Let $A$ be an $n \times n$ matrix with charachteristic polynomial
$$f(t)=(-1)^nt^n+a_{n-1}t^{n-1}+ \cdots +a_1t+a_0$$
(a) Prove that $A$ is invertible iff $a_0 \neq 0$.
(b) Prove that if $A$ is invertible, then
$$A^{-1}=\frac{-1}{a_0}[(-1)^nA^{n-1}+ \cdots +a_1I]$$
(c) Use (b) to compute $A^{-1}$ for $A=\begin{pmatrix}
1&2&1\\
0&2&3\\
0&0&-1\\
\end{pmatrix}$
This exercise provides an interesting result. We get a formula for the inverse of a matrix $A$.
$\textit{Proof}$ : Suppose to the contrary that $a_0=0$. We will show that this implies $A^{-1} \not \exists$
If $a_0=0$ then the charachteristic polynomial of $A$ becomes
$$f(t)=(-1)^nt^n+a_{n-1}t^{n-1}+ \cdots +a_1t=t[(-1)^nt^{n-1}+a_{n-1}t^{n-2}+ \cdots +a_1]$$
$\Longleftrightarrow t$ is a factor of $f(t)$ $\Longleftrightarrow t=0$ is a root of $f(t) \Longleftrightarrow 0$ is an eigenvalue of $A$ $\Longleftrightarrow$ (invertible matrix theorem) $A^{-1} \not \exists$.
Each of the above steps was "iff" so,
$\therefore$ $A$ invertible iff $a_0 \neq 0$
If $A$ is invertible, then by part (a) above, $a_0 \neq 0$ in $f(t)$. We may write the charachteristic polynomial $f(t)$ in matrix form as $f(A)$. By the Cayley-Hamilton theorem, a matrix "satisfies" its own charachteristic polynomial. So,
$$f(A)=0=(-1)^nA^n+ \cdots + a_1A+a_0I$$
$$-a_0I=(-1)^nA^n+ \cdots a_1A$$
Since $A$ is invertible, we may apply $A^{-1}$ to both sides of the above equation to get
$$-a_0IA^{-1}=[(-1)^nA^n+\cdots +a_1A]A^{-1}$$
$$-a_0A^{-1}=(-1)^nA^{n-1}+\cdots+a_1$$
$$A^{-1}=\frac{-1}{a_0}[(-1)^nA^{n-1}+ \cdots +a_1I]$$
$\blacksquare$
For part (c), we compute
$$|A-tI|=(1-t)(2-t)(-1-t)=-t^3+2t^2+t-2$$
So, $a_0=-2$ and our formula gives
$$A^{-1}=\frac{-1}{-2}(-A^2+2A+I)$$
Performing martrix operations, we get
$$A^{-1}=\begin{pmatrix}
1&-1&-2\\
0&\frac{1}{2}&\frac{3}{2}\\
0&0&-1\\
\end{pmatrix}$$
Wednesday, July 31, 2013
Tuesday, July 30, 2013
4.5.23
$|G|=462 \Longrightarrow G$ is not simple.
$\textit{Proof}$ : Note that $462=2 \cdot 3 \cdot 5 \cdot 7$. By Sylow (iii), we know that $n_{11} \equiv 1 (mod11)$ and $n_{11} \mid 2 \cdot 3 \cdot 7 = 42$. So, $$n_{11}\in{\{1,12,23,34\}}$$ But, only the number 1 from this list divides 42. So, this forces $n_{11}=1$. By the conjugacy portion (ii) of Sylow's Theorem, Sylow p-subgroups are conjugate to each other. But if there is only 1, then it is conjugate to itself, thus normal (this is corollary 20, p. 142). So, we have identified a non-trivial, normal subgroup of $G$.
$\therefore$ $G$ is not simple. $\blacksquare$
$\textit{Proof}$ : Note that $462=2 \cdot 3 \cdot 5 \cdot 7$. By Sylow (iii), we know that $n_{11} \equiv 1 (mod11)$ and $n_{11} \mid 2 \cdot 3 \cdot 7 = 42$. So, $$n_{11}\in{\{1,12,23,34\}}$$ But, only the number 1 from this list divides 42. So, this forces $n_{11}=1$. By the conjugacy portion (ii) of Sylow's Theorem, Sylow p-subgroups are conjugate to each other. But if there is only 1, then it is conjugate to itself, thus normal (this is corollary 20, p. 142). So, we have identified a non-trivial, normal subgroup of $G$.
$\therefore$ $G$ is not simple. $\blacksquare$
D+F 7.3.10
Decide which of the following are ideals in $\mathbb{Z}[x]$
(a) The set of all polynomials whose constant term is a multiple of 3.
$\textit{Answer}$ : This set, call it $J$, is a subring. $0$ is a multiple of 3 i.e. $0 \equiv 0 (mod3)$. So, $0\in{J}$. Also, polynomial subtraction leaves the difference of 2 polynomial's constant terms in the form $3n-3m=3(n-m) \equiv 0 (mod3) \quad n,m,\in{\mathbb{Z}}$.
Multiplication of polynomials is done by continued use of the distributive law. The only purely constant term in the product of 2 polynomials in $\mathbb{Z}[x]$ will be the last term. Let $f(x)$ be a polynomial with constant term $3k, k\in{\mathbb{Z}}$. Let $g(x)\in{\mathbb{Z}[x]}$. Then, the product polynomial $f(x)g(x)=g(x)f(x)$ will have constant term $3\cdot k \cdot n \equiv 0 (mod 3)$. Thus, the set of polynomials with a constant term being a multiple of 3 is closed under left and right multiplication in the ring $\mathbb{Z}[x]$.
$\therefore$ $J$ is an ideal of the ring $\mathbb{Z}[x]$.
(b) The set of all polynomials whose coefficient of $x^2$ is a multiple of 3.
$\textit{Answer}$ : Consider the polynomial $3x^2+x$ which is in the given set. Then, $(3x^2+x)\cdot (3x^2+x)=9x^4+6x^3+x^2$. Thus, this set is not even closed under multiplication by its own elements. So, it is not an ideal of $\mathbb{Z}[x]$.
(c) The set of all polynomials whose constant term, coefficient of $x$ and coefficient of $x^2$ are zero.
$\textit{Answer}$ : This set, J, is a subring since $0\in{J}$ and subtraction does not change degrees of terms.
Multiplication of polynomials only adds degree to terms. Polynomials in $J$ either have degree 0 (the zero element) or degree greater than 2. So, the product of an arbitrary polynomial with one in $J$ must have either degree 0 or have degree greater than 2. Thus, this set is closed under right and left multiplication by polynomials in $\mathbb{Z}[x]$.
$\therefore$ $J$ is an ideal of $\mathbb{Z}[x]$.
(d) $\mathbb{Z}[x^2]$ i.e. the polynomials in which only even powers of $x$ appear.
$\textit{Answer}$ : This set, $J$, is a subring since $0\in{J}$ and subtraction of polynomials does not change the degree of any individual terms. However, $J$ is not an ideal. Consider $x^2\in{J}$. $x^2 \cdot x = x^3 \not \in{J}$.
$\therefore $ $J$ is not an ideal of $\mathbb{Z}[x]$.
(e) the set of polynomials whose coefficients sum to zero.
$\textit{Answer}$ : $p(x)=0\in{J}$. If $f(x),g(x) \in{J}$ then $f(1)-g(1)=(f-g)(1)=0-0=0$. Now let $f(x)\in{J}$ and $h(x)\in{\mathbb{Z}[x]}$. Then, $f(1)h(1)=0 \cdot c = (fh)(1)= 0$ Thus, $fh \in{J}$. Polynomial multiplication is commutative, so $hf\in{J}$. So, $J$ contains the additive identity, is closed under subtraction and is closed under left and right multiplication by arbitrary elements in $\mathbb{Z}[x]$.
$\therefore \quad J$ is an ideal of $\mathbb{Z}[x]$.
(f) the set of polynomials $p(x)$ s.t. $p'(0)=0$, where $p'(x)$ is the usual first derivative of $p(x)$ with respect to $x$.
