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 $$\operatorname{ deg } \left( r \right) < \operatorname{ deg } \left( g \right)$$