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I've begun a solution by finding the orders of the elements and finding the relations between them hoping to come to some contradiction.

However I've come across this question, whcih seems to suggest there is a field of order 4? However, I haven't yet covered field extensions. Without field extensions is there no field of order 4?

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Alex
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    There is always a field of order $4$, whether or not you have covered the methods needed to construct it. – Tobias Kildetoft Jan 31 '17 at 14:34
  • Not understanding the question. Yes, the field of order $4$ is an extension of the field of order $2$. Is that what you are asking? – lulu Jan 31 '17 at 14:35
  • The first answer in the linked question tells you how you can find such a field. The second answer gives a general method to construct fields of certain orders. – Mathematician 42 Jan 31 '17 at 14:36
  • To consider fields and field extensions "without field extensions" doesn't make sense. So I don't know what you mean by asking "Without field extensions is there no field of order $4$?" – Dietrich Burde Jan 31 '17 at 14:36
  • Descartes said: "Je pense, donc je suis". In English: "I think, therefore I am". So if he hasn't thought about field extensions before, they don't exist (in his mind), nor the implications of field extensions. So philosophically, they do not exist until you think about them... I think I can prove about anything philosophically :| – Mathematician 42 Jan 31 '17 at 14:42

2 Answers2

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Consider the set $F=\{0,1,a,a+1\}$. Define in it the operations $+$ and $\times$ in the obvious way with the relations $$x+x=0\;\forall x\in F$$ and $$a^2=a+1$$ and you have the field.

ajotatxe
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There always exists a field of order $p^n$. Consider the roots of the equation $x^{p^n} -x=0$. The roots of the equation form a field of order $p^n$. You can clearly see that it has $p^n$ roots from fundamental theorem.and all are distinct.

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