$\textit{Answer}$ : Let $f(x)=x^2+1 \in{J}$ and $g(x)=x^2+x\in{\mathbb{Z}[x]}$. Then, $$(fg)'(0)=f(0)g'(0)+g(0)f'(0)=1 \cdot 1 + 0 \cdot 0 = 1$$ So, $fg \not \in{J}$.
$\therefore \quad J$ is not an ideal of $\mathbb{Z}[x]$.
(a) The set of all polynomials whose constant term is a multiple of 3.
$\textit{Answer}$ : This set, call it $J$, is a subring. $0$ is a multiple of 3 i.e. $0 \equiv 0 (mod3)$. So, $0\in{J}$. Also, polynomial subtraction leaves the difference of 2 polynomial's constant terms in the form $3n-3m=3(n-m) \equiv 0 (mod3) \quad n,m,\in{\mathbb{Z}}$.
Multiplication of polynomials is done by continued use of the distributive law. The only purely constant term in the product of 2 polynomials in $\mathbb{Z}[x]$ will be the last term. Let $f(x)$ be a polynomial with constant term $3k, k\in{\mathbb{Z}}$. Let $g(x)\in{\mathbb{Z}[x]}$. Then, the product polynomial $f(x)g(x)=g(x)f(x)$ will have constant term $3\cdot k \cdot n \equiv 0 (mod 3)$. Thus, the set of polynomials with a constant term being a multiple of 3 is closed under left and right multiplication in the ring $\mathbb{Z}[x]$.
$\therefore$ $J$ is an ideal of the ring $\mathbb{Z}[x]$.
(b) The set of all polynomials whose coefficient of $x^2$ is a multiple of 3.
$\textit{Answer}$ : Consider the polynomial $3x^2+x$ which is in the given set. Then, $(3x^2+x)\cdot (3x^2+x)=9x^4+6x^3+x^2$. Thus, this set is not even closed under multiplication by its own elements. So, it is not an ideal of $\mathbb{Z}[x]$.
(c) The set of all polynomials whose constant term, coefficient of $x$ and coefficient of $x^2$ are zero.
$\textit{Answer}$ : This set, J, is a subring since $0\in{J}$ and subtraction does not change degrees of terms.
Multiplication of polynomials only adds degree to terms. Polynomials in $J$ either have degree 0 (the zero element) or degree greater than 2. So, the product of an arbitrary polynomial with one in $J$ must have either degree 0 or have degree greater than 2. Thus, this set is closed under right and left multiplication by polynomials in $\mathbb{Z}[x]$.
$\therefore$ $J$ is an ideal of $\mathbb{Z}[x]$.
(d) $\mathbb{Z}[x^2]$ i.e. the polynomials in which only even powers of $x$ appear.
$\textit{Answer}$ : This set, $J$, is a subring since $0\in{J}$ and subtraction of polynomials does not change the degree of any individual terms. However, $J$ is not an ideal. Consider $x^2\in{J}$. $x^2 \cdot x = x^3 \not \in{J}$.
$\therefore $ $J$ is not an ideal of $\mathbb{Z}[x]$.
(e) the set of polynomials whose coefficients sum to zero.
$\textit{Answer}$ : $p(x)=0\in{J}$. If $f(x),g(x) \in{J}$ then $f(1)-g(1)=(f-g)(1)=0-0=0$. Now let $f(x)\in{J}$ and $h(x)\in{\mathbb{Z}[x]}$. Then, $f(1)h(1)=0 \cdot c = (fh)(1)= 0$ Thus, $fh \in{J}$. Polynomial multiplication is commutative, so $hf\in{J}$. So, $J$ contains the additive identity, is closed under subtraction and is closed under left and right multiplication by arbitrary elements in $\mathbb{Z}[x]$.
$\therefore \quad J$ is an ideal of $\mathbb{Z}[x]$.
(f) the set of polynomials $p(x)$ s.t. $p'(0)=0$, where $p'(x)$ is the usual first derivative of $p(x)$ with respect to $x$.
$\textit{Answer}$ : Let $f(x)=x^2+1 \in{J}$ and $g(x)=x^2+x\in{\mathbb{Z}[x]}$. Then, $$(fg)'(0)=f(0)g'(0)+g(0)f'(0)=1 \cdot 1 + 0 \cdot 0 = 1$$ So, $fg \not \in{J}$.
$\therefore \quad J$ is not an ideal of $\mathbb{Z}[x]$.
Monday, July 29, 2013
7.1.21
Let $X$ be a nonempty set and let $\mathcal{P}(X)$ be the set of all subsets of $X$ (the power set of $X$). Define addition and multiplication on $\mathcal{P}(X)$ by:
$$A+B \equiv (A-B) \cup (B-A)$$
and
$$A \cdot B \equiv A \cap B$$
i.e. addition is the symmetric difference of sets and multiplication is intersection of sets.
(a) Prove that $\mathcal{P}(X)$ is a ring under these operations. Note: $\mathcal{P}(X)$ and its subrings are often referred to as rings of sets.
(b) Prove that this ring is commutative, has an identity and is a Boolean ring.
$\textit{Proof}$ : To show that $\mathcal{P}(X)$ is a ring, it suffices to show that ($\mathcal{P}(X),+,\cdot)$ satisfies the 3 axions of rings. Using basic set theory, we see that
(i) $(\mathcal{P}(X),+)$ is an Abelian group since $$A+B=(A-B)\cup (B-A)=(B-A) \cup (A-B)=B+A$$ (ii) $\cdot$ is associative since $$A\cdot(B\cdot C)=A \cap (B \cap C)=(A \cap B) \cap C= (A \cdot B) \cdot C$$ (iii) ($\mathcal{P}(X),+,\cdot$) also satisfies distributive laws since $$(A+B) \cdot C=((A-B) \cup (B-A)) \cap C$$ $$=((A\cap C)-(B\cap C)) \cup ((B \cap C)-(A \cap C))$$ $$=(A \cap C) + (B \cap C) = A \cdot C + B \cdot C$$ Right distribution is shown similarly.
So, ($\mathcal{P}(X),+,\cdot$) is a ring.
$\cdot$ is a commutative operation since $$A \cdot B=A \cap B = B \cap A = B \cdot A$$ The universe or entire set $X$ is s.t. $$A \cdot X= A \cap X= X \cap A = A \quad \forall A \in{\mathcal{P}(X)}$$ So, $X$ is the (multiplicative) identity. As a note, $\emptyset$ is the additive identity since $$A + \emptyset = (A-\emptyset) \cup (\emptyset - A) = A \cup \emptyset = \emptyset + A=A$$ Finally, ($\mathcal{P}(X),+,\cdot$) is a Boolean ring since $$A^2=A \cdot A = A \cap A = A$$ $ \therefore$ ($\mathcal{P}(X),+,\cdot$) is a Boolean ring (automatically commutative also by exercise 7.1.15) with identity. $\blacksquare$
(a) Prove that $\mathcal{P}(X)$ is a ring under these operations. Note: $\mathcal{P}(X)$ and its subrings are often referred to as rings of sets.
(b) Prove that this ring is commutative, has an identity and is a Boolean ring.
$\textit{Proof}$ : To show that $\mathcal{P}(X)$ is a ring, it suffices to show that ($\mathcal{P}(X),+,\cdot)$ satisfies the 3 axions of rings. Using basic set theory, we see that
(i) $(\mathcal{P}(X),+)$ is an Abelian group since $$A+B=(A-B)\cup (B-A)=(B-A) \cup (A-B)=B+A$$ (ii) $\cdot$ is associative since $$A\cdot(B\cdot C)=A \cap (B \cap C)=(A \cap B) \cap C= (A \cdot B) \cdot C$$ (iii) ($\mathcal{P}(X),+,\cdot$) also satisfies distributive laws since $$(A+B) \cdot C=((A-B) \cup (B-A)) \cap C$$ $$=((A\cap C)-(B\cap C)) \cup ((B \cap C)-(A \cap C))$$ $$=(A \cap C) + (B \cap C) = A \cdot C + B \cdot C$$ Right distribution is shown similarly.
