Show that $\mathbb{Q}[\sqrt{2}] \not\cong \mathbb{Q}[\sqrt{3}]$
$\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}|a,b\in{\mathbb{Q}}\}$ and $\mathbb{Q}[\sqrt{3}]= \{c+d\sqrt{3}|c,d\in{\mathbb{Q}}\}$.
$\textit{Proof}$: Assume to the contrary that $\mathbb{Q}[\sqrt{2}] \cong \mathbb{Q}[\sqrt{3}]$. So, there exists an isomorphism between these two fields. First, we justify but do not prove that there is no problem of identification with elements between these two fields. In other words the number say 2 in $\mathbb{Q}[\sqrt{2}]$ is the same as the number 2 in $\mathbb{Q}[\sqrt{3}]$. With an isomorphism between them,
$$\varphi(1)=1$$
$$\varphi(1+1)=\varphi(1)+\varphi(1)=1+1=2$$
And so on, so the integers must be fixed by this isomorphism. With the integers fixed, then the field of fraction of the integers, which is $\mathbb{Q}$ must be fixed. In other words, the ground field of these two field extensions must be fixed. This is consistent with the idea in Galois theory that automorphisms of field extensions fix the ground field.
Isomorphisms fix structures. In $\mathbb{Q}[\sqrt{2}]$ there is an element, namely $\sqrt{2}$ such that when squared, we get the number 2. Therefore, there must exist an element in $\mathbb{Q}[\sqrt{3}]$ such that when squared, we get 3.
$(c+d\sqrt{3})^2=2$
$c^2+2cd\sqrt{3}+3d^2=2$. $(\star)$
$c,d\in{\mathbb{Q}}$ and $cd\sqrt{3} \not \in{\mathbb{Q}}$, so
$cd\sqrt{3}=0 \Longrightarrow c=0$ or $d=0$ since we are in a field.
If $c=0$ then $(\star)$ becomes
$3d^2=2 \Longrightarrow d=\pm \sqrt{\frac{2}{3}} \not \in {\mathbb{Q}}$...a contradiction
Id $d=0$, then $(\star)$ becomes
$c^2=2 \Longrightarrow c=\pm \sqrt{2} \not \in{\mathbb{Q}}$...a contradiction.
In any case, an isomorphism between $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ leads to a contradiction.
$\therefore \quad \mathbb{Q}[\sqrt{2}] \not\cong \mathbb{Q}[\sqrt{3}]$
Sunday, August 18, 2013
D+F 7.4.4
Assume $R$ is a commutative ring with $1 \neq 0$. Prove that $R$ is a field iff $0$ is a maximal ideal.
Recall that an ideal $M$ of $R$ is maximal if $M \neq R$ and the only ideals containing $M$ are $R$ and $M$.
Recall that one way of defining a field is that it is a commutative ring with identity in which every nonzero element has an inverse (with respect to both $+$ and $\times$ of course)
This problem is another way of establishing that fields have trivial ideal structures. Note the analogy here with groups: Simple groups are special types of groups that have trivial normal subgroup structures. Fields are special types of rings that have trivial ideal structures.
$\textit{Proof}$ : "$\Longrightarrow$": Let $I$ be an ideal of $R$ s.t. $(0) \subset I \subset R$. So $\exists r \neq 0$ s.t. $r\in{I}$. Since $R$ is a field, multiplicative inverses exist, so $rr^{-1}\in{I} \Longrightarrow 1 \in{I}$. With the identity being in $I$, this means that $s \cdot 1 \in{I}$ $\forall s \in{R}$ since $I$ is, in particular, a left ideal. So, $s\in{I}$. But $s$ was arbitrary, so all elements lie in $I$, i.e. $I=R$.
$\therefore \quad (0)$ is a maximal ideal of $R$.
"$\Longleftarrow$": Assume that $(0)\in{R}$ is a maximal ideal. So, the only ideals containing $(0)$ are $(0)$ and $R$. So, $\forall r \neq 0$, the ideal generated by $0$ and $r$, $(0,r)=R$. In particular, if $r \neq 0$ then $1 \in{(0,r)}$. So, $1=a \cdot 0 + b\cdot r$ for some $a,b\in{R}$.
$br=1$
$b=r^{-1}$. Since $b$ and $r$ were arbitrary, inverses exist in general in $R$. So, $R$ is a commutative ring with identity $1 \neq 0$ in which every element has an inverse.
$\therefore \quad R$ is a field. $\blacksquare$
Recall that an ideal $M$ of $R$ is maximal if $M \neq R$ and the only ideals containing $M$ are $R$ and $M$.
Recall that one way of defining a field is that it is a commutative ring with identity in which every nonzero element has an inverse (with respect to both $+$ and $\times$ of course)
This problem is another way of establishing that fields have trivial ideal structures. Note the analogy here with groups: Simple groups are special types of groups that have trivial normal subgroup structures. Fields are special types of rings that have trivial ideal structures.
$\textit{Proof}$ : "$\Longrightarrow$": Let $I$ be an ideal of $R$ s.t. $(0) \subset I \subset R$. So $\exists r \neq 0$ s.t. $r\in{I}$. Since $R$ is a field, multiplicative inverses exist, so $rr^{-1}\in{I} \Longrightarrow 1 \in{I}$. With the identity being in $I$, this means that $s \cdot 1 \in{I}$ $\forall s \in{R}$ since $I$ is, in particular, a left ideal. So, $s\in{I}$. But $s$ was arbitrary, so all elements lie in $I$, i.e. $I=R$.
$\therefore \quad (0)$ is a maximal ideal of $R$.
"$\Longleftarrow$": Assume that $(0)\in{R}$ is a maximal ideal. So, the only ideals containing $(0)$ are $(0)$ and $R$. So, $\forall r \neq 0$, the ideal generated by $0$ and $r$, $(0,r)=R$. In particular, if $r \neq 0$ then $1 \in{(0,r)}$. So, $1=a \cdot 0 + b\cdot r$ for some $a,b\in{R}$.
