Proof: Assume $F$ is finite. Then $|F^\times|=|F^+|-1$, so they cannot be in bijection. Therefore $F$ is infinite: Assume $-1=1$; then $2=0$ implying $a+a=0$ for all $a∈F^+$ so that $u^2=1$ for all $u∈F^\times$, so that there are an infinite number of roots to the equation $x^2-1$, an impossibility. Therefore $1≠-1$ so that $2a=a+a≠0$ for all nonzero $a∈F^+$, so that for all nonidentity $u∈F^\times$ we have $u^2≠1$, a contradiction by $(-1)^2=1$.$~\square$
Tuesday, May 21, 2013
Multiplicative and Additive Independence (9.5.7)
Dummit and Foote Abstract Algebra, section 9.5, exercise 7:
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For $F$ a field, prove $F^\times \not ≅ F^+$.
Proof: Assume $F$ is finite. Then $|F^\times|=|F^+|-1$, so they cannot be in bijection. Therefore $F$ is infinite: Assume $-1=1$; then $2=0$ implying $a+a=0$ for all $a∈F^+$ so that $u^2=1$ for all $u∈F^\times$, so that there are an infinite number of roots to the equation $x^2-1$, an impossibility. Therefore $1≠-1$ so that $2a=a+a≠0$ for all nonzero $a∈F^+$, so that for all nonidentity $u∈F^\times$ we have $u^2≠1$, a contradiction by $(-1)^2=1$.$~\square$
Proof: Assume $F$ is finite. Then $|F^\times|=|F^+|-1$, so they cannot be in bijection. Therefore $F$ is infinite: Assume $-1=1$; then $2=0$ implying $a+a=0$ for all $a∈F^+$ so that $u^2=1$ for all $u∈F^\times$, so that there are an infinite number of roots to the equation $x^2-1$, an impossibility. Therefore $1≠-1$ so that $2a=a+a≠0$ for all nonzero $a∈F^+$, so that for all nonidentity $u∈F^\times$ we have $u^2≠1$, a contradiction by $(-1)^2=1$.$~\square$
Saturday, May 18, 2013
Polynomial Factorization Practice #2 (9.4.1-4)
Dummit and Foote Abstract Algebra, section 9.4, exercise- 1-4:
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1. Factor the following polynomials:
(a) $x^2+x+1∈\mathbb{F}_2[x]$
(b) $x^3+x+1∈\mathbb{F}_3[x]$
(c) $x^4+1∈\mathbb{F}_5[x]$
(d) $x^4+10x+1∈\mathbb{Z}[x]$
2. Prove the following polynomials are irreducible in $\mathbb{Z}[x]$:
(a) $x^4-4x^4+6$
(b) $x^6+30x^5-15x^3+6x-120$
(c) $x^4+4x^3+6x^2+2x+1$
(d) $\dfrac{(x+2)^p-2^p}{x}$ for $p$ an odd prime.
3. Show that $f_n(x)=(x-1)(x-2)...(x-n)-1$ is irreducible over $\mathbb{Z}$ for all $n \geq 1$.
4. Show that $f_n(x)=(x-1)(x-2)...(x-n)+1$ is irreducible over $\mathbb{Z}$ for all $n \geq 1$ and $n≠4$.
Proof: (1) (a) There are no zeros to this equation, so it is irreducible.
(b) By discovering $f(1)=0$, we obtain $x^3+x+1=(x-1)(x^2+x+2)$, where the latter has no zeros.
(c) $x^4+1=x^4-4=(x^2-2)(x^2+2)$.
(d) Reduce modulo 5. If $f(x)=a(x)b(x)$, then due to monicity of $f(x)$ we have $a(x)$ and $b(x)$ are monic and not collapsing to units so that $\overline{a(x)}$ is associate to $\overline{x^2+2}$ or $\overline{x^2-2}$, say the former. Since $\overline{a(x)}$ is monic we have $\overline{a(x)}=\overline{x^2+2}$ which forces the constant term of $a(x)$ to not be $\pm 1$, a contradiction.
Alternatively, since $f(x)$ clearly has no zeros in $\mathbb{Z}$, it would have to factor as the product of two quadratics such that the following hold:$$f(x)=a(x)b(x)$$$$a(x)=x^2+a_1x+a_0$$$$b(x)=x^2+b_1x+b_0$$$$a_0b_0=1$$$$a_1b_0+a_0b_1=0$$$$a_1b_1+a_0+b_0=10$$$$a_1+b_1=0$$Solving for $b_1$ as a typical system of equations reveals$$\dfrac{1}{4}b_1^4+15b_1^2+24=0$$which is impossible.
(2) (a) Eisenstein's criterion for $p=2$ applies.
(b) Eisenstein applies for $p=3$.
(c) We generalize the technique employed in example 2. Lemma 1: Let $R$ be an integral domain with $f(x)=a_0+a_1x+...+a_nx^n∈R[x]$ and $g(x)=f(r(x))$ for $r(x)$ a nonconstant polynomial. If $g(x)$ is irreducible, then $f(x)$ is irreducible. Proof: If $f(x)=a(x)b(x)$ for some nonunit polynomials, then $g(x)=a(r(x))b(r(x))$ where necessarily $\text{deg}(a(r(x)))=\text{deg}(a(x)) \cdot \text{deg}(r(x)) \geq 1$ if $\text{deg}(a(x)) \geq 1$ and $a(r(x))=a(x)$ if $a(x)$ is a constant and similarly for $b(r(x))$, a contradiction.$~\square$
Let $g(x)=f(x-1)=x^4-2x+2$, where Eisenstein applies for $p=2$.
