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I know how to construct $\mathbb{GF}(2^m)$ from $\mathbb{GF}(2)$: First we find a primitive polynomial of degree $m$ over $\mathbb{GF}(2)$ then assuming $\alpha$ as one of it's roots, consecutive powers of $\alpha$ will be elements of $\mathbb{GF}(2^m)$.

How about constructing $\mathbb{GF}(4^2)$ from $\mathbb{GF}(4)$?

What's the procedure?

Mahdi Khosravi
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1 Answers1

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Let the elements of $GF(4)$ be $\{0,1,\alpha,\alpha^2=1+\alpha\}$. We see that the polynomial $p(x)=x^2+x$ takes the value $0$ at $x=0$ and $x=1$, but $p(\alpha)=p(\alpha+1)=1$. Therefore the polynomial $f(x)=x^2+x+\alpha$ has no zeros in $GF(4)$. Thus $f(x)$, being quadratic, is irreducible in $GF(4)[x]$.

Thus the general recipe of construction works. We get $$ GF(16)=GF(4)[x]/\langle f(x)\rangle. $$

Let $\beta\in GF(16)$ be one of the zeros of $f(x)$ in $GF(16)$. As always, $\{1,\alpha\}$ is then a basis of $GF(16)$ over $GF(4)$. Thus $$ GF(16)=\{a+b\beta\mid a,b\in GF(4)\}. $$

Jyrki Lahtonen
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  • @Mahdi: The irreducible polynomial that I use in the answer is also the minimal polynomial of a primitive element of $GF(16)$. I wrote this Q&A pair to have such calculation at hand. – Jyrki Lahtonen Jan 05 '14 at 14:27
  • Not sure if you are still around, but I do not understand the last steps. I get that $\beta$ should be a root for $f(x)$ and I believe that the your basis is chosen because it can generate all elements in $GF(4)$. But then how do you jump to your conclusion about what $GF(16)$ is? – Bram Vanbilsen Dec 29 '19 at 18:42