2

Can anybody give some intuition behind a prime field?

So the prime field of $K$ is the field that results when we intersect every subfield of $K$. But if this has characteristic $p$ for some prime, why does any multiple of $p$ in the field equal zero?

To an extent I see that every subfield contains $\mathbb{Z}/p\mathbb{Z}$, but that vision of mine is not very clear and I do not feel comfortable working with it. I do not really even know what it means that every subfield contains $\mathbb{Z}/p\mathbb{Z}$.

Would anybody be able to provide some intuition behind this? Or perhaps some resources that would provide some underpinning theory that would make this fact seem obvious?

Sandstar
  • 430
  • Think about the characteristic $0$ case instead, maybe you have more intuition there. Can you, for instance, see why every subfield of $\Bbb C$ (like $\Bbb R$, or $\Bbb Q(i)$, or more exotic subfields) contains $\Bbb Q$? – Arthur Dec 09 '19 at 11:30
  • Not really to be honest. For example it contains the subfield $2\mathbb{Z}$, but it's not even completely clear to me that that subfield "contains" $\mathbb{Q}$. – Sandstar Dec 09 '19 at 11:36
  • Is $2\Bbb Z$ a field, though? – Arthur Dec 09 '19 at 11:38
  • Oh wow that's embarrassing. It is not. Okay so I think I see that any subfield will then contain $\mathbb{Q}$. But how do I generalise this partial understand to fields of characteristic more than 0? – Sandstar Dec 09 '19 at 11:39
  • Think about why $\Bbb Q$ must be in any subfield of $\Bbb C$. Any subfield must contain $1$, and you must be able to use the standard field operations (addition, subtraction, multiplication and division by non-zero). Can you, just from $1$ and all the field operations, construct any rational number? – Arthur Dec 09 '19 at 11:43
  • Hmm yes you can. And no subfield has characteristic more than 0, so we can make any rational in any subfield. So then I can make any integer less than p in any field that has characteristic p, because once I make p by adding the unit to itself p times it will revert to 0. That is a very helpful idea! – Sandstar Dec 09 '19 at 11:46

2 Answers2

4

I think it is very helpful to understand the "minimal" example of such a field in characteristic $p>0$, namely for $p=2$ the field $\Bbb F_4$ with $4$ elements. It contains the prime field $\Bbb Z/2$ in a very natural way, but is itself of course quite different from $\Bbb Z/4$, which has zero divisors and thus is not a field. Fortunately this site has many good explanations on how to understand $\Bbb F_4$:

Can you construct a field with 4 elements?

Addition and Multiplication in $F_4$

Seeing that $\Bbb F_2[x]/(x^2+x+1)$ is a field

Why $F_2[X]/(X^2+X+1)$ has $4$ elements and what are those?

What is the difference between $Z/(4)$ and the field $F4$?

Why does $(a+1)(a+1)=a$ within a field $F_4$

Dietrich Burde
  • 130,978
  • Oh nice idea. How can I think of $\mathbb{F}_4$ though? It is not a cartesian product of $\mathbb{Z}/2\mathbb{Z}$, and it is not $\mathbb{Z}/4\mathbb{Z}$. – Sandstar Dec 09 '19 at 11:44
  • 1
    How can you think of $\Bbb F_4$ though? Exactly for this question I have added the links for you. Please have a look. – Dietrich Burde Dec 09 '19 at 11:45
2

In my opinion, you should think about the prime field $P$ of $K$ this way: it is the smallest subset of $K$ which is also a field, whose operations are inherited from $K$. Clearly, any subfield must contain the element $0$ and the element $1$. Unlike with rings, there are no subfields with an identity element different from $1$ because all elements are invertible. $1$ belongs to $P$, so $1+1,1+1+1,1+1+1+1$, etc. all belong to $P$. In the case of a field $K$ with non-zero characteristic, $1,2,3,4,\cdots ,p-1$ all belong to $P$. Also, $P$ must be closed under multiplication and have an inverse for every non-zero element. In the case of a field with characteristic $p$ this doesn't make us include any new elements in $P$, since every element in $\{ 1,2,3, \cdots ,p-1\}$ has an inverse, also in $\{ 1,2,3, \cdots ,p-1\}$. But if $\text{char}(K)=0$, then all fractions ($a/b,b\neq 0$) also belong to $P$. Nothing else is needed, so those are all possible prime fields; either $\mathbb{Z}/p\mathbb{Z}$ or $\mathbb{Q}$. The definition of the smallest field is quite abstract and not constructive, and I think that the constructive way provides more intuition.

Martín Gómez
  • 21