So, ($\mathcal{P}(X),+,\cdot$) is a ring.
$\cdot$ is a commutative operation since $$A \cdot B=A \cap B = B \cap A = B \cdot A$$ The universe or entire set $X$ is s.t. $$A \cdot X= A \cap X= X \cap A = A \quad \forall A \in{\mathcal{P}(X)}$$ So, $X$ is the (multiplicative) identity. As a note, $\emptyset$ is the additive identity since $$A + \emptyset = (A-\emptyset) \cup (\emptyset - A) = A \cup \emptyset = \emptyset + A=A$$ Finally, ($\mathcal{P}(X),+,\cdot$) is a Boolean ring since $$A^2=A \cdot A = A \cap A = A$$ $ \therefore$ ($\mathcal{P}(X),+,\cdot$) is a Boolean ring (automatically commutative also by exercise 7.1.15) with identity. $\blacksquare$
Friday, July 26, 2013
D+F 14.6.5
Determine the Galois group of $x^4+4$.
$\textit{Solution}$ : By problem 13.2.3, we know that the splitting field for $x^4+4$ is only $\mathbb{Q}[i]$, which is a degree 2 extension. The 4 roots of $x^4+4$ are $\pm 1 \pm i$ so $x^4+4$ is separable since none of the roots are repeated. By prop. 5 p.562, this means that $$|Aut(\mathbb{Q}[i])/\mathbb{Q})|=|\mathbb{Q}[i]:\mathbb{Q}|=2$$ $\Longrightarrow \mathbb{Q}[i]$ is Galois over $\mathbb{Q}$ and the Galois group is $\mathbb{Z}_2$.
$\textit{Solution}$ : By problem 13.2.3, we know that the splitting field for $x^4+4$ is only $\mathbb{Q}[i]$, which is a degree 2 extension. The 4 roots of $x^4+4$ are $\pm 1 \pm i$ so $x^4+4$ is separable since none of the roots are repeated. By prop. 5 p.562, this means that $$|Aut(\mathbb{Q}[i])/\mathbb{Q})|=|\mathbb{Q}[i]:\mathbb{Q}|=2$$ $\Longrightarrow \mathbb{Q}[i]$ is Galois over $\mathbb{Q}$ and the Galois group is $\mathbb{Z}_2$.
Thursday, July 25, 2013
D+F 13.4.3
Determine the splitting field and its degree over $\mathbb{Q}$ for $x^4+x^2+1$.
$\textit{Solution}$ : First note that $x^4+x^2+1$ partially factors over $\mathbb{Q}$ as follows: $$x^4+x^2+1=x^2(x^2+1)+(x^2+1)-x^2=(x^2+1)^2-x^2$$ $$=((x^2+1)-(x))((x^2+1)+(x))=(x^2+x+1)(x^2-x+1)$$ Solving for the roots of these 2 quaratics, we obtain 4 solutions: $$\zeta_6,\zeta_6^2,\zeta_6^4,\zeta_6^5$$ Where $\zeta_6=\dfrac{1+\sqrt{-3}}{2}$.
So, the only element that we need to adjoin to $\mathbb{Q}$ to make $x^4+x^2+1$ split completely is $\sqrt{-3}$. So, our splitting field is $\mathbb{Q}[\sqrt{-3}]$ ...a degree 2 extension.
$\textit{Solution}$ : First note that $x^4+x^2+1$ partially factors over $\mathbb{Q}$ as follows: $$x^4+x^2+1=x^2(x^2+1)+(x^2+1)-x^2=(x^2+1)^2-x^2$$ $$=((x^2+1)-(x))((x^2+1)+(x))=(x^2+x+1)(x^2-x+1)$$ Solving for the roots of these 2 quaratics, we obtain 4 solutions: $$\zeta_6,\zeta_6^2,\zeta_6^4,\zeta_6^5$$ Where $\zeta_6=\dfrac{1+\sqrt{-3}}{2}$.
So, the only element that we need to adjoin to $\mathbb{Q}$ to make $x^4+x^2+1$ split completely is $\sqrt{-3}$. So, our splitting field is $\mathbb{Q}[\sqrt{-3}]$ ...a degree 2 extension.
D+F 13.2.3
Determine the minimal polynomial over $\mathbb{Q}$ for the element $1+i$.
$\textit{Solution}$ : We want a monic, irreducible polynomial of minimal degree over $\mathbb{Q}$ s.t. $m(1+i)=0$. $$(1+i)(1+i)=2i$$ $$(2i)^2=-4$$ So, $$f(x)=x^4+4$$ satisfies $$f(1+i)=0$$ However, $x^4+4$ actually factors partially over $\mathbb{Q}$ as follows: $$x^4+4=(x^2+2)^2-4x^2=(x^2+2)^2-(2x)^2$$ $$=((x^2+2)-(2x))((x^2+2)+(2x))=(x^2-2x+2)(x^2+2x+2)$$ The first of the above quadratics has the property that $$(1+i)^2-2(1+i)+2=2i-2-2i+2=0$$ ...a smaller degree polynomial with $1+i$ as a root. This is consistent with the observation that $$\mathbb{Q}[1+i]=\{a+b(1+i):a,b\in{\mathbb{Q}}\}=\{a+b+bi:a,b\in{\mathbb{Q}}\}$$ $$=\{c+bi:c,b\in{\mathbb{Q}}=\mathbb{Q}[i]$$ ...a degree 2 extension.
By Prop. 11 p. 521, we know that the degree of our field extension should be equal to the degree of our minimal polynomial. So, we may safely conclude that the minimal polynomial for the element $1+i$ with $\mathbb{Q}$ as our base field is $m(x)=x^2-2x+2$.
$\textit{Solution}$ : We want a monic, irreducible polynomial of minimal degree over $\mathbb{Q}$ s.t. $m(1+i)=0$. $$(1+i)(1+i)=2i$$ $$(2i)^2=-4$$ So, $$f(x)=x^4+4$$ satisfies $$f(1+i)=0$$ However, $x^4+4$ actually factors partially over $\mathbb{Q}$ as follows: $$x^4+4=(x^2+2)^2-4x^2=(x^2+2)^2-(2x)^2$$ $$=((x^2+2)-(2x))((x^2+2)+(2x))=(x^2-2x+2)(x^2+2x+2)$$ The first of the above quadratics has the property that $$(1+i)^2-2(1+i)+2=2i-2-2i+2=0$$ ...a smaller degree polynomial with $1+i$ as a root. This is consistent with the observation that $$\mathbb{Q}[1+i]=\{a+b(1+i):a,b\in{\mathbb{Q}}\}=\{a+b+bi:a,b\in{\mathbb{Q}}\}$$ $$=\{c+bi:c,b\in{\mathbb{Q}}=\mathbb{Q}[i]$$ ...a degree 2 extension.
By Prop. 11 p. 521, we know that the degree of our field extension should be equal to the degree of our minimal polynomial. So, we may safely conclude that the minimal polynomial for the element $1+i$ with $\mathbb{Q}$ as our base field is $m(x)=x^2-2x+2$.
Wednesday, July 24, 2013
D+F 14.2.1
Determine the minimal polynomial over $\mathbb{Q}$ for the element $\sqrt{2} + \sqrt{5}$.
$\textit{Solution}$ : We wish to find a monic, irreducible polynomial of minimal degree denoted $m_{\alpha,\mathbb{Q}}$ s.t. $m_{\alpha,\mathbb{Q}}(\alpha)=0$ where $\alpha \in{\mathbb{Q}[\sqrt{2}+\sqrt{5}]}$.
$\mathbb{Q}[\sqrt{2}+\sqrt{5}]$ will be the splitting field for $m_{\alpha,\mathbb{Q}}$.
Label $\alpha = \sqrt{2} + \sqrt{5}$. Then, $$\alpha-\sqrt{2} = \sqrt{5} $$ Squaring both sides, $$\alpha^2-2\sqrt{2}\alpha+2=5$$ $$\alpha^2-3=2\sqrt{2}\alpha$$ Squaring both sides again, $$\alpha^4-6\alpha^2+9=8\alpha^2$$ $$\alpha^4-14\alpha^2+9=0$$ ...Is the minimal polynomial for the element $\sqrt{2} + \sqrt{5}$ over $\mathbb{Q}$. Note that it is irreducible over $\mathbb{Q}[x]$ by the rational roots theorem.
