MTH2001 Algebraic Structures Pre-Lecture 3

• Lecture: MTH2001 Algebraic Structures Lecture 3 (Provided Notes).
• Practical: MTH2001 Algebraic Structures Practical 2 (Practical Sheet).

Isomorphisms

A complex conjugation map $f:\mathbb{C}\to \mathbb{C}$ where $f(a+ib)=a-ib$ is a ring homomorphism (preserves sum and multiplication, can be proven directly). As $\mathbb{C}$ is an integral domain, the map $f$ is also an integral domain homomorphism, as well as a field homomorphism (as $\mathbb{C}$ is also a field).

In group theory, a group isomorphism is a map that is both bijective and a group homomorphism; this is similar for ring isomorphisms.

As for groups, if there’s a ring isomorphism $\varphi :R\to S$, then $R$ and $S$ are essentially the same ring - they share the same algebraic properties, e.g. cardinality, unity, commutativity.

Recall that bijectivity is something that is injective and surjective, that means that…

Definition of an Isomorphism

An isomorphism from $R$ to $S$ is a mapping $\theta :R\to S$ that is…

• Injective: $\theta (x)=\theta (y)⟹x=y$.
• Surjective: $\forall s\in S,\exists r\in R:\theta (r)=s$.
• Homomorphic: $\theta (x+y)=\theta (x)+\theta (y),\theta (xy)=\theta (x)\theta (y)$.

If such isomorphism exists, we write $R\cong S$; $R$ and $S$ are isomorphic.

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With this definition, we can show that $\mathbb{C}$ is not just a ring homomorphism, but also a ring isomorphism (and hence an integral domain and field isomorphism), by proving its bijectivity (directly). This is not always the case though, for example with $f:\mathbb{Z}/6\mathbb{Z}\to \mathbb{Z}/6\mathbb{Z}$ given by $f(x)=\bar{4}x$.

To prove something is non-isomorphic, we can use the actual definition to show that they don’t share algebraic properties:

• Commutativity,
• Unity,
• Zero divisors,
• Integral domain,
• Field,
• Finity.

Ideals

An ideal is a subring $I\subseteq R$ that “absorbs” the elements of $R$ under multiplication, which can be useful to generalise the concept of even integers, as well as provide a way to construct new rings: quotient rings.

Definition of an Ideal

A subring $I$ of a ring $R$ is an ideal of $R$ if $ar\in I$ and $ra\in I$ for all $a\in I$ and all $r\in R$.

We can then write $I⊲R$ to show that $I$ is an ideal of $R$.

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There are some trivial examples, too, that apply for every ring $R$:

• $\left\{0\right\}⊲R$ because $0r=r0=0$ for all $r\in R$.
• $R⊲R$ because $R$ is closed with respect to its multiplication.

Image and Kernel

Given a homomorphism $\theta :R\to S$, the image of $\theta$ is…

$$\text{Im}(\theta )=\left\{\theta (r):r\in R\right\}\subset S$$

and the kernel of $\theta$ is…

$$\text{Ker}(\theta )=\left\{r\in R:\theta (r)=0_s\right\}\subset R$$

Given that something is a ring homomorphism, there are theories that state…

• The image of $\theta$ is a subring of $S$.
• The kernel of $\theta$ is an ideal of $R$.
• $\theta$ is injective if and only if $\text{ker}\theta =\left\{0_R\right\}$.

These will be proven in the practical session.

Using this theorem, we can find ideals!