$br=1$
$b=r^{-1}$. Since $b$ and $r$ were arbitrary, inverses exist in general in $R$. So, $R$ is a commutative ring with identity $1 \neq 0$ in which every element has an inverse.
$\therefore \quad R$ is a field. $\blacksquare$
Thursday, August 15, 2013
D+F prop. 14.1.2
[(Prop. 14.1.2)] Given the conditions that $K/F$ is a field extension, $\alpha \in{K}$ is algebraic over $F$ and $\sigma \in{Aut(K/F)}$ then prove that any polynomial with coefficients in $F$ having $\alpha$ as a root also has $\sigma(\alpha)=\sigma\alpha$ as a root.
Some notes: $\alpha\in{K}$ albegraic over $F$ means that $\alpha$ is the root of some nonzero polynomial in $F$.
$\sigma \in{Aut(K/F)}$ means that $\sigma$ is an operation preserving permutation on the elements of $K$ and $\sigma$ fixes $F$ i.e. $\sigma(a)=a \quad \forall a\in{F}$.
$\textit{Proof}$ :
Let $\alpha$ be a root $f(x)\in{F}[x]$. Dividing a polynomial through by a nonzero constant to make it monic does not change the roots. So, we may write $$\alpha^n+a_{n-1}\alpha^{n-1}+ \cdots + a_1\alpha + a_0=0 \quad a_i\in{F}$$ Now applying $\sigma$ to both sides, we get $$\sigma(\alpha^n+a_{n-1}\alpha^{n-1}+ \cdots + a_1\alpha + a_0=\sigma(0)=0 \quad (\star)$$ $$=\sigma(\alpha^n) + \sigma(a_{n-1}\alpha^{n-1} + \cdots + \sigma(a_0)=0$$ since $\sigma$ is a homomorphism with respect to addition. $$=(\sigma(\alpha))^n+ \sigma(a_{n-1})\sigma(\alpha)^{n-1} + \cdots + \sigma(a_0)$$ since $\sigma$ is also a homomorphism with respect to multiplication. Now, since $\sigma$ fixes $F$, $$=(\sigma\alpha)^n + a_{n-1}(\sigma\alpha)^{n-1}+ \cdots + a_0$$ Which states exactly that $\sigma\alpha$ is also a root of $f(x)$.
$(\star): \sigma(0)=0$ since $\sigma$ is an automorphism: $\sigma(0)=\sigma(0+0)=\sigma(0)+\sigma(0) \Longrightarrow \sigma(0)=0$. $\blacksquare$
Some notes: $\alpha\in{K}$ albegraic over $F$ means that $\alpha$ is the root of some nonzero polynomial in $F$.
$\sigma \in{Aut(K/F)}$ means that $\sigma$ is an operation preserving permutation on the elements of $K$ and $\sigma$ fixes $F$ i.e. $\sigma(a)=a \quad \forall a\in{F}$.
$\textit{Proof}$ :
Let $\alpha$ be a root $f(x)\in{F}[x]$. Dividing a polynomial through by a nonzero constant to make it monic does not change the roots. So, we may write $$\alpha^n+a_{n-1}\alpha^{n-1}+ \cdots + a_1\alpha + a_0=0 \quad a_i\in{F}$$ Now applying $\sigma$ to both sides, we get $$\sigma(\alpha^n+a_{n-1}\alpha^{n-1}+ \cdots + a_1\alpha + a_0=\sigma(0)=0 \quad (\star)$$ $$=\sigma(\alpha^n) + \sigma(a_{n-1}\alpha^{n-1} + \cdots + \sigma(a_0)=0$$ since $\sigma$ is a homomorphism with respect to addition. $$=(\sigma(\alpha))^n+ \sigma(a_{n-1})\sigma(\alpha)^{n-1} + \cdots + \sigma(a_0)$$ since $\sigma$ is also a homomorphism with respect to multiplication. Now, since $\sigma$ fixes $F$, $$=(\sigma\alpha)^n + a_{n-1}(\sigma\alpha)^{n-1}+ \cdots + a_0$$ Which states exactly that $\sigma\alpha$ is also a root of $f(x)$.
$(\star): \sigma(0)=0$ since $\sigma$ is an automorphism: $\sigma(0)=\sigma(0+0)=\sigma(0)+\sigma(0) \Longrightarrow \sigma(0)=0$. $\blacksquare$
D+F 14.6.4
Determine the Galois group of $x^4-25$
Recall that the Galois group of a separable polynomial $f(x) \in{F[x]}$ is defined to be the Galois group of the splitting field of $f(x)$ over $F$.
$x^4-25$ partially factors over $\mathbb{Q}$ as $$x^4-25=(x^2-5)(x^2+5)$$ and the four roots are $\pm\sqrt{5} \pm\sqrt{5}i$. So, this polynomial has no repeated roots, and is thus separable. We first determine the splitting field and then use the fundamental theorem of Galois to flip the diagram of known subfields to get the Galois group lattice. We need to adjoin $\sqrt{5}$ to $\mathbb{Q}$, but notice that the two complex roots are not contained in $\mathbb{Q}[\sqrt{5}]$. So, we need to adjoin both $\sqrt{5}$ and $i$ for our splitting field to contain all 4 roots. Thus, the splitting field of $x^4-25$ should be $\mathbb{Q}[\sqrt{5},i]$. The lattice of known subfields is $\mathbb{Q}$ as the base field, then $\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[i]$ and $\mathbb{Q}[\sqrt{5}i]$ as degree 2 extensions over $\mathbb{Q}$. Then, $\mathbb{Q}[\sqrt{5},i]$ on top as a degree 2 extension of the 3 intermediate subfields. By the fundamental theorem of Galois, the flipped image of this subfield lattice is exactly the lattice of the Galois group of $x^4-25$. This lattice is the lattice for the Klein 4-group.
Recall that the Galois group of a separable polynomial $f(x) \in{F[x]}$ is defined to be the Galois group of the splitting field of $f(x)$ over $F$.
