User:J D Bowen/Math735 HW8

9.1.5) Prove $$(x,y), \ (2,x,y) \ $$ are prime ideals of $$\mathbb{Z}[x,y] \ $$, but that only the latter is maximal.

9.2.3) Let $$ f(x) \ $$ be a polynomial in $$ F[x] \ $$.  We aim to showrove that $$ F[x]/(f(x)) \ $$ is a field if and only if $$ F(x) \ $$ is irreducible.

Assume that $$ F \ $$ is a field. Then $$ F[x] \ $$ is a Euclidean Domain, which implies that $$ F[x] \ $$ is a Principal Ideal Domain. Then by proposition 7 in section 8.2 implies that prime ideals in $$ F[x] \ $$ are maximal. Proposition 14 from Section 7.4, in a commutative ring $$ R \ $$, all maximal ideals are prime. Thus, (*) in $$ F[x] \ $$ an ideal is prime $$ \iff \ $$ it is a maximal ideal.

$$ F[x]/(f(x)) \ $$ is a field $$ \iff (f(x)) \ $$ is a max ideal of $$ F[x] \ $$, by Proposition 12 from section 7.4. $$ \iff (f(x)) \ $$ is a prime ideal of $$ F[x] \ $$, by (*) $$ \iff f(x) \ $$ is prime in $$ F[x] \ $$, by Definition (2) on page 284 of the text. $$ \iff f(x) \ $$ is irreducible in $$ F[x] \ $$, by Proposition 11 in Section 8.3.

9.2.5) Exhibit all ideals in the ring $$F[x]/(p(x)) \ $$, in terms of the factorization of $$p(x) \ $$.

9.3.1) Let R be an integral domain with quotient field F and let p(x) be a monic polynomial in R[x]. Assume p(x)=a(x)b(x) where a(x)b(x) are monic polynomials in F[x] of degree smaller than p(x).  Prove that $$a(x)\not\in R[x] \implies R \ $$ is not a UFD.  Dedue $$\mathbb{Z}[2\sqrt{2}] \ $$ is not a UFD.

Arguing by contradiction, suppose that R is a Unique Factorization Domain. Since $$ p(x)=a(x)b(x) \ $$ in $$ F[x] \ $$ and p $$ \in R[x] \ $$, Gauss' Lemma implies that $$ \exists \beta \in F $$ such that $$ \beta \ $$a and $$ \beta^{-1} \ $$b are in $$ R[x] \ $$. Since b is monic, $$ \beta^{-1} \in R $$, hence $$ a= \beta^{-1} (\beta a) \in R[x] \ $$, a contradiction. It is not necessary assume that a and b have smaller degree than $$ p \ $$. For part two, note that $$ (x+\sqrt x)(x+\sqrt x)=x^2+2 \sqrt 2x +8 \ $$. The latter polynomial lies in $$ \Z [2 \sqrt{2}] \ $$, and is monic. But $$ \sqrt{2} \notin \Z [2 \sqrt{2}] \ $$. (If it were, then $$ \sqrt{2} = a+2b \sqrt{2} \ $$ for some $$ a, b \in \Z \ $$, and then $$ \sqrt{2} = a/(1-2b) \in \Q \ $$, then contradiction.) Thus $$ x+ \sqrt{2} \notin \Z [2 \sqrt{2}] \ $$, and $$ x+ \sqrt{2} \ $$ is monic. Then by the first part, $$ \Z [2 \sqrt{2}] \ $$ is not a Unique Factorization Domain.

9.3.2) We aim to show that $$f,g \ $$ polys with rational coefficients whose product $$fg \ $$ has integer coefficients, then the product of any coefficient of $$g \ $$ with any coefficient of $$f \ $$ is an integer.

By the Gauss lemma, there exists $$r, s \in \mathbb{Q} \ $$ such that $$rf\in\mathbb{Z}[x], \ sg\in\mathbb{Z}[x], rsfg=fg\in\mathbb{Z}[x] \ $$. For a fraction $$p/q \ $$, define $$\text{Num}(p/q)=p, \ \text{Den}(p/q)=q \ $$. Since $$r=s^{-1} \ $$, we can allow without loss of generality $$r\geq 1 \ $$.

Note that if $$f(x)=\Sigma a_i x^i, \ g(x)=\Sigma b_ix^i \ $$, we have

$$rf\in\mathbb{Z}[x] \implies \text{lcm}(\text{Den}(a_i)) | \text{Num}(r) \ $$,

$$sg\in\mathbb{Z}[x] \implies \text{Den}(s)|\text{lcm}(\text{Num}(b_i)) \ $$.

Together, these two facts imply

$$\text{Den}(a_i) | \text{Num}(b_j), \ \text{Den}(b_i) | \text{Num}(a_j) \ $$,

which of course imples $$a_ib_j \in\mathbb{Z} \ \forall i,j \ $$.

