Ideals

Ideal

An ideal in a crone

(R, \oplus︎, \otimes)

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 $r \otimes a \in I$

Note that sometimes we say it is a left ideal withen $r a \in I$ and a right ideal when $a r \in I$

Trivial Ideal

Suppose

R

is a crone, then

{0_{R}}

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

ϕ : R \to S

is a crone homomorphism then, then

\ker (ϕ)

is a proper ideal in

R

A Crone Homomorphism is an Injection iff

\ker (ϕ) = {0_{R}}

ϕ : R \to S

is a crone homomorphism then,

ϕ

is injective iff

\ker = {0_{R}}

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

ℐ

is a family of ideals in a crone

ℝ

then

⋂ ℐ

is an ideal in

R

Let $a, b \in ⋂ ℐ$ , then by definition $a, b \in I$ for every ideal $I \in ℐ$ , therefore $a - b \in I$ for every $I$ (because $I$ is an ideal itself), in otherwords $a - b \in ⋂ ℐ$ .

Now suppose that $a \in ⋂ ℐ$ and that $r \in R$ , then we see that $a \in I$ for every $I \in ℐ$ so that also $r \otimes a \in I$ since that's for every $I$ we can conclude that $r \otimes a \in ⋂ ℐ$ 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

ℐ_{X}

be the family of all ideals that contain

X

, then

{(X)}_{⋄} : = ⋂ ℐ_{X}

Ideal Lightened Notation

We define the notation

{(a_{1}, a_{2}, \dots, a_{n})}_{⋄} : = {({a_{1}, a_{2}, \dots, a_{n}})}_{⋄}

to lighten the notation.

Ideal Generated by a Set is an Ideal

{(X)}_{⋄}

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 {(X)}_{⋄}

and for any other ideal

J

such that

X \subseteq J

we have

{(X)}_{⋄} \subseteq J

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

Now we show it's the smallest, if $J$ is a ideal containing $X$ then by definition $J \in ℐ_{X}$ , but then $(X) = ⋂ ℐ_{X} \subseteq ⋂ {J} = J$ as needed.

Left Multiplication Yields an Ideal

Suppose that

a \in R

then

R \otimes {a}

is an ideal

Let

r \otimes a, s \otimes a \in R \otimes {a}

, then note the following:

\begin{matrix} r \otimes a - s \otimes a & a m p; = (r \otimes a) \oplus︎ - (s \otimes a) \\ a m p; = (r \otimes a) \oplus︎ ((- 1) \otimes (s \otimes a)) \\ a m p; = (r \otimes a) \oplus︎ (- s \otimes a) \\ a m p; = (r - s) \otimes a \end{matrix}

Since

R

is a ring, then

r - s \in R

which shows that

r \otimes a - s \otimes a \in R \otimes {a}

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

Finally $0_{R} \in R$ so that $0_{R} \otimes a \in R \otimes {a}$ 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 {a} = {(a)}_{⋄}

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

$({a})$ is defined as \bigcap \mathcal{ I } _ \left\{ a \right\} ParseError: Got function '\left' with no arguments as subscript at position 25: …mathcal{ I } _ \̲l̲e̲f̲t̲\{ a \right\} where \mathcal{ I } _ \left\{ a \right\} ParseError: Got function '\left' with no arguments as subscript at position 17: …mathcal{ I } _ \̲l̲e̲f̲t̲\{ a \right\} is the family of all ideals in $R$ that contain ${a}$ . From this we can straight away see that \bigcap \mathcal{ I } _ \left\{ a \right\} \subseteq R \otimes \left\{ a \right\} ParseError: Got function '\left' with no arguments as subscript at position 25: …mathcal{ I } _ \̲l̲e̲f̲t̲\{ a \right\} \… since R \otimes \left\{ a \right\} \in \mathcal{ I } _ \left\{ a \right\} ParseError: Got function '\left' with no arguments as subscript at position 50: …mathcal{ I } _ \̲l̲e̲f̲t̲\{ a \right\} and the intersection can only get smaller. Thus we've just shown that $({a}) \subseteq (a)$

We'll now prove that $(a) \subseteq ({a})$ , let $x \in (a) : = R \otimes {a}$ 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\} ParseError: Got function '\left' with no arguments as subscript at position 41: …mathcal{ I } _ \̲l̲e̲f̲t̲\{ a \right\} , that is we must show that $r \otimes a \in I$ for any I \in \bigcap \mathcal{ I } _ \left\{ a \right\} ParseError: Got function '\left' with no arguments as subscript at position 31: …mathcal{ I } _ \̲l̲e̲f̲t̲\{ 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) ParseError: Got function '\left' with no arguments as subscript at position 31: …mathcal{ I } _ \̲l̲e̲f̲t̲\{ a \right\} :…. Thus we conclude that $(a) = ({a})$ as needed.