$x^4-25$ partially factors over $\mathbb{Q}$ as $$x^4-25=(x^2-5)(x^2+5)$$ and the four roots are $\pm\sqrt{5} \pm\sqrt{5}i$. So, this polynomial has no repeated roots, and is thus separable. We first determine the splitting field and then use the fundamental theorem of Galois to flip the diagram of known subfields to get the Galois group lattice. We need to adjoin $\sqrt{5}$ to $\mathbb{Q}$, but notice that the two complex roots are not contained in $\mathbb{Q}[\sqrt{5}]$. So, we need to adjoin both $\sqrt{5}$ and $i$ for our splitting field to contain all 4 roots. Thus, the splitting field of $x^4-25$ should be $\mathbb{Q}[\sqrt{5},i]$. The lattice of known subfields is $\mathbb{Q}$ as the base field, then $\mathbb{Q}[\sqrt{5}]$, $\mathbb{Q}[i]$ and $\mathbb{Q}[\sqrt{5}i]$ as degree 2 extensions over $\mathbb{Q}$. Then, $\mathbb{Q}[\sqrt{5},i]$ on top as a degree 2 extension of the 3 intermediate subfields. By the fundamental theorem of Galois, the flipped image of this subfield lattice is exactly the lattice of the Galois group of $x^4-25$. This lattice is the lattice for the Klein 4-group.
Tuesday, August 13, 2013
D+F 13.5.8
Prove that $f(x)^p=f(x^p) \quad \forall f(x)\in{\mathbb{F}_p[x]}$.
$\textit{Proof}$ : First note that since we are in a field of charachteristic $p$, then we may use the properties of the Frobenius automorphism: $$(a+b)^p \equiv a^p + b^p \quad and \quad (ab)^p \equiv a^pb^p$$ where we have used "$\equiv$" to remind ourselves that we are working in charachteristic $p$.
Similarly for more terms, we have $$(a+b+c)^p \equiv ((a+b)+c)^p \equiv (a+b)^p+c^p \equiv a^p + b^p + c^p$$ $$(abc)^p \equiv ((ab)c)^p \equiv (ab)^pc^p \equiv a^pb^pc^p$$ Let $$f(x)=a_nx^n+ \cdots +a_0 \quad a_i\in{\mathbb{F}_p}$$ Then by induction, $$f(x)^p \equiv (a_nx^n+ \cdots + a_0)^p \equiv (a_nx^n)^p + \cdots + (a_0)^p \equiv (a_n)^p(x^n)^p + \cdots + a_0^p$$ $$ \equiv a_n(x^n)^p+ \cdots + a_0 \equiv f(x^p)$$ Since $a^p \equiv a (modp)$.
$\therefore \quad f(x)^p=f(x^p) \quad \forall f(x)\in{\mathbb{F}_p[x]} \quad \blacksquare$
$\textit{Proof}$ : First note that since we are in a field of charachteristic $p$, then we may use the properties of the Frobenius automorphism: $$(a+b)^p \equiv a^p + b^p \quad and \quad (ab)^p \equiv a^pb^p$$ where we have used "$\equiv$" to remind ourselves that we are working in charachteristic $p$.
Similarly for more terms, we have $$(a+b+c)^p \equiv ((a+b)+c)^p \equiv (a+b)^p+c^p \equiv a^p + b^p + c^p$$ $$(abc)^p \equiv ((ab)c)^p \equiv (ab)^pc^p \equiv a^pb^pc^p$$ Let $$f(x)=a_nx^n+ \cdots +a_0 \quad a_i\in{\mathbb{F}_p}$$ Then by induction, $$f(x)^p \equiv (a_nx^n+ \cdots + a_0)^p \equiv (a_nx^n)^p + \cdots + (a_0)^p \equiv (a_n)^p(x^n)^p + \cdots + a_0^p$$ $$ \equiv a_n(x^n)^p+ \cdots + a_0 \equiv f(x^p)$$ Since $a^p \equiv a (modp)$.
$\therefore \quad f(x)^p=f(x^p) \quad \forall f(x)\in{\mathbb{F}_p[x]} \quad \blacksquare$
D+F 13.4.1
Determine the splitting field and its degree over $\mathbb{Q}$ for $x^4-2$.
$\textit{Proof}$ : The four roots of $x^4-2$ can found by factoring: $$x^4-2=(x^2)^2-(\sqrt{2})^2=(x^2-\sqrt{2})(x^2+\sqrt{2})$$ $$=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-\sqrt[4]{2}i)(x+\sqrt[4]{2}i)$$ These four roots can be thought of as the 4th roots of unity scaled by $\sqrt[4]{2}$.
In this factorization, we need to adjoin $\sqrt[4]{2}$ and $i$. Now, $$\mathbb{Q}[\sqrt[4]{2}]=\{a_0+a_1\sqrt[4]{2}+a_2\sqrt{2}+a_3\sqrt[4]{2}^3|a_i\in{\mathbb{Q}}\}$$ so, $$|\mathbb{Q}[\sqrt[4]{2}]:\mathbb{Q}|=4$$ But the 2 complex roots $\sqrt[4]{2}i$ and $-\sqrt[4]{2}i$ are not in $\mathbb{Q}[\sqrt[4]{2}]$, so we need to also adjoin $i$. Now, $$|\mathbb{Q}[\sqrt[4]{2}][i]:\mathbb{Q}[\sqrt[4]{2}]|=2$$ So, $$|\mathbb{Q}[\sqrt[4]{2}][i]:\mathbb{Q}|=2 \cdot 4 = 8$$ Thus, the splitting field of $x^4-2$ is a degree 8 extension.
$\textit{Proof}$ : The four roots of $x^4-2$ can found by factoring: $$x^4-2=(x^2)^2-(\sqrt{2})^2=(x^2-\sqrt{2})(x^2+\sqrt{2})$$ $$=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-\sqrt[4]{2}i)(x+\sqrt[4]{2}i)$$ These four roots can be thought of as the 4th roots of unity scaled by $\sqrt[4]{2}$.
