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.

Let $ ab \in (x,y) $ such that $ a,b \in \Z [x,y] $. Since $ ab \in (x,y) $, each term of $ ab $ contains an x or y. Assume that $ a\notin (x,y) $ and $ b\notin (x,y) $. Then there are terms in $ a $ and $ b $ that do not contain an x or y i.e, there exists at least one term in both $ a $ and $ b $ that contains neither x nor y. Then, the product $ ab $ would have a term that contains neither x nor y, which implies that $ ab \notin (x,y) $. Thus we have reached a contradiction. Therefore, $ ab \in (x,y) $ implies that either $ a \in (x,y) $ or $ b \in (x,y) $. Thus $ (x,y) $ is a prime ideal of $\mathbb{Z}[x,y] \ $.

Now let $ ab \in (2,x,y) $ such that $ a,b \in \Z [x,y] $. Then each term of $ ab $ contains an x or y or is a constant multiple of 2. Assume that $ a\notin (2,x,y) $ and $ b\notin (2,x,y) $. Then there is a constant term in both $ a $ and $ b $ which is not a multiple of 2. Then, the product $ ab $ would have a constant term that is not a multiple of 2, because any even constant would have an even factor. This implies that $ ab \notin (2,x,y) $. Thus we have reached a contradiction. Therefore, $ ab \in (2,x,y) $ implies that either $ a \in (2,x,y) $ or $ b \in (2,x,y) $. Thus $ (2,x,y) $ is a prime ideal of $\mathbb{Z}[x,y] \ $.

Now assume that $ I $ is an ideal of $\mathbb{Z}[x,y] \ $ such that $ (2,x,y) \subset I \subset \Z [x,y] $. Then $ I $ contains some element in $\mathbb{Z}[x,y] \ $ that is not in $ (2,x,y) $. Therefore, this element has an odd constant term. Therefore, since $ I $ must be closed under addition, $ I $ contains all the elements of $ (2,x,y) $ and elements with odd constant terms. But this generates all of $\mathbb{Z}[x,y] \ $. Therefore, the ideal $ (2,x,y) $ must be maximal.

To show that $ (x,y) $ is not maximal, we need to show that there exists an ideal, $ J $ such that $ (x,y) \subset J \subset \Z [x,y] $. Let $ J = (2,x,y) $. Clearly, $ (x,y) \subset (2,x,y) $ but $ (2,x,y) $ contains elements with even constant terms, which are not in $ (x,y) $. For instance, $ 2 \in (2,x,y) $ but $ 2 \notin (x,y) $. Thus $ (x,y) \subset (2,x,y) \subset \Z [x,y] $, which implies that $ (x,y) $ is not 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) \ $ where F is a field and p(x) is a polynomial in $ F[x] \ $.

Since F is a field, by Corollary 4, $ F[x] \ $ is a Principal Ideal Domain. Thus all ideals, $ (p(x)) \subset F[x] \ $ are principal. Let the map, $ \phi \ $, from $ F[x] \implies F[x]/(p(x))\ $, where $ \phi : f(x) \implies \overline {f(x)} \ $, is a surjective homomorphism. So note the ideal $ (p(x)) \subset F[x] \ $ is mapped to $ \overline {(p(x))} \subset F[x]/(p(x)) \ $ , and generally speaking $ \phi^{-1}(I) \ $ for some ideal $ I \subset F[x]/(p(x)) \ $ is an ideal in $ F[x] \ $. We know that $ (p(x))=ker\phi \ $ So, now $ (p(x)) \subset F[x] \ $ and I is an ideal in $ F[x]/(p(x)) \ $, thus $ \phi^{-1}(I) \subset F[x] $ is an ideal. Since The preimage contains the kernel, $ \phi^{-1}(I) \supset (p(x)) $. Let $ \phi^{-1}(I)=f(x) \ $ for some $ f(x) \in F[x] \ $, then $ f(x)|p(x) \ $. Hence $ \forall (f(x)) \in F[x], \phi(f(x)) \implies \overline{(f(x))} \ $, where $ \overline{(f(x))} \ $ is an ideal in $ F[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 \ $.

Given a vector $v \ $, define $k=a_1v_1+\dots + a_nv_n \ $. Define the map $\phi(v)=v-ka \ $. This map is linear: $\phi(c_1u+c_2v)= c_1u+c_2v-k(c_1u+c_2v)a = c_1u+c_2v-ak(c_1u)-ak(c_2v) = c_1(u-ak(u))+c_2(v-ak(v)) = c_1\phi(u)+c_2\phi(v) \ $. Note that $\text{ker}(\phi)=\left\{{v:k(v)=0}\right\} = \left\{{v:v=ca, \ c\in\mathbb{R}}\right\} \ $, and so $n-1=\text{dim}(\mathbb{R}^n)-\text\text{dim}(\text{ker}(\phi))=\text{dim}(\text{im}(\phi)) \ $.

A basis can be found by taking any basis that includes $a \ $ and applying the Gram-Schmidt process; since this process will produce an orthogonal basis, that basis minus the normed $a \ $ will be contained in $a^\perp \ $, and hence be in this subspace.

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 \ $.