3

Let $F$ be a field with $4$ elements and let $\alpha \in F$ be an element $\neq 0, 1$.

a) Show that $F = \{0,1,\alpha, \alpha^2\}$ and $1 + \alpha + \alpha^2 = 0$:

I know that $F^* = F - \{0\}$ is a group with operation $(\cdot)$, so $(F^*, \cdot)$ is a group of order $3$ with identity element $1$.

Since the order of every element in $F^*$ must divide $3$ we have $\langle \alpha \rangle = \{\alpha, \alpha^2, 1\}$ is the subgroup of $F^*$ generated by $\alpha$, so $\langle \alpha \rangle = F^*$ and $F = F^* \cup\{0\} = \{0, 1,\alpha, \alpha^2\}$.

However I don't know how to show $1 + \alpha + \alpha^2 = 0$. Thus I'm sure I should use the identity $\alpha^3 = 1$.

b) Also I must show $X^2 + \alpha$ is reducible and $X^2 + X + \alpha$ is irreducible in $F[X]$, but this amounts to show that one has a root and the other not (this I've proved).

Shuzheng
  • 5,533

2 Answers2

4

If you multiply $1+\alpha+\alpha^2$ with $\alpha$ you get $\alpha+\alpha^2+1$, i.e. the same number. Which number remains unchanged when multiplied with a number $\ne 1$?

b) In th elight of the previous, you want an $x\in F$ such that $x^2+\alpha=0$, i.e. $x^2=1+\alpha^2$. Guess what is $(1+\alpha)^2$? Alternatively: what is the square of $\alpha^2$ by your cyclic group argument?

For $X^2+X+\alpha$ it suffices to show that plugging in any of the four values results in an expression that is definitely nonzero.

Hagen von Eitzen
  • 374,180
  • Thanks! - I know $0$ does, but then I must show that no other element does not equal the same element when multiplied by an element $\neq 1$. – Shuzheng Jan 06 '14 at 16:48
3

Hints:

  1. If you know that $\alpha^3=1$ and $\alpha\neq1$, then $$0=1-1=\alpha^3-1=(\alpha-1)(\alpha^2+\alpha+1).$$ Which of those factors can make the product zero?
  2. If $\alpha^3=1$, then $\alpha^4=\alpha$. Therefore $$x^2+\alpha=x^2-\alpha=x^2-\alpha^4=x^2-(\alpha^2)^2.$$ Can you see that this is reducible? For part b) see e.g. my earlier answer.
Community
  • 1
Jyrki Lahtonen
  • 133,153
  • Thanks Jyrki. We know $(\alpha -1) \neq 0$ so $\alpha^2 + \alpha + 1$ must be zero. – Shuzheng Jan 06 '14 at 19:49
  • Correct, Nicolas! Well done. – Jyrki Lahtonen Jan 06 '14 at 19:50
  • Jyrki, may I ask you: Suppose we have $\alpha, \lambda \in R$, where $R$ is a commutative ring. Also $\alpha, \lambda \neq 0,1$. Suppose $\lambda \cdot \alpha = \alpha$ - Is this possible with the assumption $\lambda \neq 1$ ? If $R$ is a domain then $\lambda \cdot \alpha = \alpha \Rightarrow \alpha \cdot (\lambda - 1) = 0$ a contradiction, since $\lambda \neq 1 \land \alpha \neq 0$. However what if $R$ is not a domain, but a general ring ? – Shuzheng Jan 06 '14 at 20:04
  • @Nicolas: In the ring $\Bbb{Z}_4$ we have $3\cdot2=2$. – Jyrki Lahtonen Jan 06 '14 at 20:05
  • Ohh so in general we cannot say $\lambda \alpha = \alpha $, where $\lambda \neq 1$ implies $\alpha = 0$. Thank you. – Shuzheng Jan 06 '14 at 20:42