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- We will prove $R$ has the ACCP then by (i) $R$ is a factorisation domain.
Suppose that $(d_n)^{∞}_{n = 0}$ has $d_{n + 1}|d_n$ for all $n ∈ ℕ_0$, so that $⟨d_0 ⟩ ⊂ ⟨d_1⟩ ⊂ ⋯$ and let$$I = ⋃_{n ∈ ℕ_0} ⟨d_n⟩$$$I$ is an ideal: If $s, t ∈ I$ then there are $n, m ∈ℕ_0$ such that $s ∈ ⟨d_n ⟩$ and $t ∈ ⟨d_m⟩$ and so $s, t ∈ ⟨d_{\max \{n, m\}} ⟩$ by nesting, and hence $s-t ∈ ⟨d_{\max \{n, m\}}⟩⊂ I$. Since $0 ∈ I$, it is a subgroup by the subgroup test, and finally if $r ∈ R$ then $rs, sr ∈ ⟨d_n⟩ ⊂ I$ as required.
Since $R$ is a PID there is some $d ∈ I$ such that $I = ⟨d⟩$. Since $d ∈ I$ there is some $N ∈ ℕ_0$ such that $d_N|d$, but then $d_n ∈ I$ for all $n ∈ ℕ_0$ and so $d_N|d|d_n$ for all $n ∈ ℕ_0$ and hence $d_n∼d_N$ for all $n ⩾ N$. The result is proved.
- We will prove $R$ has the ACCP then by (i) $R$ is a factorisation domain.
- Let $α$ be a root of a monic polynomial $p ∈ ℤ[X],\deg p(X)=d>0$. For $i⩾d$, in $ℤ[X]$ divide $X^i$ by $p(X)$ with remainder $r(x)$ then $\deg r≤d-1$. So $α^i$ is a ℤ-linear combination of $1, α, …, α^{d-1}$. Since the ℤ-module $ℤ[α]$ is generated by $1, α, α^2, …$ we have $ℤ[α]$ is generated by $1, α, …,α^d$.
If the ℤ-module $ℤ[α]$ has a basis $p_0,…,p_{d-1}$, there exists $A,B∈ℤ^{d×d}$ such that \begin{split} \pmatrix{1&α&⋯&α^{d-1}}=\pmatrix{p_0&p_1&⋯&p_{d-1}}A\\ \pmatrix{α&α^2&⋯&α^d}=\pmatrix{p_0&p_1&⋯&p_{d-1}}B \end{split}Suppose $A$ is invertible(???) we get $$α\pmatrix{1&α&⋯&α^{d-1}}=\pmatrix{1&α&⋯&α^{d-1}}A^{-1}B$$ Therefore $α$ is eigenvalue of $A^{-1}B$, $α$ is a root of its characteristic polynomial. - Since $\overlineℤ$ is the union of an accending chain of rings\[ℤ[α_1]⊂ℤ[α_1,α_2]⊂⋯⊂ℤ[α_1,…,α_n]⊂⋯\]Clearly $1∈\overlineℤ$. For any $x,y∈\overlineℤ$, there is $n$ such that $x,y∈ℤ[α_1,…,α_n]$, so $x-y,xy∈ℤ[α_1,…,α_n]$, so $\overlineℤ$ is a subring of ℂ.
- $a∈\overline{ℤ}⇒\sqrt{a}$ (a choice of root)$∈\overline{ℤ}$
$a≠0⇒a=\sqrt{a}\sqrt{a}⇒a$ is not irreducible
- Let $α$ be a root of a monic polynomial $p ∈ ℤ[X],\deg p(X)=d>0$. For $i⩾d$, in $ℤ[X]$ divide $X^i$ by $p(X)$ with remainder $r(x)$ then $\deg r≤d-1$. So $α^i$ is a ℤ-linear combination of $1, α, …, α^{d-1}$. Since the ℤ-module $ℤ[α]$ is generated by $1, α, α^2, …$ we have $ℤ[α]$ is generated by $1, α, …,α^d$.
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- MSE
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- By lemma, $A/⟨p⟩$ is a field. By the hint, $U(A)=ℝ^*$, by (i)$$U(A/⟨p⟩)=q(U(A))=q(ℝ^*)$$so $A/⟨p⟩=\{0\}∪U(A/⟨p⟩)=q(ℝ)$, so $∃a,b∈ℝ$ such that$$X+⟨p⟩=a+⟨p⟩,Y+⟨p⟩=b+⟨p⟩$$they satisfy$$a^2+b^2+1+⟨p⟩=X^2+Y^2+1+⟨p⟩=0+⟨p⟩⇒a^2+b^2+1=0$$which is impossible in ℝ.
Lemma. Let $F$ be a field and $R$ an integral domain that is a finite-dimensional $F$-vector space. Then $R$ is a field.Why we can't replace $A=ℝ[X,Y]/⟨X^2+Y^2+1⟩$ with $ℝ[X]/⟨X^2+1⟩$?
Proof. Given $0≠r∈R$ the $F$-linear map$$R→R:↦rx$$is injective ($R$ is an integral domain!), hence surjective ($R$ is finite-dimensional!). So 1 is the image of some $s∈R$, i.e. $sr=1$ and so $s=r^{-1}$ belongs to $R$.
If $A=ℝ[X]/⟨X^2+1⟩$ Because $A$ is isomorphic to ℂ, it is a field, then $U(A)=A^*$ which is strictly larger than $ℝ^*$, but in 2.(d)(ii) assumes that $U(A)=ℝ^*$.
I gathered the information:
$ℝ[X]$ is ED, so it is a PID.
$ℝ[X]/⟨X^2+1⟩$ is isomorphic to ℂ, it is a field.
$ℝ[X,Y]$ is not PID because the ideal $⟨X,Y⟩$ is not principal.
$ℝ[X,Y]/⟨X^2+Y^2+1⟩$ is a PID but not ED.
- By lemma, $A/⟨p⟩$ is a field. By the hint, $U(A)=ℝ^*$, by (i)$$U(A/⟨p⟩)=q(U(A))=q(ℝ^*)$$so $A/⟨p⟩=\{0\}∪U(A/⟨p⟩)=q(ℝ)$, so $∃a,b∈ℝ$ such that$$X+⟨p⟩=a+⟨p⟩,Y+⟨p⟩=b+⟨p⟩$$they satisfy$$a^2+b^2+1+⟨p⟩=X^2+Y^2+1+⟨p⟩=0+⟨p⟩⇒a^2+b^2+1=0$$which is impossible in ℝ.
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- For each $k$, the GCD of all the determinants of $k×k$ submatrices of $A$ is preserved under elementary operations (up to permutation of columns and rows). These invariants equal $d_1⋅…⋅d_k$ in the Smith normal form for each $k$, determining it uniquely.MSE