This is a note about abstract algebra, one of the largest branches of mathematics. I will show some concepts in abstract algebra and the power of abstraction.

*What is abstraction?*

Before getting our hand's dirty with some algebra, first let us consider the concept of abstraction. Abstraction is, loosely speaking, the process of taking a problem or a concept and removing unnecessary context. This is an important concept in mathematics. It is what allowed us to view the number \(3\) for example, as an abstract concept rather than \(3\) lots of sheep. Another context we could think about is shapes. Let's say we know a fact about a triangle and a square. We later do some observation to reveal that this also applies to a pentagon and hexagon and so on. However how do we prove that this is the case? Well we cannot prove it for every polygon we can come up with because there are infinitely many! But what if we proved the hypothesis for an \(n\)-gon. Independent of the number of vertices. Then we have automatically proven it for *all* polygons! This is the power of abstraction.

*Sets*

Sets are arguably the most important concepts in Mathematics. **All** of Mathematics can be derived from sets! However i will not show this here. I am interested in some special sets. These sets are used on a daily basis and by explaining the structures they possess, it helps give a better understanding of the. Firstly let us look at some commonly used sets:

- \(\Bbb Z\) the set of integers. \(\{\cdots, -2, -1, 0, 1, 2,\cdots\}\)
- \(\Bbb Q\) the set of rational numbers. Numbers that can be expressed as a ratio of integers \(q=a/b\) where \(q\) is rational and \(a\) and \(b\) are integers.
- \(\Bbb R\) the set of real numbers. This includes rational and irrational numbers. \(\{e, \pi, \sqrt{2}, 3, 1/2,\cdots\}\)
- \(\Bbb C\) the set of complex numbers. \(\{3+4 i, e+\pi i, 6i, 2,\cdots\}\)
- \(\Bbb Z_n\) the set of integers modulo \(n\) \(\{0,1,2,\cdots,n-1\}\)

You are probably familiar with these sets, if you are not then this post will not make much sense to you. One thing to note is that all of these sets share many algebraic properties. These properties define an abstract object called a *ring*.

**Definition 1**
A *ring* \(R\) is a set of elements on which two binary operations, addition \((+)\) and multiplication \((\cdot)\), are defined that satisfy the following properties for all \(a,b,c\in R\):

- (Addition is commutative) \(a+b=b+a\)
- (Addition is associative) \((a+b)+c=a+(b+c)\)
- (Additive identity exists) There exists an element \(0\) in \(R\) such that \(a+0=a\)
- (Additive inverses exist) For each element \(a\) in \(R\), there exists an element \(x\) such that \(a+x=0\)
- (Multiplication is associative) \((a\cdot b)\cdot c=a\cdot (b\cdot c)\)
- (Multiplication distributes over addition) \(a\cdot (b+c)=a\cdot b+a\cdot c\) and \( (b+c)\cdot a=b\cdot a+c\cdot a\)

Note that multiplication in a ring is not always commutative. A ring where multiplication is commutative is called a commutative ring. An example of a non-commutative ring would be \(M_2(\Bbb Z)\), the collection of \(2\times2\) matrices with integer entries. Remember, matrix multiplication is not commutative.

Already we can prove some things about rings,. Suppose \(R\) is a ring and \(a,b\in R\) then:

- (Additive Cancellation). If \(a+b=a+c\), then \(b=c\).
- (Solution of equations) The equation \(a+x=b\) always has a
*unique*solution in \(R\). - (Uniqueness of additive inverses) Every element of \(R\) has exactly one additive inverse.
- (Uniqueness of additive identity) There is only one element of \(R\) that satisfies the equations \(z+a=a\), for all \(a\): namely the element \(0\).

These properties of rings are not in the definition but derive directly from the ring definition. It is fun to prove these however remember that \(R\) is not necessarily commutative so the order of multiplication cannot be switched.

Now let us look at more concepts about rings.

**Definition 2**

A subset \(S\) of a ring \(R\) is said to be a **subring** of \(R\) if \(S\) is itself a ring under the operations induced from \(R\).

This is an important concept because it tells us that ring structure can be inherited for example \(\Bbb Q \subset \Bbb R\) and both are rings therefore \(\Bbb Q\) is a subring of \(\Bbb R\).

