# In abstract algebra, what is an intuitive explanation for a field?

Wikipedia has the following to say about fields.

In mathematics, a field is one of the fundamental algebraic structures used in abstract algebra. It is a nonzero commutative division ring, or equivalently a ring whose nonzero elements form an abelian group under multiplication. As such it is an algebraic structure with notions of addition, subtraction, multiplication, and division satisfying the appropriate abelian group equations and distributive law.

• What intuition do you find lacking in "an algebraic structure with notions of addition, subtraction, multiplication, and division"? – Eric Wofsey Jul 6 '16 at 20:47
• Generalization of Rationals? Usual arithmetic laws, associative, commutative, distributive,everything has an additive inverse and everything but $0$ has a multiplicative inverse. – lulu Jul 6 '16 at 20:47
• @lulu "generalization of rationals" makes me think "ordered field". – Mark S. Jul 6 '16 at 20:49
• @MarkS. Fair enough, though I think it's still the best place to start if the concept is new to you. Next step...probably finite fields. No order there! But, personally, I always start with the most familiar examples even when, as here, they come with some extra structure. – lulu Jul 6 '16 at 20:54
• @lulu, no yeah, you're right. If they haven't seen enough elementary number theory to almost know some finite fields of size p already, the rationals (and the complex numbers if they're comfortable with complex division) are a fine place to start – Mark S. Jul 6 '16 at 20:57

Stimulated by the comments, I now realize that you’ve already seen many examples of fields. If you’ve had two years of high-school algebra, you know the fields $\Bbb Q$ of all rational numbers and $\Bbb R$ of all real numbers. But you know more.

It’s universal here in the States, at least, to ask students to “rationalize the denominator” in fractions like $$\frac{1+3i}{2+4i}=\frac{(1+3i)(2-4i)}{(2+4i)(2-4i)} =\frac{14+2i}{20}=\frac{7+i}{10}=\frac7{10}+\frac1{10}i\,,$$ although teachers may not ask students to perform the last step. You see, though, that the complex numbers $\Bbb C$ of all $a+bi$ with $a,b\in\Bbb R$ are a field, and this example certainly lets you believe that the Gaussian numbers, all $a+bi$ with $a,b\in\Bbb Q$, also form a field.

I hope that in high school you also simplified fractions like $$\frac{1+3\sqrt2}{2+4\sqrt2} =\frac{(1+3\sqrt2)(2-4\sqrt2)}{(2+4\sqrt2)(2-4\sqrt2)} =\frac{-22+2\sqrt2}{-28} =\frac{11-\sqrt2}{14}=\frac{11}{14}-\frac1{14}\sqrt2\,,$$ so that you now realize that the numbers of form $a+b\sqrt2$ (where now you must restrict $a$ and $b$ to rational numbers), also play nicely together to make up a field.

It would be wrong to say that fields are everywhere in Abstract Algebra, but if you know where to look, you’ll find them surprisingly often.

Here's a simple way to think of fields: Ask yourself what's unique about a field among algebraic structures.

Addition and multiplication? Nope,all rings have these operations and most of them aren't fields.

Commutativity? Nope,there are plenty of commutative rings that aren't fields.The integers under addition and multiplication come to mind.

It helps to think about the standard number systems-$\mathbb N,\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$-and think about which ones are fields. Why are $\mathbb Q,\mathbb R$ and $\mathbb C$ fields and $\mathbb N$ and $\mathbb Z$ not fields?

Well,consider that among those systems,the "smallest" one that qualifies as a field is $\mathbb Q$. And what's different about $\mathbb Q$ from $\mathbb Z$? Think about how we "build" the rationals from the integers-it's implicit in the definition of $\mathbb Q$:

$\mathbb Q$ = { q| q =$\frac{a}{b}$ where a,b$\in \mathbb Z$ and $b\neq 0$}

We create the rationals from the integers by the operation of division-or more precisely from an algebraic standpoint, adding a unique nonzero multiplicative inverse to each rational number.More generally,for every element $x\in F$ where F is a field, there exists an element $y\neq 0 \in F$ such that xy =1. As you recall from high school algebra, division by zero isn't allowed for nonzero rational or real numbers.

It's easy to say "big deal" when you're first learning algebra and assume nonzero multiplicative inverses are more common in rings then they actually are.The number systems are rather unique in this regard-there are lots of rings that have zero divisors that are not 0-indeed, they don't even have to be unique.For example, consider the quotient ring $\mathbb{Z}/4\mathbb{Z}$. Consider the residue class $\bar 2$. Then this entire residue class is a zero divisor since $\bar 2 \times \bar 2 = \bar 4 = \bar 0$! So this is not a field.

Here's another familiar example: Consider the ring of 2 x 2 matrices over any nontrivial ring.There there are lots of zero divisors in this ring: $$\begin{pmatrix}1&1\\2&2\end{pmatrix}\begin{pmatrix}1&1\\-1&-1\end{pmatrix}=\begin{pmatrix}-2&1\\-2&1\end{pmatrix}\begin{pmatrix}1&1\\2&2\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix} \begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix} =\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix} =\begin{pmatrix}0&0\\0&0\end{pmatrix}$$.

So this is clearly not a field either.

A field is a special case of division rings i.e. rings in which a unique nonzero multiplicative inverse exists for each element. But notice we didn't specify that for a general division ring, the ring has to be commutative. What's unique about fields is that they are commutative division rings.

"Wait,so what? You've shown us commutative rings that aren't fields!"

Yes,but those rings weren't division rings. There are division rings that are not commutative and therefore not fields.The classic example is the quaternions.

In fact, division rings aren't even required to be associative in their multiplication. The classic example here is the fascinating octonions and their generalizations, the Cayley algebras.

These examples and the discussion above should help you with your intuition regarding fields. Basically, a field is any structure that lets you do division in the usual high school algebra way-and it turns out fairly strong conditions have to be applied to get those kinds of structures. There are plenty of commutative rings that are not division rings and there are plenty of division rings that are not commutative-or even associative.

So be careful and remember-never assume more then you're given in a definition about the structure you're studying.

• Boy,I'm so glad I spent an hour writing this up........lol – Mathemagician1234 Jul 10 '16 at 7:58