(d) $\dfrac{(x+2)^p-2^p}{x}=x^p+...+2p$, where by the binomial theorem the omitted terms have coefficients of multiples of $p$. Eisenstein applies for $p$.
(3) When $n=1$ we clearly have $f_n(x)$ is irreducible, so assume $n > 1$. Assuming $f_n(x)$ has a zero in $\mathbb{Z}$, then $1=(x-1)(x-2)...(x-n)$ so that in particular $x-1=\pm 1$ and $x-2=\pm 1$, a contradiction. Therefore if $f_n(x)=a(x)b(x)$ then the degrees of $a(x)$ and $b(x)$ are greater than one.
We demonstrate a basic algebraic result, first. If $R$ is an integral domain and $f(x)∈R[x]$ of degree $n \geq 1$, then there are at most $n$ zeros of $f(x)$. If not, then $f(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_{n+1})f'(x)$ entails $\text{deg}(f(x)) > n$, a contradiction.
Since $f_n(m)=-1$ for all integers $1≤m≤n$ we have $a(m)= \pm 1$ and $b(m)= \mp 1$ for at least $n$ values. We thus have $a(x)+b(x)$ has at least $n$ zeros, so that necessarily $a(x)=-b(x)$. But now the largest term's coefficient of $a(x)b(x)$ is $-1$, a contradiction.
(4) Similar as in the last problem, $f_1(x)$ is irreducible and $f_n(x)$ has no zeros for $n > 1$, so that necessarily $a(x)=b(x)=\pm 1$ for at least $n$ values, and again $a(x)=b(x)$ so that $f_n(x)$ is a perfect square. However, when $n$ is odd this is an impossibility by degree, so $n$ must be even. For $n < 4$ we can observe $f_n(4)$ is irreducible, and for $n > 4$ and $n$ even we can observe $f_n(3/2)$ is negative, a contradiction since $f_n(x)$ is a perfect square.
When $n=4$, with a process similar to that for (1)(d), we can observe $f_4(x)=(x^2-5x+5)^2$ is its decomposition.$~\square$
(a) $x^2+x+1∈\mathbb{F}_2[x]$
(b) $x^3+x+1∈\mathbb{F}_3[x]$
(c) $x^4+1∈\mathbb{F}_5[x]$
(d) $x^4+10x+1∈\mathbb{Z}[x]$
2. Prove the following polynomials are irreducible in $\mathbb{Z}[x]$:
(a) $x^4-4x^4+6$
(b) $x^6+30x^5-15x^3+6x-120$
(c) $x^4+4x^3+6x^2+2x+1$
(d) $\dfrac{(x+2)^p-2^p}{x}$ for $p$ an odd prime.
3. Show that $f_n(x)=(x-1)(x-2)...(x-n)-1$ is irreducible over $\mathbb{Z}$ for all $n \geq 1$.
4. Show that $f_n(x)=(x-1)(x-2)...(x-n)+1$ is irreducible over $\mathbb{Z}$ for all $n \geq 1$ and $n≠4$.
Proof: (1) (a) There are no zeros to this equation, so it is irreducible.
(b) By discovering $f(1)=0$, we obtain $x^3+x+1=(x-1)(x^2+x+2)$, where the latter has no zeros.
(c) $x^4+1=x^4-4=(x^2-2)(x^2+2)$.
(d) Reduce modulo 5. If $f(x)=a(x)b(x)$, then due to monicity of $f(x)$ we have $a(x)$ and $b(x)$ are monic and not collapsing to units so that $\overline{a(x)}$ is associate to $\overline{x^2+2}$ or $\overline{x^2-2}$, say the former. Since $\overline{a(x)}$ is monic we have $\overline{a(x)}=\overline{x^2+2}$ which forces the constant term of $a(x)$ to not be $\pm 1$, a contradiction.
Alternatively, since $f(x)$ clearly has no zeros in $\mathbb{Z}$, it would have to factor as the product of two quadratics such that the following hold:$$f(x)=a(x)b(x)$$$$a(x)=x^2+a_1x+a_0$$$$b(x)=x^2+b_1x+b_0$$$$a_0b_0=1$$$$a_1b_0+a_0b_1=0$$$$a_1b_1+a_0+b_0=10$$$$a_1+b_1=0$$Solving for $b_1$ as a typical system of equations reveals$$\dfrac{1}{4}b_1^4+15b_1^2+24=0$$which is impossible.
(2) (a) Eisenstein's criterion for $p=2$ applies.
(b) Eisenstein applies for $p=3$.
(c) We generalize the technique employed in example 2. Lemma 1: Let $R$ be an integral domain with $f(x)=a_0+a_1x+...+a_nx^n∈R[x]$ and $g(x)=f(r(x))$ for $r(x)$ a nonconstant polynomial. If $g(x)$ is irreducible, then $f(x)$ is irreducible. Proof: If $f(x)=a(x)b(x)$ for some nonunit polynomials, then $g(x)=a(r(x))b(r(x))$ where necessarily $\text{deg}(a(r(x)))=\text{deg}(a(x)) \cdot \text{deg}(r(x)) \geq 1$ if $\text{deg}(a(x)) \geq 1$ and $a(r(x))=a(x)$ if $a(x)$ is a constant and similarly for $b(r(x))$, a contradiction.$~\square$
Let $g(x)=f(x-1)=x^4-2x+2$, where Eisenstein applies for $p=2$.
(d) $\dfrac{(x+2)^p-2^p}{x}=x^p+...+2p$, where by the binomial theorem the omitted terms have coefficients of multiples of $p$. Eisenstein applies for $p$.
