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\begin{tabular}{c | c c c c c c c c c}
$+$ & J & J + 1 & J + 2 \
\cline{1-4}
J & J & J + 1 & J + 2 \
J + 1 & J + 1 & J + 2 & J \
J + 2 & J + 2 & J & J + 1 \
\end{tabular}
\begin{tabular}{c | c c c c c c c c c}
$\cdot$ & J & J + 1 & J + 2 \
\cline{1-4}
J & J & J & J \
J + 1 & J & J + 1 & J + 2 \
J + 2 & J & J + 2 & J + 1 \
\end{tabular}
Tables are exact same for mod 3.
Q3
$$P_2 = { \varnothing, {a}, {b}, {a, b}}$$$$P_3 = { \varnothing, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c} }$$
\begin{align*}
K = K + \varnothing &= { \varnothing, { c } } \
K + { a } &= { { a }, { a, c } } \
K + { b } &= { { b }, { b, c } } \
K + { a, b } &= { { a, b }, { a, b, c } } \
\end{align*}
$$f(X) = X \cap { a, b }$$$$f : P_3 \rightarrow P_2$$$$P_2 \cong P_3 / \langle { \varnothing, { c } } \rangle$$
\begin{tabular}{c | c c c c c c c c c}
$+$ & K & K + {a} & K + {b} & K + {a,b} \
\cline{1-5}
K & K & K + {a} & K + {b} & K + {a,b} \
K + {a} & K + {a} & K & K + {a,b} & K + {b} \
K + {b} & K + {b} & K + {a,b} & K & K + {a} \
K + {a,b} & K + {a,b} & K + {b} & K + {a} & K \
\end{tabular}
\begin{tabular}{c | c c c c c c c c c}
$\cdot$ & K & K + {a} & K + {b} & K + {a,b} \
\cline{1-5}
K & K & K & K & K \
K + {a} & K & K + {a} & K & K + {a} \
K + {b} & K & K & K + {b} & K + {b} \
K + {a,b} & K & K + {a} & K + {b} & K + {a,b} \
\end{tabular}
Let $f(x) = (a - b)x + b$, then $f \in \mathcal{F}(\mathbb{R})$, $f(0) = b$ and $f(1) = a$. Thus functions of this form can represent any value in $\mathbb{R} \times \mathbb{R}$ and so the homomorphism $\phi$ is onto$\mathbb{R} \times \mathbb{R}$.
$$K = { f \in \mathcal{F}(\mathbb{R}) : f(0) = 0 \text{ and } f(1) = 0 }$$
Q2
$$J = { f \in \mathcal{F}(\mathbb{R}) : f(0) = 0 \text{ and } f(1) = 0 }$$
Thus $J$ is the kernel of the homomorphism $\phi$. The kernel is also an ideal of $\mathcal{F}(\mathbb{R})$, so
$$\mathcal{F}(\mathbb{R}) / J \cong \mathbb{R} \times \mathbb{R}$$
Q3
$$\phi: \mathcal{F}(\mathbb{R}) \rightarrow \mathcal{F}(\mathbb{Q}, \mathbb{R})$$$$\phi(f) = f_{\mathbb{Q}} = \text{ the restriction of } f \text{ to } \mathbb{Q}$$
$\phi$ is onto because $\forall g \in \mathcal{F}(\mathbb{Q}, \mathbb{R}), \exists f \in \mathcal{F}(\mathbb{R}) : g = f_{\mathbb{Q}}$. $\phi$ is a homomorphism since $\phi(f \cdot g) = \phi(f) \cdot \phi(g)$ and $\phi(f + g) = \phi(f) + \phi(g)$.