$\textit{Solution}$ : We wish to find a monic, irreducible polynomial of minimal degree denoted $m_{\alpha,\mathbb{Q}}$ s.t. $m_{\alpha,\mathbb{Q}}(\alpha)=0$ where $\alpha \in{\mathbb{Q}[\sqrt{2}+\sqrt{5}]}$.
$\mathbb{Q}[\sqrt{2}+\sqrt{5}]$ will be the splitting field for $m_{\alpha,\mathbb{Q}}$.
Label $\alpha = \sqrt{2} + \sqrt{5}$. Then, $$\alpha-\sqrt{2} = \sqrt{5} $$ Squaring both sides, $$\alpha^2-2\sqrt{2}\alpha+2=5$$ $$\alpha^2-3=2\sqrt{2}\alpha$$ Squaring both sides again, $$\alpha^4-6\alpha^2+9=8\alpha^2$$ $$\alpha^4-14\alpha^2+9=0$$ ...Is the minimal polynomial for the element $\sqrt{2} + \sqrt{5}$ over $\mathbb{Q}$. Note that it is irreducible over $\mathbb{Q}[x]$ by the rational roots theorem.
Friday, July 19, 2013
D+F 4.5.22
$|G|=132 \Longrightarrow G$ is not simple.
$\textit{Proof}$ : The prime factorization of 132 is $132=2^2 \cdot 3 \cdot 11$. We will again use the 3rd part of Sylow's theorem to count the possible numbers of Sylow p-subgroups.
Writing $132=11^1 \cdot 12$ we note that $11 \nmid 12$ so Sylow (iii) gives $$n_{11} \equiv 1 (mod11)$$ and, $$n_{11} \mid 12$$ So, $n_{11} = 1,12$. Similarly, $$n_3 \equiv 1(mod3)$$ $$n_3 \mid 44$$ So, the only positive integers satisfying these constraints is 1 and 4, so $n_3=1,4$.
$$n_2 \equiv 1(mod2)$$ $$n_2 \mid 33$$ So, $n_3 = 1,3,11,33$
Now, suppose to the contrary that $G$ is a simple group. So, this means that $G$ has no non-trivial normal subgroups. By Corollary 20 p.142, this means that $n_p \neq 1 \quad \forall p$ in the prime factorization of $|G|=132$. So, this forces $n_{11}=12$, $n_3=4$ and $n_2=3,11,33$. The Sylow 11-subgroups are subgroups of order 11, thus isomorphic to $\mathbb{Z}_{11}$, which has 10 elements of order 11. So, with $n_{11}=12$, this means that in $G$, there are $12 \cdot 10 = 120$ elements of order 11. The Sylow 3-subgroups have order 3, so they are isomorphic to $\mathbb{Z}_3$, which has 2 elements of order 3. With $n_3 = 4$, this means that there are $4 \cdot 2 = 8$ elements of order 3 in $G$.
So far, we have counted $120+8=128$ non-identity elements of $G$. This leaves only 4 remaining. However, the Sylow 2-subgroups have order $2^2=4$. Thus, the remaining 4 elements must be exactly the Sylow 2-subgroup (which is either $\mathbb{Z}_4$ or $\mathbb{Z}_2 \bigoplus \mathbb{Z}_2$, but it doesn't actually matter at this point). So, $n_2=1$ giving us a non-trivial, normal subgroup by Corollary 20 p.142. This contradicts the assumption that $G$ was simple. In any case, we must have that $n_p=1$ for at least one $p$ in the factorization of 132.
So, any group of order 132 is not simple. $\blacksquare$
$\textit{Proof}$ : The prime factorization of 132 is $132=2^2 \cdot 3 \cdot 11$. We will again use the 3rd part of Sylow's theorem to count the possible numbers of Sylow p-subgroups.
Writing $132=11^1 \cdot 12$ we note that $11 \nmid 12$ so Sylow (iii) gives $$n_{11} \equiv 1 (mod11)$$ and, $$n_{11} \mid 12$$ So, $n_{11} = 1,12$. Similarly, $$n_3 \equiv 1(mod3)$$ $$n_3 \mid 44$$ So, the only positive integers satisfying these constraints is 1 and 4, so $n_3=1,4$.
$$n_2 \equiv 1(mod2)$$ $$n_2 \mid 33$$ So, $n_3 = 1,3,11,33$
Now, suppose to the contrary that $G$ is a simple group. So, this means that $G$ has no non-trivial normal subgroups. By Corollary 20 p.142, this means that $n_p \neq 1 \quad \forall p$ in the prime factorization of $|G|=132$. So, this forces $n_{11}=12$, $n_3=4$ and $n_2=3,11,33$. The Sylow 11-subgroups are subgroups of order 11, thus isomorphic to $\mathbb{Z}_{11}$, which has 10 elements of order 11. So, with $n_{11}=12$, this means that in $G$, there are $12 \cdot 10 = 120$ elements of order 11. The Sylow 3-subgroups have order 3, so they are isomorphic to $\mathbb{Z}_3$, which has 2 elements of order 3. With $n_3 = 4$, this means that there are $4 \cdot 2 = 8$ elements of order 3 in $G$.
So far, we have counted $120+8=128$ non-identity elements of $G$. This leaves only 4 remaining. However, the Sylow 2-subgroups have order $2^2=4$. Thus, the remaining 4 elements must be exactly the Sylow 2-subgroup (which is either $\mathbb{Z}_4$ or $\mathbb{Z}_2 \bigoplus \mathbb{Z}_2$, but it doesn't actually matter at this point). So, $n_2=1$ giving us a non-trivial, normal subgroup by Corollary 20 p.142. This contradicts the assumption that $G$ was simple. In any case, we must have that $n_p=1$ for at least one $p$ in the factorization of 132.
So, any group of order 132 is not simple. $\blacksquare$
Wednesday, July 17, 2013
D+F 7.3.18
(a) If $I$ and $J$ are ideals of $R$, prove that their intersection $I \cap J$ is also an ideal of $R$.
(b) Prove that the intersection of an arbitrary nonempty collection of ideals is again an ideal (do not assume that the collection is countable)
$\textit{Proof}$ : For (a), we wish to show that $I \cap J$ is a nonempty subring of $R$ which is closed under left and right multiplication. With $I$ and $J$ being ideals of $R$, $0\in{I}$ and $0\in{J}$ so, $0\in{I \cap J}$ so $I \cap J \neq \emptyset$.
By exercise 7.1.4, we know that the intersection of subrings is a subring. We just need to show that $I \cap J$ is closed under left and right multiplication. Let $x\in{I \cap J}$ and $r\in{R}$. We want to show that $rx,xr\in{J} \quad \forall r\in{R}$.
$$x\in{I \cap J} \Longleftrightarrow x\in{I} \wedge x\in{J}$$ $$x\in{I} \Longrightarrow rx,xr\in{I}$$ $$x\in{J} \Longrightarrow rx,xr \in{J}$$ Thus, $xr,rx \in{I \cap J}$.
$\therefore \quad I \cap J$ is an ideal of $R$
For (b) let $$\bigcap_{k\in{\Lambda}}I_k$$ be the intersection of ideals in $R$ where $\Lambda$ is an arbitrary index of ideals.
$$0\in{I_k} \quad \forall k \in{\Lambda} \Longrightarrow 0\in{\bigcap_{k\in{\Lambda}}I_k} \Longrightarrow \bigcap_{k\in{\Lambda}}I_k \neq \emptyset$$ Let $$x\in{\bigcap_{k\in{\Lambda}}I_k}$$ and $r\in{R}$. Then, $rx,xr\in{I_k}$ $\forall k\in{\Lambda}$ since each $I_k$ is an ideal. This implies that $$rx,xr \in{\bigcap_{k\in{\Lambda}}I_k}$$ $$\Longrightarrow \bigcap_{k\in{\Lambda}}I_k$$ is an ideal of $R \quad \blacksquare$
(b) Prove that the intersection of an arbitrary nonempty collection of ideals is again an ideal (do not assume that the collection is countable)
$\textit{Proof}$ : For (a), we wish to show that $I \cap J$ is a nonempty subring of $R$ which is closed under left and right multiplication. With $I$ and $J$ being ideals of $R$, $0\in{I}$ and $0\in{J}$ so, $0\in{I \cap J}$ so $I \cap J \neq \emptyset$.