(3) When $n=1$ we clearly have $f_n(x)$ is irreducible, so assume $n > 1$. Assuming $f_n(x)$ has a zero in $\mathbb{Z}$, then $1=(x-1)(x-2)...(x-n)$ so that in particular $x-1=\pm 1$ and $x-2=\pm 1$, a contradiction. Therefore if $f_n(x)=a(x)b(x)$ then the degrees of $a(x)$ and $b(x)$ are greater than one.
We demonstrate a basic algebraic result, first. If $R$ is an integral domain and $f(x)∈R[x]$ of degree $n \geq 1$, then there are at most $n$ zeros of $f(x)$. If not, then $f(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_{n+1})f'(x)$ entails $\text{deg}(f(x)) > n$, a contradiction.
Since $f_n(m)=-1$ for all integers $1≤m≤n$ we have $a(m)= \pm 1$ and $b(m)= \mp 1$ for at least $n$ values. We thus have $a(x)+b(x)$ has at least $n$ zeros, so that necessarily $a(x)=-b(x)$. But now the largest term's coefficient of $a(x)b(x)$ is $-1$, a contradiction.
(4) Similar as in the last problem, $f_1(x)$ is irreducible and $f_n(x)$ has no zeros for $n > 1$, so that necessarily $a(x)=b(x)=\pm 1$ for at least $n$ values, and again $a(x)=b(x)$ so that $f_n(x)$ is a perfect square. However, when $n$ is odd this is an impossibility by degree, so $n$ must be even. For $n < 4$ we can observe $f_n(4)$ is irreducible, and for $n > 4$ and $n$ even we can observe $f_n(3/2)$ is negative, a contradiction since $f_n(x)$ is a perfect square.
When $n=4$, with a process similar to that for (1)(d), we can observe $f_4(x)=(x^2-5x+5)^2$ is its decomposition.$~\square$
Friday, May 17, 2013
Bezout Subring of Q[x] (9.3.5)
Dummit and Foote Abstract Algebra, section 9.3, exercise 5:
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Let $R=\mathbb{Z}+x\mathbb{Q}[x] \subset \mathbb{Q}[x]$.
(a) Let $f(x),g(x)∈\mathbb{Q}[x]$ with $x^r$ the maximal power of $x$ dividing both $f(x)$ and $g(x)$, and let $f_r$ and $g_r$ be the coefficients of $x^r$ in $f(x)$ and $g(x)$ respectively. Then $\mathbb{Z}f_r+\mathbb{Z}g_r=\mathbb{Z}d_r$ for some nonzero $d_r\mathbb{Q}$. Prove there is a polynomial $d(x)∈\mathbb{Q}[x]$ such that $d(x)$ is a gcd of $f(x)$ and $g(x)$, and its minimal term is $d_rx^r$.
(b) Prove $f(x)=d(x)q_1(x)$ and $g(x)=d(x)q_2(x)$ for some $q_1(x),q_2(x)∈R$.
(c) Prove $d(x)=f(x)a(x)+g(x)b(x)$ for some $a(x),b(x)∈R$.
(d) Conclude $Rf(x)+Rg(x)=Rd(x)$ in $\mathbb{Q}[x]$, implying in part that $R$ is a Bezout domain.
(e) Show that there must be ideals of $R$ which are not finitely generated, and in fact $x\mathbb{Q}[x]$ is an ideal of $R$ that is not finitely generated.
Proof: (a) We can compute the gcd of $f(x)$ and $g(x)$ in the Euclidean domain $\mathbb{Q}[x]$. By the Euclidean algorithm we have $d(x)=v_1(x)f(x)+v_2g(x)$ for some polynomials $v_1(x),v_2(x)∈\mathbb{Q}$, so that $x^r \mid d(x)$. Furthermore, letting $f_r$ be the one of its pair that is guaranteed nonzero, by $d(x) \mid f(x)$ we must have the coefficient of $x^r$ in $d(x)$ is nonzero. Since this gcd is unique up to multiplication by a nonzero rational number, we can arrange for a gcd $d(x)$ with minimal term $d_rx^r$.
(b) Since $\mathbb{Z}f_r+\mathbb{Z}g_r=\mathbb{Z}d_r$ we have $d_rq_1=f_r$ and $d_rq_2=g_r$ for some $q_1,q_2∈\mathbb{Z}$. By definition of a gcd we have $f(x)=d(x)q_1(x)$ and $g(x)=d(x)q_2(x)$ for some $q_1(x),q_2(x)∈\mathbb{Q}[x]$, and by observation we see the constant terms of $q_1(x)$ and $q_2(x)$ are $q_1$ and $q_2$ respectively, putting these polynomials in $R$.
(c) We can observe $d(x)=f(x)a(x)+g(x)b(x)$ for some $a(x),b(x)∈\mathbb{Q}[x]$, and by 9.2.11 we have a full solution set for $a(x),b(x)$ given by$$a'(x)=a(x)+m(x) \dfrac{g(x)}{d(x)}$$$$b'(x)=b(x)-m(x) \dfrac{f(x)}{d(x)}$$as $m(x)$ ranges over $\mathbb{Q}[x]$.
Assume $f_r$ and $g_r$ are nonzero. First, observe that $d_r=f_r\alpha+g_r\beta$ for some $\alpha,\beta∈\mathbb{Z}$, so that $1=\dfrac{f_r}{d_r}\alpha+\dfrac{g_r}{d_r}\beta$, implying that $v_f=\dfrac{f_r}{d_r}$ is relatively prime to $v_g=\dfrac{g_r}{d_r}$ (which are, foremost, seen to be integers).