$J$ is also the kernel of $\mathcal{F}(\mathbb{R})$, which means it is also an ideal. Thus
$$\mathcal{F}(\mathbb{R}) / J \cong \mathcal{F}(\mathbb{Q})$$
D. Elementary Applications of the Fundamental Homomorphism Theorem
Q1
Note that ring is commutative then
$$(x + y)^2 = x(x + y) + y(x + y) = x^2 + xy + yx + y^2 = x^2 + 2xy + y^2 = x^2 + y^2$$
So $h(x) = x^2$ is a homomorphism since $h(x + y) = x^2 + y^2 = h(x) + h(y)$ and $h(xy) = x^2 y^2 = h(x) h(y)$
$$J = {x \in A : x^2 = 0 }$$$$B = { x^2 : x \in A }$$
$h$ is a homomorphism from $A$ to $B$ and the kernel is $J$$$A/J \cong B$$
Q2
$h(x) = 3x$ is a homomorphism because $h(x + y) = 3x + 3y = h(x) + h(y)$ and $h(xy) = h(x) h(y)$ because
$$h(xy) = 3xy = 6xy + 3xy = (3x)(3y) = h(x)h(y)$$
$J = { x : 3x = 0 }$ is the kernel and thus ideal of $h$. $B = { 3x : x \in A }$ is a subring of $A$ by the homomorphism shown above
$$A/J \cong B$$
Q3
$$\pi_a(xy) = axy = a^2 xy = (ax)(ay) = \pi_a(x)\pi_a(y)$$$$\pi_a(x + y) = a(x + y) = ax + ay = \pi_a(x) + \pi_a(y)$$$$I_a = { x \in A : ax = 0 } = \ker \pi_a$$$$\pi_a(1) = a$$$$\pi_a(x) = \pi_a(x \cdot 1) = a + \cdots + a \in \langle a \rangle$$$$\pi_a : A \rightarrow \langle a \rangle$$$$A / I_a \cong \langle a \rangle$$
Q4
$$\phi(ab) = \pi_{ab} = \pi_a \pi_b = \phi(a) \phi(b)$$$$\phi(a + b) = \pi_{a + b} = \pi_a + \pi_b = \phi(a) + \phi(b)$$$$I = { x \in A : ax = 0, \forall a \in A }$$$$\pi_a(x) = ax$$$$\bar{A} = { \pi_a : a \in A }$$$$\phi(a) = \pi_a$$$$\ker \phi = { x \in A : \phi(x) = \pi_0 }$$$$\forall a \in A \qquad \pi_0(a) = 0$$$$\therefore \ker \phi = I$$$$\phi : A \rightarrow \bar{A}$$$$A/I \cong \bar{A}$$
E. Properties of Quotient Rings $A/J$ in Relation to Properties of $J$
Q1
Every element of $A/J$ has a square root iff for every $x \in A$, there is some $y \in A$ such that $x - y^2 \in J$.
Let $J + x \in A/J$ then
$$J + x = (J + y)(J + y)$$
But $J$ is ideal and so absorbs products in $A$$$J + x = J + y^2$$$$J + x - y^2 = J$$$$x - y^2 \in J$$
Q2
Every element of $A/J$ is its own negative iff $x + x \in J$ for every $x \in A$.
$$\forall x \in A, x + x \in J \implies J + x + x = J$$$$\therefore \forall x \in A \qquad J + x = -(J + x)$$
Q3
$A/J$ is a boolean ring iff $x^2 - x \in J$ for every $x \in A$.
$$(J + x)^2 - (J + x) = J^2 + Jx + xJ + x^2 - J - x$$
But noting J absorbs products
$$(J+x)^2 - (J + x) = J + x^2 - x$$
But $x^2 - x \in J$ so
$$J + x^2 - x = J$$
so $A/J$ is a boolean ring.
Q4
If $J$ is the ideal of all the nilpotent elements of commutative ring $A$, then $A/J$ has no nilpotent elements (except zero).
$$a \in J \implies a^n = 0 \text{ for some } n$$
Let $x \in A : x \notin J \implies x^n \neq 0$$$(J + x)^n = J + x^n \neq J$$
Thus $\forall x \in A : x \in J$, $J + x$ is not nilpotent.
Q5
Every element of $A/J$ is nilpotent iff $J$ has the following property: for every $x \in A$, there is a positive integer $n$ such that $x^n \in J$.
$$\forall x \in A, x^n \in J$$$$(J + x)^n = J + x^n = J$$
Thus every element of $A/J$ is nilpotent.