By exercise 7.1.4, we know that the intersection of subrings is a subring. We just need to show that $I \cap J$ is closed under left and right multiplication. Let $x\in{I \cap J}$ and $r\in{R}$. We want to show that $rx,xr\in{J} \quad \forall r\in{R}$.
$$x\in{I \cap J} \Longleftrightarrow x\in{I} \wedge x\in{J}$$ $$x\in{I} \Longrightarrow rx,xr\in{I}$$ $$x\in{J} \Longrightarrow rx,xr \in{J}$$ Thus, $xr,rx \in{I \cap J}$.
$\therefore \quad I \cap J$ is an ideal of $R$
For (b) let $$\bigcap_{k\in{\Lambda}}I_k$$ be the intersection of ideals in $R$ where $\Lambda$ is an arbitrary index of ideals.
$$0\in{I_k} \quad \forall k \in{\Lambda} \Longrightarrow 0\in{\bigcap_{k\in{\Lambda}}I_k} \Longrightarrow \bigcap_{k\in{\Lambda}}I_k \neq \emptyset$$ Let $$x\in{\bigcap_{k\in{\Lambda}}I_k}$$ and $r\in{R}$. Then, $rx,xr\in{I_k}$ $\forall k\in{\Lambda}$ since each $I_k$ is an ideal. This implies that $$rx,xr \in{\bigcap_{k\in{\Lambda}}I_k}$$ $$\Longrightarrow \bigcap_{k\in{\Lambda}}I_k$$ is an ideal of $R \quad \blacksquare$
Monday, July 15, 2013
D+F 4.5.34
Let $P\in{Syl_p(G)}$ and let $N \unlhd G$.
(a) Use the conjugacy part of Sylow's Theorem to prove that $P \cap N\in{Syl_p(N)}$
(b) Prove that $PN/N\in{Syl_p(G/N)}$
The second part of Sylow's Theorem states: Let $P$ be a Sylow p-subgroup of $G$ and $Q$ a p-subgroup of $G$. Then, $\exists g\in{G}$ s.t. $Q \leq gPg^{-1}$, i.e. $Q$ is contained in some conjugate of $P$.
$\textit{Proof}$ : Let $|G|=p^am$ where $p \nmid m$. $|N|$ divides $|G|$ by Lagrange, so let $|N|=p^bn$ where $b \leq a$, $p \nmid n$ and $n \leq m$. To show that $P \cap N \in{Syl_p(N)}$ we need to show that $|P \cap N|=p^b$.
Observe that $P \cap N$ is a p-group: $$|P \cap N| \mid |P|=p^a$$ $$|P \cap N| \mid |N|=p^bn$$ Since $p$ is prime, $$\Longrightarrow |P \cap N|=p^c \quad (c \leq b)$$ We want to now show that $b \leq c$.
Corollary 15, p. 94 states that if $H,K \leq G$ and $H \leq N_G(K)$ then the subset product, $HK \leq G$. In this case, $N \lhd G \Longrightarrow N_G(N) = G$, so the subset product $PN \leq G$. By the 4th isomorphism theorem, we may mod out by $N$, so we get that $$PN/N \leq G/N \quad (\star)$$ The order of the subset product is given by the formula $$|PN| = \frac{|P| \cdot |N|}{|P \cap N|} \Longrightarrow \frac{|PN|}{|N|}=\frac{|P|}{|P \cap N|}$$ Combining with $(\star)$ we get $$\frac{|P|}{|P \cap N|} \mid \frac{|G|}{|N|}=\frac{p^am}{p^bn}=p^{a-b}\frac{m}{n}$$ $$\frac{p^a}{p^c} \mid p^{a-b}\frac{m}{n}$$ $p^{a-c}$ is a power of a prime and $p^{a-b}$ is a power of a prime, so $$p^{a-c} \mid p^{a-b} \Longrightarrow a-c \leq a-b \Longrightarrow b \leq c$$ Thus, $ |P \cap N| = p^b \quad \therefore P \cap N \in{Syl_p(N)}$.
To prove that $PN/N\in{Syl_p(G/N)}$, we make a similar argument based on subgroup orders. $$|G/N|=\frac{|G|}{|N|}=\frac{p^am}{p^bn}=p^{a-b}(\frac{m}{n})$$ Where we can say that $\frac{m}{n}$ is in lowest terms and is a positive integer.
Again, the order of subset products is given by $$|PN|=\frac{|P| \cdot |N|}{|P \cap N|}$$ Rearranging, we get $$|PN/N|=\frac{|P|}{|P \cap N|}=\frac{p^a}{p^b}=p^{a-b}$$ $$\Longrightarrow PN/N \in{Syl_p(G/N)} \quad \blacksquare$$
(a) Use the conjugacy part of Sylow's Theorem to prove that $P \cap N\in{Syl_p(N)}$
(b) Prove that $PN/N\in{Syl_p(G/N)}$
The second part of Sylow's Theorem states: Let $P$ be a Sylow p-subgroup of $G$ and $Q$ a p-subgroup of $G$. Then, $\exists g\in{G}$ s.t. $Q \leq gPg^{-1}$, i.e. $Q$ is contained in some conjugate of $P$.
$\textit{Proof}$ : Let $|G|=p^am$ where $p \nmid m$. $|N|$ divides $|G|$ by Lagrange, so let $|N|=p^bn$ where $b \leq a$, $p \nmid n$ and $n \leq m$. To show that $P \cap N \in{Syl_p(N)}$ we need to show that $|P \cap N|=p^b$.
Observe that $P \cap N$ is a p-group: $$|P \cap N| \mid |P|=p^a$$ $$|P \cap N| \mid |N|=p^bn$$ Since $p$ is prime, $$\Longrightarrow |P \cap N|=p^c \quad (c \leq b)$$ We want to now show that $b \leq c$.
Corollary 15, p. 94 states that if $H,K \leq G$ and $H \leq N_G(K)$ then the subset product, $HK \leq G$. In this case, $N \lhd G \Longrightarrow N_G(N) = G$, so the subset product $PN \leq G$. By the 4th isomorphism theorem, we may mod out by $N$, so we get that $$PN/N \leq G/N \quad (\star)$$ The order of the subset product is given by the formula $$|PN| = \frac{|P| \cdot |N|}{|P \cap N|} \Longrightarrow \frac{|PN|}{|N|}=\frac{|P|}{|P \cap N|}$$ Combining with $(\star)$ we get $$\frac{|P|}{|P \cap N|} \mid \frac{|G|}{|N|}=\frac{p^am}{p^bn}=p^{a-b}\frac{m}{n}$$ $$\frac{p^a}{p^c} \mid p^{a-b}\frac{m}{n}$$ $p^{a-c}$ is a power of a prime and $p^{a-b}$ is a power of a prime, so $$p^{a-c} \mid p^{a-b} \Longrightarrow a-c \leq a-b \Longrightarrow b \leq c$$ Thus, $ |P \cap N| = p^b \quad \therefore P \cap N \in{Syl_p(N)}$.
To prove that $PN/N\in{Syl_p(G/N)}$, we make a similar argument based on subgroup orders. $$|G/N|=\frac{|G|}{|N|}=\frac{p^am}{p^bn}=p^{a-b}(\frac{m}{n})$$ Where we can say that $\frac{m}{n}$ is in lowest terms and is a positive integer.
Again, the order of subset products is given by $$|PN|=\frac{|P| \cdot |N|}{|P \cap N|}$$ Rearranging, we get $$|PN/N|=\frac{|P|}{|P \cap N|}=\frac{p^a}{p^b}=p^{a-b}$$ $$\Longrightarrow PN/N \in{Syl_p(G/N)} \quad \blacksquare$$
Saturday, July 13, 2013
D+F 7.3.24
Let $\varphi:R \rightarrow S$ be a ring homomorphism.