In this factorization, we need to adjoin $\sqrt[4]{2}$ and $i$. Now, $$\mathbb{Q}[\sqrt[4]{2}]=\{a_0+a_1\sqrt[4]{2}+a_2\sqrt{2}+a_3\sqrt[4]{2}^3|a_i\in{\mathbb{Q}}\}$$ so, $$|\mathbb{Q}[\sqrt[4]{2}]:\mathbb{Q}|=4$$ But the 2 complex roots $\sqrt[4]{2}i$ and $-\sqrt[4]{2}i$ are not in $\mathbb{Q}[\sqrt[4]{2}]$, so we need to also adjoin $i$. Now, $$|\mathbb{Q}[\sqrt[4]{2}][i]:\mathbb{Q}[\sqrt[4]{2}]|=2$$ So, $$|\mathbb{Q}[\sqrt[4]{2}][i]:\mathbb{Q}|=2 \cdot 4 = 8$$ Thus, the splitting field of $x^4-2$ is a degree 8 extension.
Monday, August 12, 2013
D+F 13.4.4
Determine the splitting field and its degree over $\mathbb{Q}$ for $x^6-4$.
Splitting fields are the smallest field extensions which allow $f(x)$ to factor/split completely. $$x^6-4=(x^3)^2-(2)^2=(x^3+2)(x^3-2)$$ $$=(x+\sqrt[3]{2})(x-\sqrt[3]{2}\zeta_6)(x-\sqrt[3]{2}\zeta_6^5)(x-\sqrt[3]{2})(x-\sqrt[3]{2}\zeta_6^2)(x-\sqrt[3]{2}\zeta_6^4)$$ Where $\zeta_6=\frac{1+\sqrt{-3}}{2}$. These are the 6th roots of unity scaled by $\sqrt[3]{2}$. So, in order for all 6 of these roots to be contained in an extension of $\mathbb{Q}$, we need to adjoin $\sqrt{-3}$ and $\sqrt[3]{2}$. So, the splitting field of $x^6-4$ is $\mathbb{Q}[\sqrt[3]{2},\sqrt{-3}]$. Now, $|\mathbb{Q}[\sqrt{-3}]:\mathbb{Q}|=2$ and $|\mathbb{Q}[\sqrt[3]{2}]:\mathbb{Q}|=3$. By corollary 22 p. 529, $$|\mathbb{Q}[\sqrt[3]{2},\sqrt{-3}]:\mathbb{Q}|=2 \cdot 3 = 6$$
Splitting fields are the smallest field extensions which allow $f(x)$ to factor/split completely. $$x^6-4=(x^3)^2-(2)^2=(x^3+2)(x^3-2)$$ $$=(x+\sqrt[3]{2})(x-\sqrt[3]{2}\zeta_6)(x-\sqrt[3]{2}\zeta_6^5)(x-\sqrt[3]{2})(x-\sqrt[3]{2}\zeta_6^2)(x-\sqrt[3]{2}\zeta_6^4)$$ Where $\zeta_6=\frac{1+\sqrt{-3}}{2}$. These are the 6th roots of unity scaled by $\sqrt[3]{2}$. So, in order for all 6 of these roots to be contained in an extension of $\mathbb{Q}$, we need to adjoin $\sqrt{-3}$ and $\sqrt[3]{2}$. So, the splitting field of $x^6-4$ is $\mathbb{Q}[\sqrt[3]{2},\sqrt{-3}]$. Now, $|\mathbb{Q}[\sqrt{-3}]:\mathbb{Q}|=2$ and $|\mathbb{Q}[\sqrt[3]{2}]:\mathbb{Q}|=3$. By corollary 22 p. 529, $$|\mathbb{Q}[\sqrt[3]{2},\sqrt{-3}]:\mathbb{Q}|=2 \cdot 3 = 6$$
D+F 7.4.9
Let $R$ be the ring of continuous functions on $[0,1]$ and let $I$ be the collection of functions $f(x)\in{R}$ s.t.
$$f(\frac{1}{3})=f(\frac{1}{2})=0$$
Prove that $I$ is an ideal of $R$ but not a prime ideal.
$\textit{Proof}$ : To show that $I$ is an ideal, we need to show that $I \neq \emptyset$, $I$ is closed under subtraction and $I$ is closed under left and right multiplication by arbitrary elements of $R$.
$0$, the zero function is in $I$, so $I \neq \emptyset$
Let $f(x),g(x) \in{I}$. Then $f(\frac{1}{3})=f(\frac{1}{2})=g(\frac{1}{3})=g(\frac{1}{2})=0$. $$(f-g)(\frac{1}{3})=f(\frac{1}{3})-g(\frac{1}{3})=0-0=0$$ and $$(f-g)(\frac{1}{2})=f(\frac{1}{2})-g(\frac{1}{2})=0-0=0$$ Or also $$f(\frac{1}{2})-g(\frac{1}{3})=0-0=0$$ $$f(\frac{1}{3})-g(\frac{1}{2})=0-0=0$$ So, $I$ is closed under subtraction.
Now let $f(x)\in{I}$ and $h(x)\in{R}$. $$(fh)(\frac{1}{2})=f(\frac{1}{2}) \cdot h(\frac{1}{2})=0 \cdot a = 0$$ Similarly, $(hf)(\frac{1}{2})=0=(fh)(\frac{1}{3})=(hf)(\frac{1}{3})=0$.
So, $I$ is an ideal of $R$.
Since the ring of continuous functions is a commutative ring, we may have the notion of a prime ideal. An ideal $P$ is prime in the commutative ring $R$ if $P \neq R$ and $$fg \in{P} \Longrightarrow f\in{P} \quad or \quad g\in{P}$$ We would like to find $f,g \in{I}$ s.t. $fg\in{I}$ but $f \not \in{I}$ and $g \not \in {I}$.