Q6
$A/J$ has a unity element iff there exists an element $a \in A$ such that $ax - x \in J$ and $xa - x \in J$ for every $x \in A$.
\begin{align*}
(J + a)(J + x) &= J + x \
&= J + ax \
(J + x)(J + a) &= J + x \
&= J + xa
\end{align*}
$$J + x = J + ax$$$$J + ax - x = J$$
So $ax - x \in J$
Likewise
$$J + xa = J + x$$$$J + xa - x = J$$$$\implies xa - x \in J$$
F. Prime and Maximal Ideals
Let $A$ be a commutative ring with unity, and $J$ an ideal of $A$. Prove the following:
$J$ is a prime ideal iff $A/J$ is an integral domain.
Assume $J$ is a prime ideal.
$$ab \in J \implies a \in J \text{ or } b \in J$$$$J + ab = J + ac \implies J + b = J + c$$
Let $J + ab = J + ac$$$J + ab - ac = J$$$$a(b - c) \in J$$
But $a \notin J, b \notin J$ and $c \notin J$$$\implies b - c \in J$$$$J + b = J + c$$
Thus
$$J + ab = J + ac \implies J + b = J + c$$
For the converse, assume $a \notin J$, if $a \in J$, then we are done, otherwise
$$J + ab = J + a0 \implies J + b = J + 0$$$$J + ab = J \implies J + b = J$$$$a(b - c) \in J \implies b - c \in J$$$$ab \in J \implies b \in J$$
Q3
Every maximal ideal of $A$ is a prime ideal.
Let $J$ be a maximal ideal of $A$.
Then $A/J$ is a field.
Every field is an integral domain, so $A/J$ is an integral domain.
Since $A/J$ is an integral domain, so $J$ is a prime ideal.
Q4
If $A/J$ is a field, then $J$ is a maximal ideal.
$$\phi(x) = J + x$$$$j \in J, \phi(j) = J$$$A/J$ is a field, so ideal is $J$, which is maximal.
G. Further Properties of Quotient Rings in Relation to Their Ideals
Q1
Prove that $A/J$ is a field iff for every element $a \in A$, where $a \notin J$, there is some $b \in A$ such that $ab - 1 \in J.$
$$(J + a)(J + b) = J + ab = J + 1 \implies ab - 1 \in J$$
Q2
Prove that every nonzero element of $A/J$ is either invertible or a divisor of zero iff the following property holds, where $a, x \in A$: For every $a \notin J$, there is some $x \notin J$ such that either $ax \in J$ or $ax - 1 \in J$.
$$ax \in J \implies (J + a)(J + x) = J$$
and so $J + a$ is a divisor of zero.
Q3
An ideal $J$ of a ring $A$ is called primary iff for all $a, b \in A$, if $ab \in J$, then either $a \in J$ or $b^n \in J$ for some positive integer $n$. Prove that every zero divisor in $A/J$ is nilpotent iff $J$ is primary.
Nilpotent means $(J + x)^n = J$, but $(J + x)^n = J + x^n$, that is $x^n \in J$. Every zero divisor in $A/J$ means $(J + a)(J + b) = J + ab = J$ or that $ab \in J$. Thus either $a^1 \in J$ or $b^n \in J$. Thus we can say that $J + b$ is nilpotent since $(J + b)^n = J$.
Q4
An ideal $J$ of a ring $A$ is called semiprime iff it has the following property: For every $a \in A$, if $a^n \in J$ for some positive integer $n$, then necessarily $a \in J$. Prove that $J$ is semiprime iff $A/J$ has no nilpotent elements (except zero).
$A/J$ has no nilpotent elements means that $J + x^n \neq J$ for any integer $n$. Thus for every $a \in A: a \notin J$, then $a^n \notin J$. If $A/J$ has a nilpotent element, then $J$ cannot be semiprime because $a \notin J$ and $a^n \in J$ is a contradiction. This also holds true in reverse since $a^n \in J$ where $a \notin J$ would imply $J$ is not semiprime.
Q5
Prove that an integral domain can have no nonzero nilpotent elements. Then use part 4, together with Exercise F2, to prove that every prime ideal in a commutative ring is semiprime.
Nilpotent elements are also zero divisors since $a^n = 0 = a \cdot a^{n - 1}$. So an integral domain cannot have nilpotent elements.
From F2, we learn that if $J$ is a prime ideal, then $A/J$ is an integral domain (no nilpotent elements). From the last exercise, we see that if $A/J$ has no nilpotent elements, then $J$ is semiprime.
H. $\mathbb{Z}_n$ as a Homomorphic Image of $\mathbb{Z}$