We have $k_1:= \mathbb F_7(\alpha)$ and $k_2 := \mathbb F_7(\beta)$ where $\alpha^2 = 3$ and $\beta^2 = -1$ in $\mathbb F_7$. I have to show that these two are isomorphic.

Let $\phi:k_1 \rightarrow k_2$ be a homomorphism which preserves $1 \in k_1$. Then

$$\phi(\alpha^2)= \phi(3) = 3 = \phi(\alpha)^2$$


$$\phi(\alpha) = x + y\beta\;,\;\;x,y \in \mathbb F_7$$


$$(x+y\beta)^2 = \phi(\alpha)^2 = x^2 + 2xy\beta +y ^2 \beta^2 = x^2 +2xy\beta -y^2=3$$

So $x$ or $y$ must be $0$. But $y$ can't be zero because $3$ has no root in $\mathbb F_7$. So $x = 0$ such that

$$-y^2 = 3 \rightarrow y \in \{-2, 2\}$$

Is the function $\phi$ with $\phi(x) = x\;$ for $\;x \in \mathbb F_7\;$ and $\;\phi(\alpha) = 2\beta\;$ then an isomorphism ?

  • $\begingroup$ I assume you are not allowed to use that any two finite fields of the same order are isomorphic? $\endgroup$ – Tobias Kildetoft May 29 '13 at 17:36
  • $\begingroup$ Nope. I have to construct an explicit function $\phi$. $\endgroup$ – user42761 May 29 '13 at 17:37
  • $\begingroup$ Yes. The same approach works for general $p$ instead of $7$ as follows. If $a$ and $b$ are two quadratic non-residues modulo $p$, then their ratio $a/b$ is a quadratic residue. Therefore the exists an element $c\in \mathbb{F}_p$ such that $a=bc^2$. Consequently $\sqrt{a}=\pm c\sqrt{b}$, and the field extensions $\mathbb{F}_p[\sqrt{a}]$ and $\mathbb{F}_p[\sqrt{b}]$ are isomorphic. You can also deduce the same from the uniqueness of the field of $p^2$ elements. $\endgroup$ – Jyrki Lahtonen May 30 '13 at 11:30

Well, you have a candidate; why don't you check whether $\phi((a+b\alpha)(c+d\alpha)) = \phi(a+b\alpha)\phi(c+d\alpha)$, for all $a,b,c,d \in \mathbb{F}_7$? For sums it's trivial, as you already define it as a linear map over the base field.

  • $\begingroup$ Thanks. This was indeed the only thing missing. The rest follows easily :) I have chosen $\phi(\alpha) = 5\beta = -2 \beta$. I did not check if $2\beta$ also works. $\endgroup$ – user42761 May 29 '13 at 17:48

Hint: $ $ mod $\,7\!:\ \alpha^2 \equiv 3 \equiv -4 = 4\beta^2\! = (2\beta)^2,\,$ so check $\,\alpha \to 2\beta\,$ preserves sums and products.


An idea: since

$$k_1=\Bbb F_7[x]/\langle\;x^2-3\;\rangle\;,\;\;k_2=\Bbb F_7[x]/\langle\;x^2+1\;\rangle$$

Now, in both cases there exists a primitive element, i.e.: an element that generates the multiplicative group $\,k_i^*\;$ , so if you find both elements in both fields you shall be done...

For example: let $\,\alpha\in\overline{\Bbb F_7}\;$ be a root of $\,x^2-3\in\Bbb F_7[x]\,$ , then in $\,k_1\,$ we have $$\alpha^2=3\implies k_1:=\{p(\alpha)\;;\;p(x)\in\Bbb F_7[x]\;,\;\;\deg p\le 1\}$$

For example, we have that $\,\alpha +5\;,\;3\alpha\;,\;2\alpha -5\,$ , etc. are element in $\,k_1\,$ .

Since $\,|k_1^*|=48\;$ ,every non-zero element in the field has multiplicative order a divisor of $\,48\,$ , so we can try the following: pick (nd remember: we work modulo $\,7\,$ all the time!)

$$\begin{align*}r:&=\alpha+1\\ r^2&=\alpha^2+2\alpha+1=3+2\alpha+1=2\alpha+4\\ r^4&=4\alpha^2+2\alpha+2=2\alpha\\ r^6&=(2\alpha)^3=\alpha(3)=3\alpha\\ r^{12}&=(3\alpha)^2=2\cdot 3=-1\end{align*}$$

and we've found that $\;r\;$ is not a generator of $\,k_1^*\,$ (why? check what happens with $\,r^{24}\,$ ...) . Try other elements until you find an generator and then try to do the same with $\,k_2\,$ and then you'll have an easy choice for your isomorphism...


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