Letting $a$ and $b$ be the constant terms of $a(x)$ and $b(x)$ respectively, we can see by our mentioned equation that $(f_rx^r)a+(g_rx^r)b=d_rx^r$, so that $f_ra+g_rb=d_r$ implying $v_fa+v_gb=1$. As such, we can see $v_fz_1+v_gz_2=v_fa+v_gb$ for some $z_1,z_2∈\mathbb{Z}$, so that $v_fz_1-v_fa=-v_gz_2+v_gb$ and we can set $$m=\dfrac{z_1-a}{v_g}=\dfrac{-z_2+b}{v_f}$$so that with some manipulation and replacement we have$$z_1=a+m\dfrac{g_r}{d_r}$$$$z_2=b-m\dfrac{f_r}{d_r}$$Now, by letting $m(x)=m$ and taking the solutions $a'(x)$ and $b'(x)$ as above, by noting that the constant terms of $\dfrac{f(x)}{d(x)}$ and $\dfrac{g(x)}{d(x)}$ must be $\dfrac{f_r}{d_r}$ and $\dfrac{g_r}{d_r}$ respectively we see that the equations above actually describe the constant terms of $a'(x)$ and $b'(x)$, which implies $a'(x),b'(x)∈R$.
Assume one of $f_r$ or $g_r$ is zero, letting $f_r$ be the nonzero. Since now $d_r \mid f_r$ and $f_r \mid d_r$ we have $f_r=\pm d_r$, so that in the equation $d(x)=f(x)a(x)+g(x)b(x)$ we can see the constant term of $a(x)$ must be $\pm 1$ as the latter summand doesn't affect the sum coefficient of $x^r$. Continuing the terminology of the last procession we still have the same complete set of solutions, although observing the constant terms $z_1$ and $z_2$ of $a'(x)$ and $b'(x)$ simplifying to$$z_1=a+m\dfrac{g_r}{d_r}=a=\pm 1$$$$z_2=b-m\dfrac{f_r}{d_r}=b-m$$So choose $m(x)=b$ so that $a'(x),b'(x)∈R$ are valid solutions.
(d) This is evident from parts (a) and (c) as greatest common divisors in $R$ both exist and are expressible as $R$-linear combinations of elements from $R$, so that $(~f(x),g(x)~)=(~d(x)~)$. This implies every finitely generated ideal of $R$ is principle, i.e. $R$ is a Bezout domain.
(e) Since $R$ has been shown in the previous exercise to not be a UFD, it cannot be a PID, so that $R$ must contain infinitely generated nonprinciple ideals. Now, assume $x\mathbb{Q}[x]=(a_1,...,a_n)$ is finitely generated; then $x\mathbb{Q} \subseteq x\mathbb{Q}[x]=a_1R+...+a_nR$. By observing the first degree terms, we have $\mathbb{Q}=a_1\mathbb{Z}+...+a_n\mathbb{Z}$, which translates to saying that $\mathbb{Q}$ as an additive group is finitely generated by the $a_i$, a contradiction by 2.4.14(d).$~\square$
(a) Let $f(x),g(x)∈\mathbb{Q}[x]$ with $x^r$ the maximal power of $x$ dividing both $f(x)$ and $g(x)$, and let $f_r$ and $g_r$ be the coefficients of $x^r$ in $f(x)$ and $g(x)$ respectively. Then $\mathbb{Z}f_r+\mathbb{Z}g_r=\mathbb{Z}d_r$ for some nonzero $d_r\mathbb{Q}$. Prove there is a polynomial $d(x)∈\mathbb{Q}[x]$ such that $d(x)$ is a gcd of $f(x)$ and $g(x)$, and its minimal term is $d_rx^r$.
(b) Prove $f(x)=d(x)q_1(x)$ and $g(x)=d(x)q_2(x)$ for some $q_1(x),q_2(x)∈R$.
(c) Prove $d(x)=f(x)a(x)+g(x)b(x)$ for some $a(x),b(x)∈R$.
(d) Conclude $Rf(x)+Rg(x)=Rd(x)$ in $\mathbb{Q}[x]$, implying in part that $R$ is a Bezout domain.
(e) Show that there must be ideals of $R$ which are not finitely generated, and in fact $x\mathbb{Q}[x]$ is an ideal of $R$ that is not finitely generated.
Proof: (a) We can compute the gcd of $f(x)$ and $g(x)$ in the Euclidean domain $\mathbb{Q}[x]$. By the Euclidean algorithm we have $d(x)=v_1(x)f(x)+v_2g(x)$ for some polynomials $v_1(x),v_2(x)∈\mathbb{Q}$, so that $x^r \mid d(x)$. Furthermore, letting $f_r$ be the one of its pair that is guaranteed nonzero, by $d(x) \mid f(x)$ we must have the coefficient of $x^r$ in $d(x)$ is nonzero. Since this gcd is unique up to multiplication by a nonzero rational number, we can arrange for a gcd $d(x)$ with minimal term $d_rx^r$.
(b) Since $\mathbb{Z}f_r+\mathbb{Z}g_r=\mathbb{Z}d_r$ we have $d_rq_1=f_r$ and $d_rq_2=g_r$ for some $q_1,q_2∈\mathbb{Z}$. By definition of a gcd we have $f(x)=d(x)q_1(x)$ and $g(x)=d(x)q_2(x)$ for some $q_1(x),q_2(x)∈\mathbb{Q}[x]$, and by observation we see the constant terms of $q_1(x)$ and $q_2(x)$ are $q_1$ and $q_2$ respectively, putting these polynomials in $R$.