11.1.1) Let $$(a_1, a_2, ..., a_n) \ $$ be a vector in $$\mathbb{R}^n \ $$. We aim to show the collection of vectors $$\left\{{x:x=(x_1,...,x_n) }\right\} \ $$ such that $$a_1x_1+...+a_nx_n=0 \ $$ is a subspace of $$\mathbb{R}^n \ $$.

Observe that the vector $$0=(0,\dots,0) \ $$ satisfies the condition trivially. Now suppose two vectors $$x, y \ $$ satisfy the condition. Then

$$(x+y)\cdot a = (x_1+y_1)a_1+\dots+(x_n+y_n)a_n = (x_1a_1+\dots+x_na_n)+(y_1a_1+\dots+y_na_n)=0+0=0 \ $$,

and so the set is closed under addition. Further observe that

$$(-x_1)a_1+\dots+(-x_n)a_n = -(x_1a_1+\dots+x_na_n)=-0=0 \ $$

and so if a vector satisfies the condition, its additive inverse also satisfies the condition. Finally, note that

$$(cx_1)a_1+\dots+(cx_n)a_n = c(x_1a_1+\dots+x_na_n)=c0=0 \ $$.

Hence, this set is a subspace of $$\mathbb{R}^n \ $$.

11.1.2) Let $$P^5 \ $$ be the collection of polynomials with $$\mathbb{Q} \ $$ coefficients, degree at most 5. We aim to show this is a space of dimension 6, with $$1, x, \dots, x^5 \ $$ as a basis.  We also wish to show $$1,1+x,1+x+x^2, \dots, 1+x+x^2+x^3+x^4+x^5$$ is a basis.

Note that $$q_1 x^j + q_2 x^k = 0, \ 0\leq j<k\leq 5 \ \implies q_1=-q_2 x^{k-j} \implies q_1=q_1=0 \ $$, and so the set $$1, x, \dots, x^5 \ $$ is a linearly independent set in $$P^5 \ $$.

By definition of polynomial, all polynomials are certainly in the span of this set, and since the span of this set is contained in the space, and it is linearly independent, it is a basis, implying $$\text{dim}(P_5)=6 \ $$

Now define $$f_0=1, \ f_1=1+x, \ \dots, f_5 = 1+x+x^2+x^3+x^4+x^5 \ $$.

For any function

$$f=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5\in P_5 \ $$,

define

$$c_5= a_5, \ c_4=a_4-a_5, \ c_3=a_3-a_4, \ c_2=a_2-a_3, \ c_1=a_1-a_2, \ c_0=a_0-a_1 \ $$.

Then

$$c_0f_0+c_1f_1+c_2f_2+c_3f_3+c_4f_4+c_5f_5 \ $$

$$=(c_0)+(c_1+c_1x)+(c_2+c_2x+c_2x^2)+(c_3+c_3x+c_3x^2+c_3x^3)+(c_4+c_4x+c_4x^2+c_4x^3+c_4x^4)+(c_5+c_5x+c_5x^3+c_5x^3+c_5x^4+c_5x^5) \ $$

$$=(c_0+c_1+c_2+c_3+c_4+c_5)+(c_1+c_2+c_3+c_4+c_5)x+(c_2+c_3+c_4+c_5)x^2+(c_3+c_4+c_5)x^3+(c_4+c_5)x^4+c_5x^5 \ $$

$$=a_0+a_1x+a_2x^2+a_3x^4+a_5x^5 = f \ $$.

Hence $$f\in P_5 \implies f\in \text{span}(f_0, \dots, f_5) \ $$. So $$P_5 \subset \text{span}(f_0, \dots, f_5) \ $$

We certainly have $$\text{span}(f_0, \dots, f_5) \subset P_5 \ $$, since $$\forall i, f_i \in P_5 \ $$. So $$\text{span}(f_0, \dots, f_5)=P_5 \ $$.

Now suppose we have $$\Sigma a_i f_i(z) = 0 \ \forall z\in\mathbb{C} \ $$. Then we have

$$\Sigma a_if_i(z) \ $$

$$=(a_0+a_1+a_2+a_3+a_4+a_5)+(a_1+a_2+a_3+a_4+a_5)z+(a_2+a_3+a_4+a_5)z^{[2]}+(a_3+a_4+a_5)z^{[3]}+(a_4+a_5)z^{[4]}+a_5z^{[5]}=0 \ $$

Of course, this implies $$a_5=0 \ $$, which causes a chain of implications $$\implies a_4=0 \implies a_3=0 \implies a_2=0 \implies a_1=0 \implies a_0=0 \ $$. Hence, the $$f_i \ $$ are linearly independent. Therefore, they constitute a basis for $$P_5 \ $$.