(a) Prove that if $J$ is an ideal of $S$ then $\varphi^{-1}(J)$ is an ideal of $R$. In other words, the preimage of an ideal is an ideal.
$\textit{Proof}$ : Ideals are subrings that are closed under left and right multiplication by arbitrary ring elements. To show that $\varphi^{-1}(J)$ is a subring, we show that it is nonempty and closed under both ring operations. Since $J$ is an ideal in $S$, $$0\in{J} \Longrightarrow \varphi^{-1}(0)\in{\varphi^{-1}(J)} \Longrightarrow \varphi^{-1}(J) \neq \emptyset$$ Let $r_1$, $r_2 \in{\varphi^{-1}(J)}$. So, $\varphi(r_1)$, $\varphi(r_2) \in{J}$.
$J$ is an ideal, so $$\varphi(r_1)\varphi(r_2)=\varphi(r_1r_2)\in{J} \Longrightarrow r_1r_2\in{\varphi^{-1}(J)}$$ So, $\varphi^{-1}(J)$ is closed under multiplication making $\varphi^{-1}(J)$ a subring of $R$.
To show that $\varphi^{-1}(J)$ is an ideal, we show that it is closed under multiplication by all elements of $R$. Let $x\in{R}$. $$\varphi(xr)=\varphi(x)\varphi(r)$$ $J$ is an ideal, so, $$\varphi(r)\in{J} \Longrightarrow \varphi(x)\varphi(r)\in{J}$$ So, $\varphi^{-1}(J)$ is closed under left multiplication. Similarly, $$\varphi(rx)=\varphi(r)\varphi(x)$$ $$\varphi(r)\in{J} \Longrightarrow \varphi(x)\varphi(r)\in{J}$$ So, $\varphi^{-1}(J)$ is closed under right multiplication. $\therefore \varphi^{-1}(J)$ is an ideal of $R$ $\blacksquare$
Note: we could have skipped showing that $\varphi^{-1}(J)$ is closed under multiplication by its own elements, since this is taken care of in showing that $\varphi^{-1}(J)$ is closed with left and right multiplication by arbitrary elements in $R$.
(b) Prove that if $\varphi$ is surjective (onto) and $I$ is an ideal of $R$, then $\varphi(I)$ is an ideal of $S$. Give an example where this fails if $\varphi$ is not surjective. In other words, images of ideals are ideals under surjective ring homomorphisms.
$\textit{Proof}$ : Since $I$ is an ideal of $R$, $$0\in{I} \Longrightarrow \varphi(0)\in{\varphi(I)} \Longrightarrow \varphi(I) \neq \emptyset$$ Let $a,b\in{I}$. $I$ is an ideal, so $a+b\in{I}$. $$\varphi(a+b)=\varphi(a)+\varphi(b)\in{\varphi(I)}$$ So, $\varphi(I)$ is closed under addition.
Now, let $y\in{I}$ and $s\in{S}$ . Since $\varphi$ is surjective, $\exists r\in{R}$ s.t. $\varphi(r)=s$ and also $\varphi(x)=y$ with $x\in{I}$. $$xr\in{I} \Longrightarrow \varphi(xr)=\varphi(x)\varphi(r)\in{\varphi(I)}$$ $$rx\in{I} \Longrightarrow \varphi(rx)=\varphi(r)\varphi(x)\in{\varphi(I)}$$ So, $\varphi(I)$ is a left and right ideal, and thus an ideal of $S$.
If we relax the condition that $\varphi$ is surjective, consider the following example:
Let $\phi:\mathbb{Z} \rightarrow \mathbb{Q}$ be the identity map. $\mathbb{Z}$ is trivially an ideal of itself, but its image , which is still just $\mathbb{Z}$ is not an ideal of $\mathbb{Q}$ since $\frac{1}{2} \cdot 1 = \frac{1}{2} \not \in{\mathbb{Z}}$ i.e. $\mathbb{Z}$ is not closed under multiplication by elements in $\mathbb{Q}$.
(a) Prove that if $J$ is an ideal of $S$ then $\varphi^{-1}(J)$ is an ideal of $R$. In other words, the preimage of an ideal is an ideal.
$\textit{Proof}$ : Ideals are subrings that are closed under left and right multiplication by arbitrary ring elements. To show that $\varphi^{-1}(J)$ is a subring, we show that it is nonempty and closed under both ring operations. Since $J$ is an ideal in $S$, $$0\in{J} \Longrightarrow \varphi^{-1}(0)\in{\varphi^{-1}(J)} \Longrightarrow \varphi^{-1}(J) \neq \emptyset$$ Let $r_1$, $r_2 \in{\varphi^{-1}(J)}$. So, $\varphi(r_1)$, $\varphi(r_2) \in{J}$.
$J$ is an ideal, so $$\varphi(r_1)\varphi(r_2)=\varphi(r_1r_2)\in{J} \Longrightarrow r_1r_2\in{\varphi^{-1}(J)}$$ So, $\varphi^{-1}(J)$ is closed under multiplication making $\varphi^{-1}(J)$ a subring of $R$.
To show that $\varphi^{-1}(J)$ is an ideal, we show that it is closed under multiplication by all elements of $R$. Let $x\in{R}$. $$\varphi(xr)=\varphi(x)\varphi(r)$$ $J$ is an ideal, so, $$\varphi(r)\in{J} \Longrightarrow \varphi(x)\varphi(r)\in{J}$$ So, $\varphi^{-1}(J)$ is closed under left multiplication. Similarly, $$\varphi(rx)=\varphi(r)\varphi(x)$$ $$\varphi(r)\in{J} \Longrightarrow \varphi(x)\varphi(r)\in{J}$$ So, $\varphi^{-1}(J)$ is closed under right multiplication. $\therefore \varphi^{-1}(J)$ is an ideal of $R$ $\blacksquare$
Note: we could have skipped showing that $\varphi^{-1}(J)$ is closed under multiplication by its own elements, since this is taken care of in showing that $\varphi^{-1}(J)$ is closed with left and right multiplication by arbitrary elements in $R$.
(b) Prove that if $\varphi$ is surjective (onto) and $I$ is an ideal of $R$, then $\varphi(I)$ is an ideal of $S$. Give an example where this fails if $\varphi$ is not surjective. In other words, images of ideals are ideals under surjective ring homomorphisms.
$\textit{Proof}$ : Since $I$ is an ideal of $R$, $$0\in{I} \Longrightarrow \varphi(0)\in{\varphi(I)} \Longrightarrow \varphi(I) \neq \emptyset$$ Let $a,b\in{I}$. $I$ is an ideal, so $a+b\in{I}$. $$\varphi(a+b)=\varphi(a)+\varphi(b)\in{\varphi(I)}$$ So, $\varphi(I)$ is closed under addition.
Now, let $y\in{I}$ and $s\in{S}$ . Since $\varphi$ is surjective, $\exists r\in{R}$ s.t. $\varphi(r)=s$ and also $\varphi(x)=y$ with $x\in{I}$. $$xr\in{I} \Longrightarrow \varphi(xr)=\varphi(x)\varphi(r)\in{\varphi(I)}$$ $$rx\in{I} \Longrightarrow \varphi(rx)=\varphi(r)\varphi(x)\in{\varphi(I)}$$ So, $\varphi(I)$ is a left and right ideal, and thus an ideal of $S$.
If we relax the condition that $\varphi$ is surjective, consider the following example:
Let $\phi:\mathbb{Z} \rightarrow \mathbb{Q}$ be the identity map. $\mathbb{Z}$ is trivially an ideal of itself, but its image , which is still just $\mathbb{Z}$ is not an ideal of $\mathbb{Q}$ since $\frac{1}{2} \cdot 1 = \frac{1}{2} \not \in{\mathbb{Z}}$ i.e. $\mathbb{Z}$ is not closed under multiplication by elements in $\mathbb{Q}$.