Ideal Generated by a Finite Set is their Linear Combinations

Let

A = {a_{1}, a_{2}, \dots a_{n}}

For any

n \in ℕ_{1}

we have

{(A)}_{⋄} = {\sum_{i = 1}^{n} r_{i} a_{i} : r_{i} \in R, i \in [1... n]}

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 $(A) \subseteq I$ , this is because $(A)$ is the smallest ideal containing $A$

Now we'll prove that $I \subseteq (A)$ let $s = \sum_{i = 1}^{n} r_{i} a_{i} \in I$ , to do so we recognize that $(A) : = ⋂ ℐ_{A}$ so that we must show that $s \in I$ for every $I \in ℐ_{A}$ .

This $(A)$ is an ideal that contains $A$ therefore note that $r_{i} a_{i}$ is part of $(A)$ , therefore we can see that each $a_{i} r_{i} \in (A)$ ,

Since $(A)$ is an ideal then we know it's closed under finite addition meaning that $\sum_{i = 1}^{n} x_{i} \in (A)$ for any $x_{i} \in (A)$ , since we know each $a_{i} r_{i} \in (A)$ then by setting $x_{i} = a_{i} r_{i}$ we have shown that $\sum_{i = 1}^{n} a_{i} r_{i} \in (A)$ as needed and thus $I = (A)$

Product of Ideals is Contained in their Intersection

Let

I, J

be two ideals and define their product

I J : = {\sum_{i = 1}^{n} r_{i} a_{i} b_{i} : a_{i} \in I, b_{i} \in J, r_{i} \in R, n \in ℕ_{1}}

Prove that

I J \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

(b_{i} \in R)

, 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

ℤ [x]

generated by

{2, x}

, prove that

I^{2} : = I I

contains elements not of the form

a b

for

a, b \in I

An Ideal of Continuous Functions

Let

R : = C ([0, 1])

be the set of continuous functions

f : [0, 1] \to ℝ

. For any

c \in ℝ

we define

I_{c} : = {f \in R : f (c) = 0}

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 $R / I_{c} ≅ ℝ$ Hint: consider the map $ϕ (f + I_{c}) = f (c)$

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 $(f + g) (x) = f (x) + g (x)$ but addition and multiplication commute in $ℝ$ 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 $- f (x) = (- 1) f (x)$ has the property that $f + (- f) = 0$ 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 $(f - g) (c) : = f (c) - g (c) = 0$ , so we know that $f - g \in I_{c}$ . Now let $h : [0, 1] \to ℝ$ be any continuous function, and suppose that $k \in I_{c}$ then $(h \cdot k) (c) : = h (c) \cdot k (c) = h (c) \cdot 0 = 0$ , 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 $f (c_{1}) = 0$ but $f (c_{2}) = 1$ and $g (c_{1}) = 1$ and $g (c_{2}) = 0$ , then if we consider $(f - g) (c_{1}) : = f (c_{1}) - g (c_{1}) = 0 + 1 \neq 0$ and then also $(g - f) (c_{1}) : = g (c_{1}) - f (c_{1}) = 1 + 0 \neq 0$ 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 $f - g (c_{1}) = 0 = f - g (c_{2})$ 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 $h k (c_{1}) = 0 = h k (c_{2})$ as needed, so we've shown this is an ideal.

We need to show that $R / I_{c}$ is ismorphic to $ℝ$ , we take the hint and will do it through the function $ϕ (f + I_{c}) = f (c)$ . Recall that $R / I_{c} = {f + I_{c} : f \in R} = {{f + g : g \in I_{c}} : f \in R}$ . 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 $ϕ ((f + I_{c}) + (g + I_{c})) = ϕ ((f + g) + I_{c}) = (f + g) (c) = f (c) + g (c) = ϕ (f + I_{c}) + ϕ (g + I_{c})$ and also $ϕ ((f + I_{c}) (g + I_{c})) = ϕ ((f \cdot g) + I_{c}) = f (c) g (c) = ϕ (f + I_{c}) \cdot ϕ (g + I_{c})$ and suppose that $1$ is the constant function that sends every value to one, then $ϕ (1 + I_{c}) = 1 (c) = 1$ therefore $ϕ$ is a crone homomorphism, now we want to prove that it is bijective, let $r \in ℝ$ , then there is a constant function $𝐫$ , so that $ϕ (𝐫 + I_{c}) = 𝐫 (c) = r$ , so we've shown that $ϕ$ is surjective, now suppose that $a \neq b \in ℝ$ , then if we have $ϕ (j + I_{c}) = a$ and $ϕ (i + I_{c}) = b$ , 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 $m (c) = j (c) = a$ , if $m$ were to also be in $i + I_{c}$ that would mean that $m = i + q$ for some $q \in I_{c}$ therefore $m (c) = i (c) = b$ which is a contradiction because $a \neq b$ , so we've just shown that $i + I_{c} \neq j + I_{c}$ so that $ϕ$ is a bijection, therefore it is an isomorphism.

Quotient Remainder For Polynomials

Suppose that

R

is a domain, and that

f (x), g (x) \in R [x]

, then there exists unique polynomials

q (x), r (x) \in R [x]

such that

f (x) = q (x) g (x) + r (x)

where either

r (x) = 0_{R}

\deg (r) \leq \deg (g)