# Exercise in arithmetic of a finite field

I am in difficult in resolving this exercise in Galois Theory : "in $GF(2^5)$ calculates the product $(1,1,1,0,1)(0,1,0,1,0)$ , generator of $GF(2^5)^*$ ". I don't know how to proceed.. thank you

• What is GF(2^5)? I've never seen this notation. – Sempliner Aug 22 '15 at 8:27
• In Italy we use this notation : p^n is the order ... in this case 2^5 are 32 elements – Alberto Aug 22 '15 at 8:31
• Oh so this is the finite field $\mathbb{F}_{2^5}$? And you are trying to calculate the product of the two field elements $(1,1,1,0,1), (0, 1, 0, 1, 0)$ written as vectors over $\mathbb{F}_2$? – Sempliner Aug 22 '15 at 8:33
• @Sempliner yes, $\mathrm{GF}(2^5)$ and $\mathbb{F}_{2^5}$ are the same thing – Alessandro Codenotti Aug 22 '15 at 8:37
• Looks like you are writing the elements in terms of a monomial basis. The answer depends on the minimal polynomial. I hazard a guess that we are to assume that $t^5+t^2+1=0$, because that is the most common choice. Please confirm or correct and clarify! – Jyrki Lahtonen Aug 22 '15 at 13:31

Hint: Do you know what the multiplication by $(0, 1, 0, 1, 0) = (0, 1, 0, 0, 0) + (0, 0, 0, 1, 0)$ is given as a matrix with coefficients in $\mathbb{F}_2$? From this it should be fairly easy.
I just know that $(0,1,0,1,0)$ represent $t^3+t$ polynomial .. Ps: I m new here, is it possible upload image ? And how to write correctly the mathematical formulas?