Friday, July 12, 2013
D+F 7.3.17
Let $R$ and $S$ be nonzero rings with identities $1_R$ and $1_S$. Let $\varphi:R \rightarrow S$ be a nonzero ring homomorphism.
(a) Prove that if $\varphi(1_R) \neq 1_S$ then $\varphi(1_R)$ is a zero divisor in $S$. Deduce that if $S$ is an integral domain, then every ring homomorphism from $R$ to $S$ sends the identity of $R$ to the identity of $S$.
(b) Prove that if $u$ is a unit in $R$ and $\varphi(1_R)=1_S$ then $\varphi(u)$ is a unit in $S$ and $\varphi(u^{-1})=\varphi(u)^{-1}$.
$\textit{Proof}$ :
(a): Let $\varphi(1_R)=y\in{S}-\{0\}$. (The identity in $R$ must be sent to a nonzero element of $S$ for $\varphi$ to be nonzero) $$y=\varphi(1_R)=\varphi(1_R1_R)=\varphi(1_R)\varphi(1_R)=y^2$$ $y=y^2$
$y-y^2=0$ (additive inverses exist because rings with respect to the addition operation are abelian groups)
$y(1_S-y)=0$ (distributive law for rings)
$ \therefore $ $y$ is a zero divisor.
If $S$ is an integral domain, then $S$ contains no zero divisors. So, in particular, $\varphi(1_R)$ is not a zero divisor. By the contrapositive of part (a), $\varphi(1_R) = 1_S$
(b) Assume that $u$ is a unit in $R$ and $\varphi(1_R)=1_S$. With $u$ being a unit of $R$, this means that
$\exists v \in{R}$ s.t. $uv=vu=1_R$. Since $\varphi$ is a ring homomorphism, $$\varphi(uv)=\varphi(vu)=\varphi(1_R)=1_S$$ $$\Longrightarrow \varphi(u)\varphi(v)=\varphi(v)\varphi(u)=1_S$$ Therefore, $ \varphi(u)$ is a unit in $S$.
We now want to show that $\varphi(u^{-1})=\varphi(u)^{-1}$. Note that $$1_S=\varphi(1_R)=\varphi(uu^{-1})=\varphi(u)\varphi(u^{-1})$$ $$\varphi(u)\varphi(u^{-1})=1_S$$ Since we have established that $\varphi(u)$ is a unit, it has an inverse. So, $$\varphi(u)^{-1}\varphi(u)\varphi(u^{-1})=\varphi(u)^{-1} \cdot 1_S$$ $$\therefore \quad \varphi(u^{-1})=\varphi(u)^{-1} \quad \blacksquare$$
(a) Prove that if $\varphi(1_R) \neq 1_S$ then $\varphi(1_R)$ is a zero divisor in $S$. Deduce that if $S$ is an integral domain, then every ring homomorphism from $R$ to $S$ sends the identity of $R$ to the identity of $S$.
(b) Prove that if $u$ is a unit in $R$ and $\varphi(1_R)=1_S$ then $\varphi(u)$ is a unit in $S$ and $\varphi(u^{-1})=\varphi(u)^{-1}$.
$\textit{Proof}$ :
(a): Let $\varphi(1_R)=y\in{S}-\{0\}$. (The identity in $R$ must be sent to a nonzero element of $S$ for $\varphi$ to be nonzero) $$y=\varphi(1_R)=\varphi(1_R1_R)=\varphi(1_R)\varphi(1_R)=y^2$$ $y=y^2$
$y-y^2=0$ (additive inverses exist because rings with respect to the addition operation are abelian groups)
$y(1_S-y)=0$ (distributive law for rings)
$ \therefore $ $y$ is a zero divisor.
If $S$ is an integral domain, then $S$ contains no zero divisors. So, in particular, $\varphi(1_R)$ is not a zero divisor. By the contrapositive of part (a), $\varphi(1_R) = 1_S$
(b) Assume that $u$ is a unit in $R$ and $\varphi(1_R)=1_S$. With $u$ being a unit of $R$, this means that
$\exists v \in{R}$ s.t. $uv=vu=1_R$. Since $\varphi$ is a ring homomorphism, $$\varphi(uv)=\varphi(vu)=\varphi(1_R)=1_S$$ $$\Longrightarrow \varphi(u)\varphi(v)=\varphi(v)\varphi(u)=1_S$$ Therefore, $ \varphi(u)$ is a unit in $S$.
We now want to show that $\varphi(u^{-1})=\varphi(u)^{-1}$. Note that $$1_S=\varphi(1_R)=\varphi(uu^{-1})=\varphi(u)\varphi(u^{-1})$$ $$\varphi(u)\varphi(u^{-1})=1_S$$ Since we have established that $\varphi(u)$ is a unit, it has an inverse. So, $$\varphi(u)^{-1}\varphi(u)\varphi(u^{-1})=\varphi(u)^{-1} \cdot 1_S$$ $$\therefore \quad \varphi(u^{-1})=\varphi(u)^{-1} \quad \blacksquare$$
Wednesday, July 10, 2013
D+F 7.3.16
Given $\varphi: R \rightarrow S$ be a surjective ring homomorphism. Prove that the image of the center of $R$ is
contained in the center of $S$.
$\textit{Proof}$ : Denote the center of $R$ as $Z(R)$ and the center of $S$ as $Z(S)$. Let $x \in{Z(R)}$. Then, $xr=rx \quad \forall r\in{Z(R)}$.
We want to show that $\varphi(x)s=s\varphi(x) \quad \forall s\in{S}$. With $s\in{S}$, then since $\varphi$ is surjective, $\exists a\in{R}$ s.t. $\varphi(a)=s$. Then, $$s\varphi(x)=\varphi(a)\varphi(x)=\varphi(ax)=\varphi(xa)$$ since $x\in{Z(R)}$. $$\varphi(xa)=\varphi(x)\varphi(a)=\varphi(x)s \Longrightarrow \varphi(x)\in{Z(S)}$$ We have shown that the image of an arbitrary element in $Z(R)$ commutes with any element in $S$.
$\therefore \varphi(Z(R)) \subseteq Z(S) \quad \blacksquare$
$\textit{Proof}$ : Denote the center of $R$ as $Z(R)$ and the center of $S$ as $Z(S)$. Let $x \in{Z(R)}$. Then, $xr=rx \quad \forall r\in{Z(R)}$.
We want to show that $\varphi(x)s=s\varphi(x) \quad \forall s\in{S}$. With $s\in{S}$, then since $\varphi$ is surjective, $\exists a\in{R}$ s.t. $\varphi(a)=s$. Then, $$s\varphi(x)=\varphi(a)\varphi(x)=\varphi(ax)=\varphi(xa)$$ since $x\in{Z(R)}$. $$\varphi(xa)=\varphi(x)\varphi(a)=\varphi(x)s \Longrightarrow \varphi(x)\in{Z(S)}$$ We have shown that the image of an arbitrary element in $Z(R)$ commutes with any element in $S$.
$\therefore \varphi(Z(R)) \subseteq Z(S) \quad \blacksquare$
Tuesday, July 9, 2013
commutative rings with prime ideals are fields
Let $R$ be a commutative ring. If every ideal of $R$ is prime, then $R$ is a field.
A commutative ring with identity is an integral domain if it has no zero divisors. We proceed by establishing that $R$ is an integral domain and then showing that inverses of elements exist, this $R$ is a field. Recall that an ideal $P$ is a prime ideal if $P \neq R$ and $ab \in{P} \Longrightarrow a \in{P}$ or $b \in{P}$.
$\textit{Proof}$ : Consider the ideal generated by the additive identity $(0)=0 \subseteq R$. $ab \in{(0)} \Longleftrightarrow ab=0 \Longrightarrow a=0$ or $b=0$. This is exactly the definition of not having zero divisors. So, $R$ is an integral domain.