Consider the linear functions $f(x)=x-\frac{1}{3}$ and $g(x)=x-\frac{1}{2}$.
$$f(\frac{1}{2}) \neq 0 \Longrightarrow f\not \in {I}$$ $$g(\frac{1}{3}) \neq 0 \Longrightarrow g \not \in{I}$$ However, $$(fg)(x)=(x-\frac{1}{3})(x-\frac{1}{2})$$ and so $$(fg)(\frac{1}{3})=(fg)(\frac{1}{2})=0 \Longrightarrow fg\in{I}$$ which amounts to a counterexample to $I$ being prime. So, $I$ is an ideal of $R$ but not a prime ideal. $\blacksquare$
$\textit{Proof}$ : To show that $I$ is an ideal, we need to show that $I \neq \emptyset$, $I$ is closed under subtraction and $I$ is closed under left and right multiplication by arbitrary elements of $R$.
$0$, the zero function is in $I$, so $I \neq \emptyset$
Let $f(x),g(x) \in{I}$. Then $f(\frac{1}{3})=f(\frac{1}{2})=g(\frac{1}{3})=g(\frac{1}{2})=0$. $$(f-g)(\frac{1}{3})=f(\frac{1}{3})-g(\frac{1}{3})=0-0=0$$ and $$(f-g)(\frac{1}{2})=f(\frac{1}{2})-g(\frac{1}{2})=0-0=0$$ Or also $$f(\frac{1}{2})-g(\frac{1}{3})=0-0=0$$ $$f(\frac{1}{3})-g(\frac{1}{2})=0-0=0$$ So, $I$ is closed under subtraction.
Now let $f(x)\in{I}$ and $h(x)\in{R}$. $$(fh)(\frac{1}{2})=f(\frac{1}{2}) \cdot h(\frac{1}{2})=0 \cdot a = 0$$ Similarly, $(hf)(\frac{1}{2})=0=(fh)(\frac{1}{3})=(hf)(\frac{1}{3})=0$.
So, $I$ is an ideal of $R$.
Since the ring of continuous functions is a commutative ring, we may have the notion of a prime ideal. An ideal $P$ is prime in the commutative ring $R$ if $P \neq R$ and $$fg \in{P} \Longrightarrow f\in{P} \quad or \quad g\in{P}$$ We would like to find $f,g \in{I}$ s.t. $fg\in{I}$ but $f \not \in{I}$ and $g \not \in {I}$.
Consider the linear functions $f(x)=x-\frac{1}{3}$ and $g(x)=x-\frac{1}{2}$.
$$f(\frac{1}{2}) \neq 0 \Longrightarrow f\not \in {I}$$ $$g(\frac{1}{3}) \neq 0 \Longrightarrow g \not \in{I}$$ However, $$(fg)(x)=(x-\frac{1}{3})(x-\frac{1}{2})$$ and so $$(fg)(\frac{1}{3})=(fg)(\frac{1}{2})=0 \Longrightarrow fg\in{I}$$ which amounts to a counterexample to $I$ being prime. So, $I$ is an ideal of $R$ but not a prime ideal. $\blacksquare$
Friday, August 9, 2013
Prime implies Irreducible in integral domains
Proposition 8.3.10:
In an integral domain, a prime element is always irreducible.
First recall some definitions:
Definition: An integral domain is a commutative ring with no zero divisors.
The lack of zero divisors gives integral domains a useful cancellation property:
$$ab=ac \Longrightarrow b=c \quad or \quad a=0$$ A nonzero elemnent $p$ is prime in the ring $R$ if the ideal generated by $p$, $(p)$ is a prime ideal.
An ideal $P$ is prime if $P \neq R$ and $ab\in{P} \Longrightarrow a\in{P}$ or $b\in{P}$.
An element $r\in{R}-\{0\}-R^\times$ is irreducible if $r=xy \Longrightarrow x\in{R^\times}$ or $y\in{R^\times}$ where $R^\times$ is the group of units of $R$.
$\textit{Proof}$ : Let $R$ be an integral domain and let $p\in{R}$ be a prime element. So, $(p)$ is a nonzero prime ideal generated by $p$ and $p=ab$. We want to show that $a\in{R^\times}$ or $b\in{R^\times}$.
So, $ab=p\in{(p)}$. Since $(p)$ is a prime ideal this means that $(p) \neq R$ and $ab\in{(p)} \Longrightarrow a\in{(p)}$ or $b\in{(p)}$. Say $a\in{(p)}$. Then, $\exists r\in{R}$ s.t. $a=pr$. So, we have:
$$p=ab=pr \cdot b$$ $$p=p \cdot rb$$ $$p \cdot 1 = p \cdot rb$$ Since $R$ is an integral domain, this imples
$$p=0 \quad or \quad 1 = rb$$ But $p \neq 0$ since prime elements are nonzero. Thus, $1=rb \Longrightarrow b\in{R^\times}$.
Swapping the roles of $a$ and $b$, if $b\in{(p)}$ then $a\in{R^\times}$. In either case, $p$ is irreducible. $\blacksquare$
In an integral domain, a prime element is always irreducible.
First recall some definitions:
Definition: An integral domain is a commutative ring with no zero divisors.
The lack of zero divisors gives integral domains a useful cancellation property:
$$ab=ac \Longrightarrow b=c \quad or \quad a=0$$ A nonzero elemnent $p$ is prime in the ring $R$ if the ideal generated by $p$, $(p)$ is a prime ideal.
An ideal $P$ is prime if $P \neq R$ and $ab\in{P} \Longrightarrow a\in{P}$ or $b\in{P}$.
An element $r\in{R}-\{0\}-R^\times$ is irreducible if $r=xy \Longrightarrow x\in{R^\times}$ or $y\in{R^\times}$ where $R^\times$ is the group of units of $R$.
$\textit{Proof}$ : Let $R$ be an integral domain and let $p\in{R}$ be a prime element. So, $(p)$ is a nonzero prime ideal generated by $p$ and $p=ab$. We want to show that $a\in{R^\times}$ or $b\in{R^\times}$.
So, $ab=p\in{(p)}$. Since $(p)$ is a prime ideal this means that $(p) \neq R$ and $ab\in{(p)} \Longrightarrow a\in{(p)}$ or $b\in{(p)}$. Say $a\in{(p)}$. Then, $\exists r\in{R}$ s.t. $a=pr$. So, we have:
$$p=ab=pr \cdot b$$ $$p=p \cdot rb$$ $$p \cdot 1 = p \cdot rb$$ Since $R$ is an integral domain, this imples
$$p=0 \quad or \quad 1 = rb$$ But $p \neq 0$ since prime elements are nonzero. Thus, $1=rb \Longrightarrow b\in{R^\times}$.