(c) We can observe $d(x)=f(x)a(x)+g(x)b(x)$ for some $a(x),b(x)∈\mathbb{Q}[x]$, and by 9.2.11 we have a full solution set for $a(x),b(x)$ given by$$a'(x)=a(x)+m(x) \dfrac{g(x)}{d(x)}$$$$b'(x)=b(x)-m(x) \dfrac{f(x)}{d(x)}$$as $m(x)$ ranges over $\mathbb{Q}[x]$.
Assume $f_r$ and $g_r$ are nonzero. First, observe that $d_r=f_r\alpha+g_r\beta$ for some $\alpha,\beta∈\mathbb{Z}$, so that $1=\dfrac{f_r}{d_r}\alpha+\dfrac{g_r}{d_r}\beta$, implying that $v_f=\dfrac{f_r}{d_r}$ is relatively prime to $v_g=\dfrac{g_r}{d_r}$ (which are, foremost, seen to be integers).
Letting $a$ and $b$ be the constant terms of $a(x)$ and $b(x)$ respectively, we can see by our mentioned equation that $(f_rx^r)a+(g_rx^r)b=d_rx^r$, so that $f_ra+g_rb=d_r$ implying $v_fa+v_gb=1$. As such, we can see $v_fz_1+v_gz_2=v_fa+v_gb$ for some $z_1,z_2∈\mathbb{Z}$, so that $v_fz_1-v_fa=-v_gz_2+v_gb$ and we can set $$m=\dfrac{z_1-a}{v_g}=\dfrac{-z_2+b}{v_f}$$so that with some manipulation and replacement we have$$z_1=a+m\dfrac{g_r}{d_r}$$$$z_2=b-m\dfrac{f_r}{d_r}$$Now, by letting $m(x)=m$ and taking the solutions $a'(x)$ and $b'(x)$ as above, by noting that the constant terms of $\dfrac{f(x)}{d(x)}$ and $\dfrac{g(x)}{d(x)}$ must be $\dfrac{f_r}{d_r}$ and $\dfrac{g_r}{d_r}$ respectively we see that the equations above actually describe the constant terms of $a'(x)$ and $b'(x)$, which implies $a'(x),b'(x)∈R$.
Assume one of $f_r$ or $g_r$ is zero, letting $f_r$ be the nonzero. Since now $d_r \mid f_r$ and $f_r \mid d_r$ we have $f_r=\pm d_r$, so that in the equation $d(x)=f(x)a(x)+g(x)b(x)$ we can see the constant term of $a(x)$ must be $\pm 1$ as the latter summand doesn't affect the sum coefficient of $x^r$. Continuing the terminology of the last procession we still have the same complete set of solutions, although observing the constant terms $z_1$ and $z_2$ of $a'(x)$ and $b'(x)$ simplifying to$$z_1=a+m\dfrac{g_r}{d_r}=a=\pm 1$$$$z_2=b-m\dfrac{f_r}{d_r}=b-m$$So choose $m(x)=b$ so that $a'(x),b'(x)∈R$ are valid solutions.
(d) This is evident from parts (a) and (c) as greatest common divisors in $R$ both exist and are expressible as $R$-linear combinations of elements from $R$, so that $(~f(x),g(x)~)=(~d(x)~)$. This implies every finitely generated ideal of $R$ is principle, i.e. $R$ is a Bezout domain.
(e) Since $R$ has been shown in the previous exercise to not be a UFD, it cannot be a PID, so that $R$ must contain infinitely generated nonprinciple ideals. Now, assume $x\mathbb{Q}[x]=(a_1,...,a_n)$ is finitely generated; then $x\mathbb{Q} \subseteq x\mathbb{Q}[x]=a_1R+...+a_nR$. By observing the first degree terms, we have $\mathbb{Q}=a_1\mathbb{Z}+...+a_n\mathbb{Z}$, which translates to saying that $\mathbb{Q}$ as an additive group is finitely generated by the $a_i$, a contradiction by 2.4.14(d).$~\square$
Thursday, May 16, 2013
Polynomial Factorization Practice (9.3.4)
Dummit and Foote Abstract Algebra, section 9.3, exercise 4:
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Let $R=\mathbb{Z}+x\mathbb{Q}[x] \subset \mathbb{Q}[x]$.
(a) Prove $R$ is an integral domain and $R^\times = \pm 1$.
(b) Show that the irreducibles of $R$ are the irreducibles of $\mathbb{Z}$ together with the irreducible polynomials of $\mathbb{Q}[x]$ with constant term $\pm 1$. Prove these irreducibles are prime.
(c) Show $x$ cannot be written as the product of irreducibles in $R$ (in particular, $x$ is not irreducible) and conclude that $R$ is not a UFD.
(d) Show $x$ is not prime in $R$ and describe $R/(x)$.
Proof: (a) Since this set is clearly closed under subtraction and multiplication, it is a subring, and by its containment in $\mathbb{Q}[x]$ it is an integral domain. Note that the degree norm $N$ has the property $N(\alpha \beta)=N(\alpha)+N(\beta)$ as in a typical polynomial ring, so that the only possible units are within $\mathbb{Z}$, and as such the only units are $\pm 1$.
