# Number of elements of a field

Consider a field $F_{16} = F_{2}(a)$ where minimal polynomial of element $a$ is $m_{a,F_{2}} = a^4+a+1$. How many elements are in the field: $$F_{2}(a^2+a)$$

My idea was to find minimal polynomial of $a^2+a$ in $F_{16}$ which I understand as $F_2/(a^4+a+1)$ - meaning operations mod $a^4+a+1$. That gave me $m_{a^2+a,F_{16}} = x^2+x+1$. And than use relation between degree of a minimal polynomial and dimension of an extension of a fild.

But that does not seem right. Any ideas how to start?

• You can find related things in a recent question : (math.stackexchange.com/q/2297568) – Jean Marie Jun 9 '17 at 9:24
• Note that we clearly have $F_2(a^2+a)\subseteq F_{16}=F_2(a)$, so the answer is at most $16$. – Arthur Jun 9 '17 at 9:27
• The minimal polynomial of $a^2+a$ over $F_{16}$ is $x-(a^2+a)$, which is of first degree. – Arthur Jun 9 '17 at 11:08

Step one: Don't write the minimal polynomial of $a^2+a$ in terms of $a$. The minimal polynomial of $a^2+a$ is $x^2+x+1$, as we have that $$(a^2+a)^2 + (a^2+a)+1 = 0.$$ If you use $a$ as the variable here, things get really confusing. Next, this is the minimal polynomial over $\mathbb{F}_2$, not over $\mathbb{F}_{16}$. As this is a polynomial of degree two, you should get some ideas on how big the field is.
• Why is it minimal polynomial over $F_2$? Because $(a^2+a)^2 + (a^2+a) + 1 = a^4+2a^3+a^2+a^2+1=a^4+a+1$, and that is not zero in $F_2$ (the residual after division by $m_a,F_2$ is). – Rosta655321 Jun 9 '17 at 9:42
• @Rosta655321: I'm sorry, but $a^4+a+1$ is $0$, by hypothesis. The $0$ element is the same in $\mathbf F_2$ and in $\mathbf F_{16}$. – Bernard Jun 9 '17 at 10:21