Now let $x \neq 0$. $x^2 \in (x^2) \Longrightarrow x \cdot x \in{(x^2)}$. So, if $x \in{(x^2)} \Longrightarrow \exists a\in{R}$ s.t. $x=ax^2$. Since we have cancellation laws in an integral domain, then $xx^{-1}=ax^2x^{-1}$ so $1=ax$. So, $x$ has an inverse. $\therefore R$ is a field. $\blacksquare$
A commutative ring with identity is an integral domain if it has no zero divisors. We proceed by establishing that $R$ is an integral domain and then showing that inverses of elements exist, this $R$ is a field. Recall that an ideal $P$ is a prime ideal if $P \neq R$ and $ab \in{P} \Longrightarrow a \in{P}$ or $b \in{P}$.
$\textit{Proof}$ : Consider the ideal generated by the additive identity $(0)=0 \subseteq R$. $ab \in{(0)} \Longleftrightarrow ab=0 \Longrightarrow a=0$ or $b=0$. This is exactly the definition of not having zero divisors. So, $R$ is an integral domain.
Now let $x \neq 0$. $x^2 \in (x^2) \Longrightarrow x \cdot x \in{(x^2)}$. So, if $x \in{(x^2)} \Longrightarrow \exists a\in{R}$ s.t. $x=ax^2$. Since we have cancellation laws in an integral domain, then $xx^{-1}=ax^2x^{-1}$ so $1=ax$. So, $x$ has an inverse. $\therefore R$ is a field. $\blacksquare$
Tuesday, July 2, 2013
D+F 6.2.3
$|G|=380 \Longrightarrow $ $G$ is not simple.
$\textit{Proof}$ : First note that $380=2^2 \cdot 5 \cdot 19$. By Sylow's theorem (iii), we get that for $p=19$, $n_{19}=1$(mod 19) and also $n_{19} \mid 2^2 \cdot 5 = 20$. So, the possible values of $n_{19}$ are $1,20$.
For $p=5$, we get that $n_5=1$(mod 5) and also $n_5 \mid 2^2 \cdot 19 = 76$. Given these two constraints, the possible values for $n_5$ are $n_5=1,76$.
For $G$ to be simple, we cannot have any non-trivial normal subgroups. This means that by corollary 20 p.142, we must have $n_p=1$ for all $p$ in the factorization of $|G|$.
Suppose that it is the case that $n_{19}=20$ and $n_5=76$. The Sylow-19 subgroups have order 19, (the 19 factor in 380 is square-free) and the only group of order 19 is $\mathbb{Z}_{19}$, which has 18 elements of order 19. With $n_{19}=20$, this would mean that our entire group $G$ has $20 \cdot 18=360$ elements of order 19. The Sylow 5-subgroups have order 5, and the only group of order 5 is $\mathbb{Z}_5$, which has 4 elements of order 5. With $n_5=76$, this means our entire group $G$ would have $76 \cdot 4 = 304$ elements of order 5. This is an immediate contradiction on the size of $G$. Thus, it must be the case that either $n_5=1$ or $n_{19}=1$. In each case, we would have a non-trivial normal subgroup making $G$ not simple. $\blacksquare$
$\textit{Proof}$ : First note that $380=2^2 \cdot 5 \cdot 19$. By Sylow's theorem (iii), we get that for $p=19$, $n_{19}=1$(mod 19) and also $n_{19} \mid 2^2 \cdot 5 = 20$. So, the possible values of $n_{19}$ are $1,20$.
For $p=5$, we get that $n_5=1$(mod 5) and also $n_5 \mid 2^2 \cdot 19 = 76$. Given these two constraints, the possible values for $n_5$ are $n_5=1,76$.
For $G$ to be simple, we cannot have any non-trivial normal subgroups. This means that by corollary 20 p.142, we must have $n_p=1$ for all $p$ in the factorization of $|G|$.
Suppose that it is the case that $n_{19}=20$ and $n_5=76$. The Sylow-19 subgroups have order 19, (the 19 factor in 380 is square-free) and the only group of order 19 is $\mathbb{Z}_{19}$, which has 18 elements of order 19. With $n_{19}=20$, this would mean that our entire group $G$ has $20 \cdot 18=360$ elements of order 19. The Sylow 5-subgroups have order 5, and the only group of order 5 is $\mathbb{Z}_5$, which has 4 elements of order 5. With $n_5=76$, this means our entire group $G$ would have $76 \cdot 4 = 304$ elements of order 5. This is an immediate contradiction on the size of $G$. Thus, it must be the case that either $n_5=1$ or $n_{19}=1$. In each case, we would have a non-trivial normal subgroup making $G$ not simple. $\blacksquare$
D+F 4.5.24
Prove that if $G$ is a group of order 231 then $Z(G)$ contains a Sylow 11-subgroup of $G$. Also, show that there is a normal Sylow 7-subgroup in $G$.
$\textit{Proof}$ : Begin by showing there is a normal Sylow 7-subgroup in $G$: By Sylow's Theorem (iii), $$n_p \equiv 1(modp)$$ and $$n_p \mid m$$ when $|G| = p^am$.
With $p=7$, we can write $|G|=231=7 \cdot 33$. The possibilities for $n_7$ are 1,8,15,22, and 29. Only the number 1 divides 33, thus $$n_7=1$$ Since Sylow p-subgroups are conjugate to each other, the Sylow subgroup of order 7 (231 has only 1 factor of 7) is normal in $G$ by corollary 20 p.142.
Now, to show that $Z(G)$ contains a Sylow-11 subgroup we initially proceed as above. The possibilities for $n_{11}$ are 1,12, and 23. But, only the number 1 also divides 21. Therefore, there is a unique Sylow 11 subgroup, $H$ in $G$ that is also normal. By Corollary 15 p.134, $$N_G(H)/C_G(H) \lesssim Aut(H) \quad (\star)$$ Since $H \lhd G$, $N_G(H)=G$. Also, $Aut(H)=Aut(\mathbb{Z}_{11})$ since $\mathbb{Z}_{11}$ is the only group of order 11. So $(\star)$ above becomes $$G/C_G(H) \lesssim Aut(\mathbb{Z}_{11}) \cong \mathbb{Z}_{11}^* \cong \mathbb{Z}_{10} $$ In terms of group orders, this means that $$|G/C_G(H)| \large{\mid} 10$$ So, $$\frac{231}{|C_G(H)|} \mid 10 \Longrightarrow \frac{231}{|C_G(H)|}=1,2,5,10 \Longrightarrow |C_G(H)|=231$$ So, $H$ commutes with all 231 of the elements of $G$. In other words, $$C_G(H) \leq Z(G) \quad \blacksquare$$
$\textit{Proof}$ : Begin by showing there is a normal Sylow 7-subgroup in $G$: By Sylow's Theorem (iii), $$n_p \equiv 1(modp)$$ and $$n_p \mid m$$ when $|G| = p^am$.
With $p=7$, we can write $|G|=231=7 \cdot 33$. The possibilities for $n_7$ are 1,8,15,22, and 29. Only the number 1 divides 33, thus $$n_7=1$$ Since Sylow p-subgroups are conjugate to each other, the Sylow subgroup of order 7 (231 has only 1 factor of 7) is normal in $G$ by corollary 20 p.142.
Now, to show that $Z(G)$ contains a Sylow-11 subgroup we initially proceed as above. The possibilities for $n_{11}$ are 1,12, and 23. But, only the number 1 also divides 21. Therefore, there is a unique Sylow 11 subgroup, $H$ in $G$ that is also normal. By Corollary 15 p.134, $$N_G(H)/C_G(H) \lesssim Aut(H) \quad (\star)$$ Since $H \lhd G$, $N_G(H)=G$. Also, $Aut(H)=Aut(\mathbb{Z}_{11})$ since $\mathbb{Z}_{11}$ is the only group of order 11. So $(\star)$ above becomes $$G/C_G(H) \lesssim Aut(\mathbb{Z}_{11}) \cong \mathbb{Z}_{11}^* \cong \mathbb{Z}_{10} $$ In terms of group orders, this means that $$|G/C_G(H)| \large{\mid} 10$$ So, $$\frac{231}{|C_G(H)|} \mid 10 \Longrightarrow \frac{231}{|C_G(H)|}=1,2,5,10 \Longrightarrow |C_G(H)|=231$$ So, $H$ commutes with all 231 of the elements of $G$. In other words, $$C_G(H) \leq Z(G) \quad \blacksquare$$
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