Swapping the roles of $a$ and $b$, if $b\in{(p)}$ then $a\in{R^\times}$. In either case, $p$ is irreducible. $\blacksquare$
Wednesday, August 7, 2013
D+F 7.4.19
Let $R$ be a finite commutative ring with identity. Prove that every prime ideal of $R$ is a maximal ideal.
$\textit{Proof}$ : Let $I$ be an ideal of $R$.
$I$ maximal $\Longleftrightarrow$ $R/I$ is a field (prop. 12)
$R/I$ a field $\Longrightarrow$ $R/I$ an integral domain. Conversely,
$R/I$ an integral domain $\Longrightarrow$ $R/I$ a field since $|R| < \infty$. (cor. 3 p. 228)
$\Longleftrightarrow$ $I$ is a prime ideal (prop. 13)
$\therefore$ in finite commutative rings with identity, maximal iff prime. $\blacksquare$
$\textit{Proof}$ : Let $I$ be an ideal of $R$.
$I$ maximal $\Longleftrightarrow$ $R/I$ is a field (prop. 12)
$R/I$ a field $\Longrightarrow$ $R/I$ an integral domain. Conversely,
$R/I$ an integral domain $\Longrightarrow$ $R/I$ a field since $|R| < \infty$. (cor. 3 p. 228)
$\Longleftrightarrow$ $I$ is a prime ideal (prop. 13)
$\therefore$ in finite commutative rings with identity, maximal iff prime. $\blacksquare$
D+F 7.4.13
Let $\varphi:R \rightarrow S$ be a homomorphism of commutative rings.
(a) Prove that if $P$ is a prime ideal of $S$ then either $\varphi^{-1}(P)=R$ or $\varphi^{-1}(P)$ is a prime ideal of $R$.
(b) Prove that if $M$ is a maximal ideal of $S$ and $\varphi$ is surjective then $\varphi^{-1}(M)$ is a maximal ideal of $R$. Give an example to show that this may not be true if $\varphi$ is not surjective.
$\textit{Proof}$ : Exercise 7.3.24 establishes that the preimage or pullback of an ideal is indeed an ideal. So $\varphi^{-1}(P)$ is an ideal of $R$.
Recall that an ideal of a commutative ring is called a prime ideal if $P \neq R$ and $ab\in{P} \Longrightarrow a\in{P}$ or $b\in{P}$.
Let $r_1r_2 \in{\varphi^{-1}(P)} \Longrightarrow \varphi(r_1r_2)\in{P}$. $\varphi(r_1r_2)=\varphi(r_1)\varphi(r_2)\in{P}$ since $\varphi$ is a ring homomorphism. Since $P$ is a prime ideal in $R$, this means $\varphi(r_1)\in{P}$ or $\varphi(r_2)\in{P}$ which implies
$r_1\in{\varphi^{-1}(P)}$ or $r_2\in{\varphi^{-1}(P)}$ So, $\varphi^{-1}(P)$ is a prime ideal in $R$. Now, if $\varphi^{-1}(P)$ happened to contain a unit (and there is no apparent reason why it cannot), then by proposition 9 . 253, $\varphi^{-1}(P)=R$. In any case, $\varphi^{-1}(P)$ is either a prime ideal or all of $R$.
For part (b), recall that an ideal $M$ of $S$ is a maximal ideal if $M \neq S$ and the only ideals containing $M$ are $M$ and $S$.
Let $M$ be a maximal ideal of $S$. Since $M$ is not all of $S$, then $\exists s \in{S}$ and $s\not \in{M}$. Since $\varphi$ is surjective, $\exists r\in{R}$ s.t. $\varphi(r)=s$ and $r\not \in \varphi^{-1}(M)$. Thus, $\varphi^{-1}(M) \neq R$.
Let $L \subset R$ be an ideal of $R$ containing $\varphi^{-1}(M)$. We want to show that $L=R$ or $L=\varphi^{-1}(M)$.
$L$ contains $\varphi^{-1}(M) \Longrightarrow \varphi(L)$ contains $M$. Then, $\varphi(L)=M$ or $S$ since $M$ is maximal in $S$. Therefore, $L=\varphi^{-1}(M)$ or $R$. $\blacksquare$.
If we relax the condition that $\varphi$ is surjective, then consider the following example:
$\phi: R \rightarrow \mathbb{F}_2$
$\phi: r \mapsto 0$
$(0)=0$ is maximal in $\mathbb{F}_2$. But, $\phi^{-1}(\mathbb{F}_2)=R$, the entire ring, so is not maximal.
(a) Prove that if $P$ is a prime ideal of $S$ then either $\varphi^{-1}(P)=R$ or $\varphi^{-1}(P)$ is a prime ideal of $R$.
(b) Prove that if $M$ is a maximal ideal of $S$ and $\varphi$ is surjective then $\varphi^{-1}(M)$ is a maximal ideal of $R$. Give an example to show that this may not be true if $\varphi$ is not surjective.
$\textit{Proof}$ : Exercise 7.3.24 establishes that the preimage or pullback of an ideal is indeed an ideal. So $\varphi^{-1}(P)$ is an ideal of $R$.
Recall that an ideal of a commutative ring is called a prime ideal if $P \neq R$ and $ab\in{P} \Longrightarrow a\in{P}$ or $b\in{P}$.
Let $r_1r_2 \in{\varphi^{-1}(P)} \Longrightarrow \varphi(r_1r_2)\in{P}$. $\varphi(r_1r_2)=\varphi(r_1)\varphi(r_2)\in{P}$ since $\varphi$ is a ring homomorphism. Since $P$ is a prime ideal in $R$, this means $\varphi(r_1)\in{P}$ or $\varphi(r_2)\in{P}$ which implies
$r_1\in{\varphi^{-1}(P)}$ or $r_2\in{\varphi^{-1}(P)}$ So, $\varphi^{-1}(P)$ is a prime ideal in $R$. Now, if $\varphi^{-1}(P)$ happened to contain a unit (and there is no apparent reason why it cannot), then by proposition 9 . 253, $\varphi^{-1}(P)=R$. In any case, $\varphi^{-1}(P)$ is either a prime ideal or all of $R$.