(b) Should $p=\alpha \beta$ we have $0=N(p)=N(\alpha)+N(\beta)$, so that $\alpha,\beta∈\mathbb{Z}$ and thus by the primality of $p$ we have one of the two is equal to $\pm 1$, i.e. is a unit. Letting $p(x)=q_nx^n + ... + q_1x \pm 1$ be an irreducible element of $\mathbb{Q}[x]$ with constant term $\pm 1$, assuming $p(x)=\alpha \beta$ for nonunit $\alpha,\beta∈R$, by necessity $\alpha$ and $\beta$ have constant term $\pm 1$ so that $N(\alpha),N(\beta) > 0$. But now $\alpha$ and $\beta$ are nonunits in $\mathbb{Q}[x]$, violating the irreducibility of $p(x)$. Finally, assuming $a(x)$ is an irreducible element of $R$ by necessity we have its constant term is $\pm 1$ else for a prime divisor $p$ of the term we have $a(x)=p \dfrac{a(x)}{p}$ is factorization when $a(x)≠\pm p$. Assuming $a(x)=b(x)c(x)$ for some nonunit $b(x),c(x)∈\mathbb{Q}[x]$, we have the constant term of $b(x)$ is $\dfrac{m}{n}$ and necessarily the constant term of $c(x)$ is $\dfrac{n}{m}$ so that $b(x)=\dfrac{m}{n}b'(x)$ and $c(x)=\dfrac{n}{m}c'(x)$ for $b'(x),c'(x)∈R$ and now $a(x)=b'(x)c'(x)$ is a valid factorization of $a(x)$ in $R$ by $0 < N(b(x))=N(b'(x))$ and $0 < N(c(x))=N(c'(x))$ so that $b'(x),c'(x)∉R^\times$.
For prime $p$ the ideal $(p) \subseteq R$ is the kernel of the surjective homomorphism $φ : R → \mathbb{Z}/p\mathbb{Z}$ by $q_nx^n + ... + q_1x + z \mapsto z + p\mathbb{Z}$. For irreducible $q∈\mathbb{Q}[x]$ with constant term $\pm 1$, we have $q$ is prime and thus $R/(q)$ is contained in the integral domain $\mathbb{Q}/(q)$ and is therefore an integral domain itself, proving $q$ is prime.
(c) Note that any $vx∈R$ for $v∈\mathbb{Q}$ is not irreducible, since $vx=(2)(\dfrac{v}{2}x)$ is a valid factorization. Now, if $x=\pi_1...\pi_n$ for irreducible $\pi_i$, then $1=N(\pi_1)+...+N(\pi_n)$ so that exactly one $\pi_k$ is of the form $qx+z$ and for $i≠k$ we have $\pi_i∈\mathbb{Z}$. By our remark above $z≠0$, but now the product $\pi_1...\pi_n$ has a nonzero constant term, a contradiction.
(d) If $x$ were prime in $R$, it would be irreducible, a contradiction by the above. We now claim that $R/(x)=\{\overline{a+bx}~|~a∈\mathbb{Z},b∈\mathbb{Q} \land 0≤b<1\}$. Since clearly $\supseteq$, we must show the reverse: For arbitrary $a(x)∈R$, we have $$a(x)=q_nx^n + ... + q_1x+z=$$$$(q_nx^{n-1}+...q_2x)x+(w+q_1')x+z=(q_nx^{n-1}+...q_2x+w)x+q_1'x+z$$ for some $w∈\mathbb{Z}$ and $0≤q_1'<1$, so that $\overline{a(x)}∈\{\overline{a+bx}~|~a∈\mathbb{Z},b∈\mathbb{Q} \land 0≤b<1\}$. From here, the isomorphism from $R/(x)$ onto $\mathbb{Z}+(\mathbb{Q}/\mathbb{Z})x$ is evident, where $\mathbb{Q}/\mathbb{Z}$ is the ring extension of $\mathbb{Q}/\mathbb{Z}$ as additive groups by standard multiplication.$~\square$
(a) Prove $R$ is an integral domain and $R^\times = \pm 1$.
(b) Show that the irreducibles of $R$ are the irreducibles of $\mathbb{Z}$ together with the irreducible polynomials of $\mathbb{Q}[x]$ with constant term $\pm 1$. Prove these irreducibles are prime.
(c) Show $x$ cannot be written as the product of irreducibles in $R$ (in particular, $x$ is not irreducible) and conclude that $R$ is not a UFD.
(d) Show $x$ is not prime in $R$ and describe $R/(x)$.
Proof: (a) Since this set is clearly closed under subtraction and multiplication, it is a subring, and by its containment in $\mathbb{Q}[x]$ it is an integral domain. Note that the degree norm $N$ has the property $N(\alpha \beta)=N(\alpha)+N(\beta)$ as in a typical polynomial ring, so that the only possible units are within $\mathbb{Z}$, and as such the only units are $\pm 1$.
(b) Should $p=\alpha \beta$ we have $0=N(p)=N(\alpha)+N(\beta)$, so that $\alpha,\beta∈\mathbb{Z}$ and thus by the primality of $p$ we have one of the two is equal to $\pm 1$, i.e. is a unit. Letting $p(x)=q_nx^n + ... + q_1x \pm 1$ be an irreducible element of $\mathbb{Q}[x]$ with constant term $\pm 1$, assuming $p(x)=\alpha \beta$ for nonunit $\alpha,\beta∈R$, by necessity $\alpha$ and $\beta$ have constant term $\pm 1$ so that $N(\alpha),N(\beta) > 0$. But now $\alpha$ and $\beta$ are nonunits in $\mathbb{Q}[x]$, violating the irreducibility of $p(x)$. Finally, assuming $a(x)$ is an irreducible element of $R$ by necessity we have its constant term is $\pm 1$ else for a prime divisor $p$ of the term we have $a(x)=p \dfrac{a(x)}{p}$ is factorization when $a(x)≠\pm p$. Assuming $a(x)=b(x)c(x)$ for some nonunit $b(x),c(x)∈\mathbb{Q}[x]$, we have the constant term of $b(x)$ is $\dfrac{m}{n}$ and necessarily the constant term of $c(x)$ is $\dfrac{n}{m}$ so that $b(x)=\dfrac{m}{n}b'(x)$ and $c(x)=\dfrac{n}{m}c'(x)$ for $b'(x),c'(x)∈R$ and now $a(x)=b'(x)c'(x)$ is a valid factorization of $a(x)$ in $R$ by $0 < N(b(x))=N(b'(x))$ and $0 < N(c(x))=N(c'(x))$ so that $b'(x),c'(x)∉R^\times$.
