Rings and modules exercises

1. Consider the rings $$ R_1=\frac{ℤ_2[x]}{⟨x^2+1⟩}, \quad R_2=\frac{ℤ_3[x]}{⟨x^2+1⟩}, \quad R_3=\frac{ℤ_4[x]}{⟨x^2+1⟩} . $$ Are any of these rings integral domains?

   For which $n$, $ℤ_n[x]/⟨x^2+1⟩$ is an integral domain?
   If $ℤ_n[x]/⟨x^2+1⟩$ is ID, then $ℤ_n[x]$ is ID, then $ℤ_n$ is ID (as $ℤ_n$ is a subring of $ℤ_n[x]$), so $n$ is prime.
   $∃a:n|a^2+1⟹n≡1\pmod4⟹n∈\tt\href{https://oeis.org/A002144}{A002144}$
   There is a Wikipedia entry Pythagorean prime

   In total, how many ideals does $R_1$ have?

   [In fact $R_1$ is the field $𝔽_4$] Let $I$ be an ideal of $R_1=\{0,1,x,1+x\}$.
   If $x∉I$, then $I⊂\{0,1,1+x\}$.

   • $I=\{0,1,1+x\}$ is not an ideal, as $1+(1+x)=x∉I$.

   • $\{0,1\}$ is not an ideal, as $x⋅1=x∉I$.

   So $I=\{0\}$ or $\{0,1+x\}$.

   If $x∈I$, then $x^2=1∈I$, so $I=R_1$.
   So $R_1$ has 3 ideals in total.

2. By considering the image of 1, or otherwise, determine explicitly all the ring homomorphisms from $R_1$ to $R_3$.

Let ϕ be a ring homomorphisms from $R_1$ to $R_3$. By definition $ϕ(1)=1$.\begin{array}l ϕ(2)=2ϕ(1)=2\\ ϕ(2)=ϕ(0)=0\end{array} 2 and 0 not equal in $R_3$, so no such ϕ.

1. If $E$ and $F$ are subfields of ℂ the complex numbers, and $F \subseteq E$, define the degree of the field extension $E: F$.

   Suppose that $E/F$ is a field extension. Then $E$ may be considered as a vector space over $F$ (the field of scalars). The dimension of this vector space is called the degree of the field extension, and it is denoted by $[E:F]$.

2. If $α∈ℂ$, define the field $ℚ(α)$ generated by $α$. Define what it means for an element $α∈ℂ$ to be algebraic over $ℚ$, and define the minimal polynomial $f_α∈ℚ[t]$ of α.

   ℚ(α) is the intersection of all subfields containing α.
   α is algebraic over ℚ iff ℚ(α):ℚ is finite.
   The minimal polynomial $f_α$ is the monic polynomial of minimal degree such that $f_α(α)=0$.
   $f_α$ is the (unique) monic generator of $\{f∈ℚ[X]:f(α)=0\}$.

3. If $\operatorname{deg}(f)=d$, show that $\left\{1,α,…,α^{d-1}\right\}$ is a ℚ-basis of $ℚ(α)$, and hence $[ℚ(α): ℚ]=d$.

   Concrete solution

   An element in $ℚ(α)$ has the form $g(α)\over f(α)$ for $f(α)≠0$. Using Bezout to write $1\over f(α)$ as linear combination of $\left\{1,α,…,α^{d-1}\right\}$

   Abstract solution

   $I=\ker(ℚ[t]\xrightarrow{\text{ev}_α}ℂ)$ \[\operatorname{Im}(\text{ev}_α)≅ℚ[t]/I⊂ℂ\] ℂ is integral domain, so $ℚ[t]/I$ is integral domain, so $I$ is prime ideal.
   Recall: primes ideals in PID are maximal.
   So $ℚ[t]/I$ is a field, $ℚ[α]⊂ℚ[t]/I$ so $ℚ[α]=ℚ[t]/I$ and $ℚ[α]$ has the required basis.

4. State and prove the tower law.

   Suppose that \(\mathbb{K}\) is a field extension of \(\mathbb{L}\) and \(\mathbb{L}\) is a field extension of \(\mathbb{F}\). Then \(\mathbb{K}\) is a field extension of \(\mathbb{F}\), and if either \(| \mathbb{L}: \mathbb{F} | < \infty\)\(\) or \(| \mathbb{L}: \mathbb{K} | | \mathbb{K}: \mathbb{F} | < \infty\) then \(| \mathbb{L}: \mathbb{F} | = {| \mathbb{L}: \mathbb{K} |} {| \mathbb{K}: \mathbb{F} |}\).