For part (b), recall that an ideal $M$ of $S$ is a maximal ideal if $M \neq S$ and the only ideals containing $M$ are $M$ and $S$.
Let $M$ be a maximal ideal of $S$. Since $M$ is not all of $S$, then $\exists s \in{S}$ and $s\not \in{M}$. Since $\varphi$ is surjective, $\exists r\in{R}$ s.t. $\varphi(r)=s$ and $r\not \in \varphi^{-1}(M)$. Thus, $\varphi^{-1}(M) \neq R$.
Let $L \subset R$ be an ideal of $R$ containing $\varphi^{-1}(M)$. We want to show that $L=R$ or $L=\varphi^{-1}(M)$.
$L$ contains $\varphi^{-1}(M) \Longrightarrow \varphi(L)$ contains $M$. Then, $\varphi(L)=M$ or $S$ since $M$ is maximal in $S$. Therefore, $L=\varphi^{-1}(M)$ or $R$. $\blacksquare$.
If we relax the condition that $\varphi$ is surjective, then consider the following example:
$\phi: R \rightarrow \mathbb{F}_2$
$\phi: r \mapsto 0$
$(0)=0$ is maximal in $\mathbb{F}_2$. But, $\phi^{-1}(\mathbb{F}_2)=R$, the entire ring, so is not maximal.
Monday, August 5, 2013
D+F 7.3.2
Prove that the rings $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.
$\textit{Proof}$ : In order for $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ to be isomorphic, they must have exactly the same algebraic structure. Now, the units of $\mathbb{Z}$ are $$\mathbb{Z}^*=\{1,-1\}$$ The units of $\mathbb{Q}$ are $$\mathbb{Q}^*=\mathbb{Q}-\{0\}$$ By proposition 4, p. 235, the above are the units of $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$, respectively. With completely different subsets of units, $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ cannot be isomorphic. $\blacksquare$
$\textit{Proof}$ : In order for $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ to be isomorphic, they must have exactly the same algebraic structure. Now, the units of $\mathbb{Z}$ are $$\mathbb{Z}^*=\{1,-1\}$$ The units of $\mathbb{Q}$ are $$\mathbb{Q}^*=\mathbb{Q}-\{0\}$$ By proposition 4, p. 235, the above are the units of $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$, respectively. With completely different subsets of units, $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ cannot be isomorphic. $\blacksquare$
7.3.1
Prove that $2\mathbb{Z} \not\cong 3\mathbb{Z}$.
$\textit{Proof}$ : Suppose to the contrary that $2\mathbb{Z} \cong 3\mathbb{Z}$. Then $\exists \varphi: 2\mathbb{Z} \rightarrow 3\mathbb{Z}$ s.t. $\varphi$ is a ring isomorphism. $$\varphi(0)=\varphi(0+0)=\varphi(0)+\varphi(0)$$ We can subtract $\varphi(0)$ from both sides since $(3 \mathbb{Z},+)$ is an Abelian group. So, $$\varphi(0)=0$$ Now, $$\varphi(0)= \varphi(1+1)=\varphi(1)+\varphi(1)=0$$ What is $\varphi(1)$? In $3 \mathbb{Z}$, $0+0=0$, $1+1=1$ and $2+2 \equiv 1$. From above, $\varphi(1)+\varphi(1)=0$ so the only possibility is $\varphi(1)=0$.
$\varphi(0)=0$ and $\varphi(1)=0 \Longrightarrow \varphi$ is not an injective map, which contradicts the assumption that $\varphi$ was an isomorphism.
$\therefore$ $2 \mathbb{Z} \not \cong 3 \mathbb{Z}$ $\blacksquare$
$\textit{Proof}$ : Suppose to the contrary that $2\mathbb{Z} \cong 3\mathbb{Z}$. Then $\exists \varphi: 2\mathbb{Z} \rightarrow 3\mathbb{Z}$ s.t. $\varphi$ is a ring isomorphism. $$\varphi(0)=\varphi(0+0)=\varphi(0)+\varphi(0)$$ We can subtract $\varphi(0)$ from both sides since $(3 \mathbb{Z},+)$ is an Abelian group. So, $$\varphi(0)=0$$ Now, $$\varphi(0)= \varphi(1+1)=\varphi(1)+\varphi(1)=0$$ What is $\varphi(1)$? In $3 \mathbb{Z}$, $0+0=0$, $1+1=1$ and $2+2 \equiv 1$. From above, $\varphi(1)+\varphi(1)=0$ so the only possibility is $\varphi(1)=0$.
$\varphi(0)=0$ and $\varphi(1)=0 \Longrightarrow \varphi$ is not an injective map, which contradicts the assumption that $\varphi$ was an isomorphism.
$\therefore$ $2 \mathbb{Z} \not \cong 3 \mathbb{Z}$ $\blacksquare$
Sunday, August 4, 2013
D+F 4.5.18
Prove that a group of order 200 has a normal Sylow 5-subgroup.
