MTH2001 Algebraic Structures Practical 2

• Practical Sheet.

Question 1

Using the Definition of a Field, we can check if the following are fields…

1. $M=\left\{\frac{a}{b}:a,b\in \mathbb{Z},a\text{ is odd}\right\}$ - not a field as it’s not closed under addition; let $a_1=2m+1,a_2=2n+1$ such that $\frac{a_1}{b}+\frac{a_2}{b}=\frac{a_1+a_2}{b}=\frac{2m+1+2n+1}{a}=2(m+n+1)$, which is an even number.
2. $N=\left\{\frac{a}{b}:a,b\in \mathbb{Z},a\text{ is even}\right\}$ - not a field as it lacks multiplicative inverses; the inverse of $\frac{a}{b}$ is $\frac{b}{a}$, where if $b$ is odd then $\frac{b}{a}$ does not exist in $N$.
3. ${\mathbb{R}}_0^{+}=\left\{x\in \mathbb{R}:x\ge 0\right\}$ - not a field as it lacks additive inverses as it’s restricted to positive only.

Question 2

Using the Definition of a Ring Homomorphism we can test to see which of the following maps are ring homomorphisms, and then find their kernels (values that map to zero).

1. $\theta :\mathbb{Z}\to \mathbb{Z}$ defined by $\theta (a)=3a$: Doesn’t conserve multiplication…
   • $\theta (ab)=3ab\ne 9ab=\theta (a)\theta (b)\square$.
2. $\theta :{\mathbb{Z}}_6\to {\mathbb{Z}}_3$ defined by $\theta (\bar{a})=\bar{a}$: Conserves operations…
   • $\theta (\bar{a}+\bar{b})=\overline{a+b}=\bar{a}+\bar{b}=\theta (\bar{a})+\theta (\bar{b})$ in ${\mathbb{Z}}_3$.
   • $\theta (\bar{a}\cdot \bar{b})=\overline{a\cdot b}=\bar{a}\cdot \bar{b}=\theta (\bar{a})\cdot \theta (\bar{b})$.
   • $\boxed{\text{A homomorphism with kernel }\{\overline{0},\overline{3}\}}$.
3. $\theta :\mathbb{C}\to \mathbb{R}$ defined by $\theta (z)=|z|$: Doesn’t conserve addition…
   • $\theta (z_1+z_2)=|1+(-1)|=0\text{vs.}\theta (z_1)+\theta (z_2)=1+1=2$.
4. $\theta :\mathbb{C}\to \mathbb{C}$ defined by $\theta (z)=iz$: $\theta (z_1{z}_2)=i{z}_1{z}_2\ne -z_1{z}_2=\theta (z_1)\theta (z_2)\square$.
5. $\varphi :\mathbb{Z}[x]\to \mathbb{Z}$ defined by $\varphi (f(x))=f(3)\forall f(x)\in \mathbb{Z}[x]$: preserves operations…
   • $\varphi (f+g)=(f+g)(3)=f(3)+g(3)=\varphi (f)+\varphi (g)$.
   • $\varphi (f\cdot g)=(f\cdot g)(3)=f(3)\cdot g(3)=\varphi (f)\cdot \varphi (g)$.
   • $\boxed{\text{A homomorphism with kernel }⟨x-3⟩}$.

Question 3

Given the following subrings, we can use the Definition of an Ideal to see if they’re ideals:

1. The subset $\mathcal{M}$ of $M(2,\mathbb{R})$ given by $\mathcal{M}=\left\{\left[\begin{array}{cc}x& y\\ 0& 0\end{array}\right]:x,y\in \mathbb{R}\right\}$: isn’t closed under left multiplication…
   • Let $A=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$ and $B=\left[\begin{array}{cc}x& y\\ 0& 0\end{array}\right]\in \mathcal{M}$. Then: $AB=\left[\begin{array}{cc}0& 0\\ x& y\end{array}\right]\notin \mathcal{M}\square$.
2. The subset of $\mathbb{Z}[x]$ given by $\mathcal{N}=\left\{x\cdot f(x):f(x)\in \mathbb{Z}[x]\right\}$: closed under addition and multiplication…
   • Closed under addition: $x\cdot f(x)+x\cdot g(x)=x\cdot (f(x)+g(x))\in \mathcal{N}$.
   • Closed under multiplication by $\mathbb{Z}[x]$: For $h(x)\in \mathbb{Z}[x]$, $h(x)\cdot x\cdot f(x)=x\cdot (h(x)f(x))\in \mathcal{N}$.
   • $\boxed{\text{Is an ideal}}$.

Question 4

Given that $\theta :R\to S$ is a ring homomorphism, we can show that its image $\text{Im}(\theta )$ is a subring of $S$ and its kernel $\text{ker}(\theta )$ is a subring of $R$ as follows…

1. $\text{Im}(\theta )$ is a subring of $S$:

   • Contains $0_S$: $\theta (0_R)=0_S$.
   • Closed under addition: $\theta (a)+\theta (b)=\theta (a+b)$.
   • Closed under multiplication: $\theta (a)\cdot \theta (b)=\theta (ab)$.
   • Contains additive inverses: $-\theta (a)=\theta (-a)$.

2. $\ker (\theta )$ is a subring of $R$:

   • Contains $0_R$: $\theta (0_R)=0_S$.
   • Closed under addition: $\theta (a+b)=0_S$ if $a,b\in \ker (\theta )$.
   • Closed under additive inverses: $\theta (-a)=-\theta (a)=0_S$.
   • Closed under multiplication: $\theta (ab)=\theta (a)\theta (b)=0_S$.

We can then also that the kernel $\text{ker}(\theta )$ is an ideal of $R$…

3. $\ker (\theta )$ is an ideal of $R$:

• Absorbs multiplication: For $r\in R$ and $a\in \ker (\theta )$:

$$\theta (ra)=\theta (r)\theta (a)=\theta (r)\cdot 0_S=0_S\text{and}\theta (ar)=0_S.$$

$\boxed{\text{Im}(\theta )\text{ is a subring of }S\text{ and }\ker (\theta )\text{ is an ideal of }R}$