**The Subring theorem**
A non-empty subset of a ring is a subring under the same operations if and only if it is closed under multiplication and subtraction.

This is an important theorem because it allows us to check if a subset is indeed a subring without checking every property of a ring. This can be proven however it will not be proven here.

Some rings \(R\) have a *unity*, or *multiplicative identity*; that is, an element \(u\in R\) where \(au=ua=a\) for all \(a \in R\). If a unity exists, it is unique.

For example the number \(1\) is the unity in \(\Bbb Z\), \(\Bbb Q\), \(\Bbb R\) and \(\Bbb C\). The matrix \(\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\) is the unity in \(M_2(\Bbb Z\)). But \(2\Bbb Z\), the set of even integers \(\{\cdots, -4, -2, 0, 2, 4,\cdots\}\), has no unity.

An element \(a\ne0\) is a *zero divisor* if there is an element \(b\ne0\) with \(ab=0\). For example in \(\Bbb Z_6,\quad 2\cdot3=0\)

A commutative ring with unity is an **integral domain** if it has no zero divisors. \(\Bbb Z\), \(\Bbb Q\), \(\Bbb R\) and \(\Bbb C\) are all **integral domains**. The integers modulo \(n\), \(\Bbb Z_n\), is only an **integral domain** if and only if \(n\) is prime.

Integral domains have the nice property of multiplicative cancellation. We can state this as a theorem:

If \(R\) is an integral domain and \(a,b,c\in R\) with \(a\ne0\), then \(ab=ac\) implies that \(b=c\).

If \(R\) is a ring with unity \(1\), then an element \(a\in R\) is a *unit* if there exists a \(b\in R\) such that \(ab=1\). In this case, \(b\) is said to be the *multiplicative inverse of* \(a\). If all the non-zero elements of a commutative ring with unity are units, then we say the ring is a *field*. The rings \(\Bbb Q\), \(\Bbb R\) and \(\Bbb C\) are all fields but \(\Bbb Z\) is not. All fields are integral domains. \(\Bbb Z_p\) is a field, for a prime \(p\). Indeed, all finite integral domains are fields. The fact that \(\Bbb Z_p\) is a field is an important setting for proving Fermat's little theorem which states that:

If \(p\) is prime and \(0<x<p\), then \(x^p\equiv x\pmod p\).

Finally, consider \(F[x]\), the set of polynomials over an arbitrary field \(F\) with coefficients in \(F\). \(F[x]\) is an integral domain. The *Division Theorem*, the *Root Theorem*, *Euclid's Algorithm*, the *GCD identity* and *Unique Factorization* all hold for \(F[x]\). A particularly important example is \(F=\Bbb C\), the field of complex numbers. \(\Bbb C[x]\) satisfies the *Fundamental Theorem of Algebra*: Every non-constant polynomial in \(C\) has a root.

There are a lot of theorems mentioned earlier. You may recognise some of them but as you can see the power of abstract algebra has generalised them for the field of polynomials \(F[x]\) rather than for integers.

I hope this post has enlightened you about the power of abstract algebra and given you a taste of higher Mathematics. The last paragraph about polynomials over fields is an interesting thing that has many fun proofs to try. But an interesting note to take is that all of these things that we do with integers, arithmetic, GCD and factorisation and so on can be generalised to polynomials and other sets.

## Comments

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TopNewestBrilliant seriously needs wikis on higher pure mathematics, which I believe to contain the true flavour of mathematics. Anyways, a nice post. – Kuldeep Guha Mazumder · 1 year, 5 months ago

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– Guido Barta · 1 year, 4 months ago

yes it would be great also to have exercises about abstract algebraLog in to reply

– Kuldeep Guha Mazumder · 1 year, 4 months ago

No not only exercises, theory is of utmost necessity. Though Brilliant focusses primarily on problem solving but I being essentially a proponent of theory developing (I don't underrate problem solving by any means), think some necessary focus must be put on the theory of these topics of pure mathematics.Log in to reply

Very nice post, Ali! Very helpful! – Cody Johnson · 3 years, 1 month ago

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Nice brief intro to abstract algebraic structures – Ethan W · 2 years, 3 months ago

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