$\textit{Proof}$ : Note that $200=2^3 \cdot 5^2$. By Sylow (iii), we know that $$n_5 \equiv 1(mod5)$$ and $$n_5 | 2^3$$ So the possible values of $n_5$ are $n_5=1,6$. But $6 \nmid 8$. So, we are forced that $$n_5=1$$ By Sylow (ii), Sylow p-subgroups are conjugate to each other. So, this Sylow 5-subgroup, $P$, which has order $5^2=25$ is conjugate to itself i.e. $$gPg^{-1}=P \quad \forall g\in{G} \Longrightarrow P \unlhd G$$ This is corollary 20, p. 142. So, for groups of order 200, they must always contain a normal Sylow 5-subgroup. $\blacksquare$
$\textit{Proof}$ : Note that $200=2^3 \cdot 5^2$. By Sylow (iii), we know that $$n_5 \equiv 1(mod5)$$ and $$n_5 | 2^3$$ So the possible values of $n_5$ are $n_5=1,6$. But $6 \nmid 8$. So, we are forced that $$n_5=1$$ By Sylow (ii), Sylow p-subgroups are conjugate to each other. So, this Sylow 5-subgroup, $P$, which has order $5^2=25$ is conjugate to itself i.e. $$gPg^{-1}=P \quad \forall g\in{G} \Longrightarrow P \unlhd G$$ This is corollary 20, p. 142. So, for groups of order 200, they must always contain a normal Sylow 5-subgroup. $\blacksquare$
Thursday, August 1, 2013
D+F 4.5.36
Prove that if $N \unlhd G$ then $n_p(G) \geq n_p(G/N)$.
In this proof, we are comparing sizes of sets. We are comparing the number of Sylow $p$-subgroups of $G$ to the number of Sylow $p$-subgroups of the quotient group $G/N$. To do this, we create a map $\varphi: Syl_p(G) \longrightarrow Syl_p(G/N)$. To show that $n_p(G) \geq n_p(G/N)$, we proceed by showing that $\varphi$ is a surjective (onto) map. This implies that the domain of $\varphi$ has cardinality greater than or equal to the cardinality of its codomain $Syl_p(G/N)$.
$\textit{Proof}$ : Let $|G|=p^am$, $|N|=p^bn$. Then $|G/N|=p^{a-b}(\frac{m}{n})$.
By exercise 4.5.34, we are going to choose our target elements in $Syl_p(G/N)$ to be of the form $PN/N$. So, $$\varphi: Syl_p(G) \longrightarrow Syl_p(G/N)$$ $$P \mapsto PN/N$$ This will allow us to use the 2nd and 4th isomorphism theorems easily.
Pick $R/N\in{Syl_p(G/N)}$. We want to show $\exists P \in{Syl_p(G)}$ s.t. $\varphi(P)=R/N$.
If $R/N \in{Syl_p(G/N)}$ then $R/N \leq G/N$ and $|R/N|=p^{a-b}$.
By the 4th isomorphism theorem, $$R/N \leq G/N$$ where $N \leq R\leq G$
Now, $|R|=|R/N| \cdot |N|=p^{a-b} \cdot p^b\cdot n = p^a \cdot n$. So, applying Sylow's theorem to $|R|$, $\exists P \leq R$ s.t. $P\in{Syl_p(R)}$ where $|P|=p^a$.
$|P|=p^a \Longrightarrow P\in{Syl_p(G)}$.
Also, by the 2nd isomorphism theorem, $$PN/N \cong R/N$$ which can also be viewed as Sylow p-subgroups of $G/N$ being conjugate to each other, and thus isomorphic since conjugation is a group automorphism. Combining the conditions:
$N \leq R$
$P \leq R$
$P\in{Syl_p(R)}$
$PN/N \cong R/N$
we get that $$PN \leq R$$ By the 4th isomorphism theorem, $$PN/N \leq R/N$$ And since $|PN/N|=|R/N|$, then $$PN/N = R/N$$ Given arbitrary $R/N \in{Syl_p(G/N)}$ we have identified $P\in{Syl_p(G)}$ s.t. $\varphi(P)=R/N$. Thus, $\varphi$ is a surjective mapping.
$\therefore$ $n_p(G) \geq n_p(G/N)$ $\blacksquare$
In this proof, we are comparing sizes of sets. We are comparing the number of Sylow $p$-subgroups of $G$ to the number of Sylow $p$-subgroups of the quotient group $G/N$. To do this, we create a map $\varphi: Syl_p(G) \longrightarrow Syl_p(G/N)$. To show that $n_p(G) \geq n_p(G/N)$, we proceed by showing that $\varphi$ is a surjective (onto) map. This implies that the domain of $\varphi$ has cardinality greater than or equal to the cardinality of its codomain $Syl_p(G/N)$.
$\textit{Proof}$ : Let $|G|=p^am$, $|N|=p^bn$. Then $|G/N|=p^{a-b}(\frac{m}{n})$.
By exercise 4.5.34, we are going to choose our target elements in $Syl_p(G/N)$ to be of the form $PN/N$. So, $$\varphi: Syl_p(G) \longrightarrow Syl_p(G/N)$$ $$P \mapsto PN/N$$ This will allow us to use the 2nd and 4th isomorphism theorems easily.
Pick $R/N\in{Syl_p(G/N)}$. We want to show $\exists P \in{Syl_p(G)}$ s.t. $\varphi(P)=R/N$.
If $R/N \in{Syl_p(G/N)}$ then $R/N \leq G/N$ and $|R/N|=p^{a-b}$.
By the 4th isomorphism theorem, $$R/N \leq G/N$$ where $N \leq R\leq G$
Now, $|R|=|R/N| \cdot |N|=p^{a-b} \cdot p^b\cdot n = p^a \cdot n$. So, applying Sylow's theorem to $|R|$, $\exists P \leq R$ s.t. $P\in{Syl_p(R)}$ where $|P|=p^a$.
$|P|=p^a \Longrightarrow P\in{Syl_p(G)}$.
Also, by the 2nd isomorphism theorem, $$PN/N \cong R/N$$ which can also be viewed as Sylow p-subgroups of $G/N$ being conjugate to each other, and thus isomorphic since conjugation is a group automorphism. Combining the conditions:
$N \leq R$
$P \leq R$
$P\in{Syl_p(R)}$
$PN/N \cong R/N$
we get that $$PN \leq R$$ By the 4th isomorphism theorem, $$PN/N \leq R/N$$ And since $|PN/N|=|R/N|$, then $$PN/N = R/N$$ Given arbitrary $R/N \in{Syl_p(G/N)}$ we have identified $P\in{Syl_p(G)}$ s.t. $\varphi(P)=R/N$. Thus, $\varphi$ is a surjective mapping.
$\therefore$ $n_p(G) \geq n_p(G/N)$ $\blacksquare$
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