   Proof. The first part is immediate because the relation ‘is a subfield of' is transitive. Let \(e_1, \ldots, e_n\) be a basis for \(\mathbb{L}\) as a vector space over \(\mathbb{K}\), and let \(f_1, \ldots, f_m\) be a basis for \(\mathbb{K}\) as a vector space over \(\mathbb{F}\). Now, for \(x∈\mathbb{L}\) there are scalars \(\lambda_1, \ldots, \lambda_n∈\mathbb{K}\) such that \(x = \lambda_1 e_1 + \cdots + \lambda_n e_n\), and since \(f_1, \ldots, f_m\) is spanning, for each \(1 \leqslant j \leqslant n\) there are scalars \(\mu_{1, j}, \ldots, \mu_{m, j} \in \mathbb{F}\) such that \(\lambda_j = \mu_{1, j} f_1 + \cdots + \mu_{m, j} f_m\). Hence \(x = \sum^n_{j = 1} \sum^m_{i = 1} \mu_{i, j} f_i e_j\), so we have that \((f_i e_j)^{m, n}_{i = 1, j = 1}\) is an \(\mathbb{F}\)-spanning subset of \(\mathbb{K}\). Now suppose \(\mu_{1, 1}, \ldots, \mu_{m, n}∈\mathbb{F}\) are such that \(\sum^n_{j = 1} \sum^m_{i = 1} \mu_{i, j} f_i e_j = 0_{\mathbb{L}}\). Then \(\sum^n_{j = 1} (\sum^m_{i = 1} \mu_{i, j} f_i) e_j = 0_{\mathbb{L}}\), but \(\sum^m_{i = 1} \mu_{i, j} f_i∈\mathbb{K}\) for each \(1 \leqslant j \leqslant n\) and since \(e_1, \ldots, e_n\) are \(\mathbb{K}\)-linearly independent we have \(\sum^m_{i = 1} \mu_{i, j} f_i = 0_{\mathbb{K}}\) for all \(1 \leqslant j \leqslant n\). But now \(f_1, \ldots, f_m\) are \(\mathbb{F}\)-linearly independent and so \(\mu_{i, j} = 0_{\mathbb{F}}\) for all \(1 \leqslant i \leqslant m\) and \(1 \leqslant j \leqslant n\). It follows that \((f_i e_j)^{m, n}_{i = 1, j = 1}\) is a basis for \(\mathbb{L}\) as an \(\mathbb{F}\)-vector space and the result follows.\(\Box\)

5. Find the degree of the following field extensions:
   1. $ℚ(\sqrt{2}): ℚ$
   2. $ℚ(\sqrt{2}, \sqrt{3}): ℚ$ Degree of field extension

   $[ℚ(\sqrt2, \sqrt3):ℚ(\sqrt3)]≤[ℚ(\sqrt2):ℚ]=2$ because so $[ℚ(\sqrt2, \sqrt3):ℚ(\sqrt2)] = 2$.

1. Define what it means for an $R$-module $M$ to be free?

   A free module is a module that has a basis – that is, a generating set consisting of linearly independent elements.

   What does it mean for $x∈M$ to be a torsion element?

   An element $m$ of a module $M$ over a ring $R$ is called a torsion element of the module if there exists a regular element $r$ of the ring (an element that is neither a left nor a right zero divisor) that annihilates $m$, i.e., $r.m=0$.

   An $R$-module is said to be torsion-free if the only torsion element is $0_M$.
   Show that if $T=\{x∈M: x\text{ is a torsion element}\}$ then $T$ is a submodule of $M$ and that $M/T$ is torsion-free.
2. Now suppose that $R$ is a Euclidean domain and suppose that $M$ is a finitely generated $R$-module.
   Show that if $N<M$ is a submudule of $M$ then $N$ is also finitely generated. Q3 on sheet 3
3. Let $R=𝒞[0,1]$ be the ring of continuous functions on the unit interval $[0,1]$. Let $M=\left\{f∈R: ∃ N∈ℤ_{>0}, f(1/n)=0 \quad ∀n≥N\right\}$. Show that $M$ is a submodule of $R$ (where as $R$ is viewed as an $R$-module over itself via the multiplication action).

   By submodule test, we check $∀f_1,f_2∈M:$ 1) $f_1+f_2∈M$; 2) $∀f∈M:ff_1∈M$.
   1) $∃N_1,N_2∈ℤ_{>0}$ such that $f_1(1/n)=0\ ∀n≥N_1$ and $f_2(1/n)=0\ ∀n≥N_2$. Let $N=\max\{N_1,N_2\}$ then $(f_1+f_2)(1/n)=0\ ∀n≥N$.
   2) $f_1(1/n)=0\ ∀n≥N_1⇒f(1/n)f_1(1/n)=0\ ∀n≥N_1$

4. Hence or otherwise show that for a general ring, a submodule of a finitely generated module need not be finitely generated.

$R$ is a finitely generated $R$-module [since $R=⟨1⟩$]. $M⩽R$ and we'll prove $M$ is not finitely generated:
Let $f_1,…f_k∈M$, then $∃N_1,…N_k∈ℤ_{>0}$ such that $f_i(1/n)=0\ ∀n≥N_i$. Let $N_\max=\max\{N_1,…N_k\}$, $∃f∈R$ such that $f(1/n)=0\ ∀n>N_\max$ and $f(1/N_\max)≠0$. \begin{eqnarray*}f(1/n)=0\ ∀n>N_\max&⇒&f∈M\\ f(1/N_\max)≠0&⇒&f∉⟨f_1,…f_k⟩ \end{eqnarray*}