 2 \right) \) as needed, so we've shown this is an ideal.

We need to show that \( R / I _ c \) is ismorphic to \( \mathbb{ R } \), we take the hint and will do it through the function \( \phi \left( f + I _ c \right) = f \left( c \right) \). Recall that \( R / I _ c = \left\{ f + I _ c : f \in R \right\} = \left\{ \left\{ f + g : g \in I _ c \right\} : f \in R \right\} \). For two crones to be ismorophic it means that they form a crone homomorphism which is a bijection, so we have to verify a few things \[ \phi \left( \left( f + I _ c \right) + \left( g + I _ c \right) \right) = \phi \left( \left( f + g \right) + I _ c \right) = \left( f + g \right) \left( c \right) = f \left( c \right) + g \left( c \right) = \phi \left( f + I _ c \right) + \phi \left( g + I _ c \right) \] and also \[ \phi \left( \left( f + I _ c \right) \left( g + I _ c \right) \right) = \phi \left( \left( f \cdot g \right) + I _ c \right) = f \left( c \right) g \left( c \right) = \phi \left( f + I _ c \right) \cdot \phi \left( g + I _ c \right) \] and suppose that \( 1 \) is the constant function that sends every value to one, then \[ \phi \left( 1 + I _ c \right) = 1 \left( c \right) = 1 \] therefore \( \phi \) is a crone homomorphism, now we want to prove that it is bijective, let \( r \in \mathbb{ R } \), then there is a constant function \( \textbf{r} \), so that \( \phi \left( \textbf{r} + I _ c \right) = \textbf{ r } \left( c \right) = r \), so we've shown that \( \phi \) is surjective, now suppose that \( a \neq b \in \mathbb{ R } \), then if we have \( \phi \left( j + I _ c \right) = a \) and \( \phi \left( i + I _ c \right) = b \), we'd like to prove that \( j + I _ c \neq i + I _ c \), we can easily show that \( j + I _ c \cap i + I _ n = \emptyset \), for this suppose that \( m \in j + I _ c \) so that \( m = j + p \) where \( p \in I _ c \), then note that \( m \left( c \right) = j \left( c \right) = a \), if \( m \) were to also be in \( i + I _ c \) that would mean that \( m = i + q \) for some \( q \in I _ c \) therefore \( m \left( c \right) = i \left( c \right) = b \) which is a contradiction because \( a \neq b \), so we've just shown that \( i + I _ c \neq j + I _ c \) so that \( \phi \) is a bijection, therefore it is an isomorphism.

Quotient Remainder For Polynomials

Suppose that \( R \) is a domain, and that \( f \left( x \right) , g \left( x \right) \in R [ x ] \), then there exists unique polynomials \( q \left( x \right) , r \left( x \right) \in R [ x ] \) such that
\[
f \left( x \right) = q \left( x \right) g \left( x \right) + r \left( x \right)
\]
where either \( r \left( x \right) = 0 _ R\) or \( \operatorname{ deg } \left( r \right) < \operatorname{ deg } \left( g \right) \)