multiplication over gf(16)

Question

Can some one show me how to do multiplication over gf(16) step by step

I found this example online, http://userpages.umbc.edu/~rcampbel/Math413Spr05/Notes/12-13_Finite_Fields.html#An_Example. An Example:

$x^4 = x+1$

$x^5 = x^2+x$

$(x^2+x+1) (x^3+x^2+1)$ $= x^5 + 2x^4 + 2x^3 + 2x^2 + x + 1$ $= x^2 + 1$

I know $x^4 = x+1$, because $x^4 mod (x^4 + x + 1)$, but how does $x^5 = x^2+x$?

thanks! I still confused on how to multiply two polynomials, if anyone can give me an example with step. — puppylord, Oct 25 '15 at 17:58
You just have to use the minimal polynomial repeatedly: it is only a modular computation. I'll add an example in a moment. — Bernard, Oct 25 '15 at 18:00
Using the discrete logarithm table (see the middle section of that answer), where I denote by $\gamma$ the coset $x+(x^4+x+1)$, the calculation would go as follows. Reading the table from left to right or right to left as need may be: $$(\gamma^2+\gamma+1)(\gamma^3+\gamma^2+1)=\gamma^{10}\gamma^{13}=\gamma^{23}=\ \gamma^8=\gamma^2+1,$$ The same answer :-) — Jyrki Lahtonen, Oct 25 '15 at 18:45
That log table basically lists all the powers Bernard (+1) was calculating for you. A point is that the powers repeat in cycles of fifteen. They are useful for doing multiplication in fields like this. — Jyrki Lahtonen, Oct 25 '15 at 18:46

score 1 · Accepted Answer · answered Oct 25 '15 at 18:12

1

By distributivity, all you actually In the general casehave to know is how to multiply $1, x,x^2,x^3$ with each other, i.e. express $x^4,x^5,x^6$. In the general case, here is how it goes: \begin{align*} x^4&=\color{red}{x+1} &x^5&=x\cdot x^4=x(x+1)=\color{red}{x^2+x}\\ x^6&=x\cdot x^5=\color{red}{x^3+x^2}&x^7&=x\cdot x^6=x^4+x^3=\color{red}{x^3+x+1} \\ x^8&=(x^4)^2=(x+1)^2=\color{red}{x^2+1}&&\&c. \end{align*}

answered Oct 25 '15 at 18:12

Bernard

179,256

can you also show me how to $x^5 + 2x^4 + 2x^3 + 2x^2 + x + 1 = x^5 + x + 1$? – puppylord Oct 25 '15 at 18:21
$\mathbf F_{16}$ has characteristic $2$! – Bernard Oct 25 '15 at 18:23
still confused :. Maybe I am missing something basic ..so 2 (x^4 +x^3+x^2) = 0 ? – puppylord Oct 25 '15 at 18:28
In a field of characteristic $2$, one has, by definition $2\cdot 1=0$. I used that for the computation of $x^8$. – Bernard Oct 25 '15 at 18:29
sorry I think I am confused on something. I have to think about it. – puppylord Oct 25 '15 at 18:32
is $5x^2 = x^2$, cause 5 mod 2 = 1? – puppylord Oct 25 '15 at 18:43
Yoou didn't say you have a $5x^2$ term, but yes, it's true. As a vector space over $\mathbf F_2$, the only possible coefficients are $0$ and $1$. – Bernard Oct 25 '15 at 18:46

score 0 · Answer 2 · answered Oct 26 '15 at 14:31

A better way to do the multiplication (not necessarily more efficient, but a better way to think about it) is by using remainders of polynomial division. So for example, $x^{5} = x(x^{4}+x+1) + x + 1$, and since $x^{4}+x+1 = 0$, $x^{5} = x+1$. You really are looking at polynomials $\bmod{(x^{4}+x+1)}$.

multiplication over gf(16)

2 Answers2