For prime $p$ the ideal $(p) \subseteq R$ is the kernel of the surjective homomorphism $φ : R → \mathbb{Z}/p\mathbb{Z}$ by $q_nx^n + ... + q_1x + z \mapsto z + p\mathbb{Z}$. For irreducible $q∈\mathbb{Q}[x]$ with constant term $\pm 1$, we have $q$ is prime and thus $R/(q)$ is contained in the integral domain $\mathbb{Q}/(q)$ and is therefore an integral domain itself, proving $q$ is prime.
(c) Note that any $vx∈R$ for $v∈\mathbb{Q}$ is not irreducible, since $vx=(2)(\dfrac{v}{2}x)$ is a valid factorization. Now, if $x=\pi_1...\pi_n$ for irreducible $\pi_i$, then $1=N(\pi_1)+...+N(\pi_n)$ so that exactly one $\pi_k$ is of the form $qx+z$ and for $i≠k$ we have $\pi_i∈\mathbb{Z}$. By our remark above $z≠0$, but now the product $\pi_1...\pi_n$ has a nonzero constant term, a contradiction.
(d) If $x$ were prime in $R$, it would be irreducible, a contradiction by the above. We now claim that $R/(x)=\{\overline{a+bx}~|~a∈\mathbb{Z},b∈\mathbb{Q} \land 0≤b<1\}$. Since clearly $\supseteq$, we must show the reverse: For arbitrary $a(x)∈R$, we have $$a(x)=q_nx^n + ... + q_1x+z=$$$$(q_nx^{n-1}+...q_2x)x+(w+q_1')x+z=(q_nx^{n-1}+...q_2x+w)x+q_1'x+z$$ for some $w∈\mathbb{Z}$ and $0≤q_1'<1$, so that $\overline{a(x)}∈\{\overline{a+bx}~|~a∈\mathbb{Z},b∈\mathbb{Q} \land 0≤b<1\}$. From here, the isomorphism from $R/(x)$ onto $\mathbb{Z}+(\mathbb{Q}/\mathbb{Z})x$ is evident, where $\mathbb{Q}/\mathbb{Z}$ is the ring extension of $\mathbb{Q}/\mathbb{Z}$ as additive groups by standard multiplication.$~\square$
Tuesday, May 14, 2013
Polynomial Ideals and Primality (9.1.13)
Dummit and Foote Abstract Algebra, section 9.1, exercise 13:
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Letting $F$ be a field, prove $F[x,y]/(y^2-x) \not ≅ F[x,y]/(y^2-x^2)$.
Proof: We show something a bit stronger, that $F[x,y]/(y^n-x) \not ≅ F[x,y]/(x^2-y^2)$ for any $n$, with $F$ only an integral domain.
First we show $y-x,y+x∉(y^2-x^2)=(y-x)(y+x)$ so that there are zero divisors in $F[x,y]/(y^2-x^2)$. Assuming the contrary we have for $f$ some polynomial $y-x=(y-x)(y+x)f⇒(y+x)f=1$, an impossibility. Similarly for $y+x∉(y^2-x^2)$.
Now we show $F[x,y]/(y^n-x)=F[y]/(y^n-x)≅F[y]$ is an integral domain, which will complete the proof. To see the first equality, notice that $\supseteq$ is clear and any polynomial of $F[x,y]/(y^n-x)$ can be represented by the image of a polynomial devoid of variables in $x$ since $\overline{y}^n=\overline{x}$. Letting $φ$ be the natural surjective homomorphism of $F[y]$ onto $F[y]/(y^n-x)$, we can observe that if $\text{ker}~φ≠0$ then $0≠(y^n-x)f∈F[y]$ for some $f∈F[x,y]$, an impossibility by $(y^n-x)f=y^nf-xf$ and observing the result of the $xf$ term by choosing the result of the multiplication monomial with maximal degree of $x$ in $f$, which cannot be affected by any terms in $xf$ (else the $f$ monomials' $y$ and $x$ degrees are the same and can be rewritten as one) nor in $y^nf$.$~\square$
Proof: We show something a bit stronger, that $F[x,y]/(y^n-x) \not ≅ F[x,y]/(x^2-y^2)$ for any $n$, with $F$ only an integral domain.
First we show $y-x,y+x∉(y^2-x^2)=(y-x)(y+x)$ so that there are zero divisors in $F[x,y]/(y^2-x^2)$. Assuming the contrary we have for $f$ some polynomial $y-x=(y-x)(y+x)f⇒(y+x)f=1$, an impossibility. Similarly for $y+x∉(y^2-x^2)$.
Now we show $F[x,y]/(y^n-x)=F[y]/(y^n-x)≅F[y]$ is an integral domain, which will complete the proof. To see the first equality, notice that $\supseteq$ is clear and any polynomial of $F[x,y]/(y^n-x)$ can be represented by the image of a polynomial devoid of variables in $x$ since $\overline{y}^n=\overline{x}$. Letting $φ$ be the natural surjective homomorphism of $F[y]$ onto $F[y]/(y^n-x)$, we can observe that if $\text{ker}~φ≠0$ then $0≠(y^n-x)f∈F[y]$ for some $f∈F[x,y]$, an impossibility by $(y^n-x)f=y^nf-xf$ and observing the result of the $xf$ term by choosing the result of the multiplication monomial with maximal degree of $x$ in $f$, which cannot be affected by any terms in $xf$ (else the $f$ monomials' $y$ and $x$ degrees are the same and can be rewritten as one) nor in $y^nf$.$~\square$
Monday, May 13, 2013
Prime Ideal Squared (9.1.12)
Dummit and Foote Abstract Algebra, section 9.1, exercise 12:
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Let the overbar denote passage from $\mathbb{Q}[x,y,z]$ into $\mathbb{Q}/(xy-z^2)$. Prove that $\overline{P}=(\overline{x},\overline{z})$ is a prime ideal. Show that $\overline{xy}∈\overline{P}^2$ but that no power of $\overline{y}$ lies in $\overline{P}^2$, showing that $\overline{P}$ is a prime ideal whose square is not a primary ideal.
Proof: It is easy to check that for a surjective homomorphism $φ$, we have $(φ(a_1),φ(a_2),...)=φ(a_1,a_2,...)$, so that $(\overline{x},\overline{z})=(x,z)/(xy-z^2)$. Evidently $(xy-z^2) \subseteq (x,z)$ so that $\overline{\mathbb{Q}[x,y,z]}/\overline{(x,z)}≅\mathbb{Q}[x,y,z]/(x,z)≅\mathbb{Q}[y]$ is an integral domain, so that $\overline{P}$ is a prime ideal. Furthermore, $\overline{xy-z^2}=\overline{0}$ so that $\overline{xy}=\overline{z}^2∈\overline{P}^2$ as claimed. Now, if $\overline{y}^n∈\overline{P}^2 \subseteq \overline{P}$ for some $n∈\mathbb{Z}^+$ then we have the natural image of $\overline{y}^n$ in the previously mentioned quotient isomorphic to $\mathbb{Q}[y]$ is zero, an impossibility by the commutativity of natural projections.$~\square$
Proof: It is easy to check that for a surjective homomorphism $φ$, we have $(φ(a_1),φ(a_2),...)=φ(a_1,a_2,...)$, so that $(\overline{x},\overline{z})=(x,z)/(xy-z^2)$. Evidently $(xy-z^2) \subseteq (x,z)$ so that $\overline{\mathbb{Q}[x,y,z]}/\overline{(x,z)}≅\mathbb{Q}[x,y,z]/(x,z)≅\mathbb{Q}[y]$ is an integral domain, so that $\overline{P}$ is a prime ideal. Furthermore, $\overline{xy-z^2}=\overline{0}$ so that $\overline{xy}=\overline{z}^2∈\overline{P}^2$ as claimed. Now, if $\overline{y}^n∈\overline{P}^2 \subseteq \overline{P}$ for some $n∈\mathbb{Z}^+$ then we have the natural image of $\overline{y}^n$ in the previously mentioned quotient isomorphic to $\mathbb{Q}[y]$ is zero, an impossibility by the commutativity of natural projections.$~\square$
Sunday, May 12, 2013
Characterization of PIDs (8.3.11)
Dummit and Foote Abstract Algebra, section 8.3, exercise 11:
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Prove$$R~\text{a Principle Ideal Domain}~⇔R~\text{a UFD}~\land R~\text{a Bezout Domain}$$Proof: The $⇒$ direction is evident, so it suffices to demonstrate the converse. Observe a typical ideal $I \subseteq R$, and choose arbitrary $x∈I$ with $x=u\pi_1^{\alpha_1}...\pi_n^{\alpha_n}$ being its unique decomposition up to units. For any $e∈\mathbb{Z}^+ \cup \{0\}$ and $i$ with $1≤i≤n$ let $A_{i,e}=\{r∈I ~ | ~ \pi_i^e \mid r\}$. For each $i$ let $\beta_i$ be the largest integer such that $A_{i,\beta_i}=I$, which is guaranteed to exist for arbitrary $i$ as the exponent $\alpha_i$ of $\pi_i$ is finite so that as a bound $x∉A_{i,\alpha_i+1}$. For each $i$, choose $a_i$ such that $a_i ∈ I \setminus A_{i,\beta_i+1}$, so that $a_i=\pi_i^{\beta_i}n_i$ with $\pi_i \not \mid n_i$. Letting $A=\bigcup_{1≤i≤n} a_i$ and taking successive greatest common divisors of the $a_i$ and then taking a greatest common divisor of this and $x$, we have a final common divisor $d=\pi_1^{\beta_1}...\pi_n^{\beta_n}$ (since these are the only irreducibles dividing $x$, and for each $i$ there is an element $a_i$ with its minimal $\pi_i$ exponent $\beta_i$, and all the exponents of the $a_j$ and $x$ of $\pi_i$ is at least $\beta_i$ as $a_j,x∈I=A_{i,\beta_i}$), writable as an $R$-linear combination of elements from $A \cup \{x\} \subseteq I$ by 7.2.7. We claim $(d)=I$: Evidently $(d) \subseteq I$, and for the reverse containment, for any $r∈I$ we have $\pi_i^{\beta_i} \mid r$ for applicable $i$ since $r∈I=A_{i,\beta_i}$, implying $I \subseteq (d)$ and now $I=(d)$ is a principle ideal.$~\square$
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