Question

Show that the Freshman’s Dream may be false if the characteristic p is not prime or if R is noncommutative.

Short answer

${\left(a+b\right)}^{P^n}\ne a^{p^n}+b^{P^n}$

Step-by-step solution

Definition of Freshman Dream

Let p be a prime and R a commutative ring with identity of characteristic p . Then for every $a,b\in R$and every positive integer n such that ${\left(a+b\right)}^{P^n}\ne a^{p^n}+b^{P^n}$.

Showing that (a+b)pn≠apn+bpn  

Freshman Dream may be false if the characteristic p is not prime or if R is noncommutative. Now prove the claim by taking example.

Consider $Z_6$. Here $char{Z}_6\left(p\right)=6$is not prime number. Let $a=1,b=2$then for $n=1$. Take left hand side of ${(a+b)}^{p^n}=a^{pn}+b^{p^n}$.

$$\begin{gathered} {\left(1+2\right)}^{6^1}={\left(1+2\right)}^6 \\ =3^6 \\ =729 \end{gathered}$$

Simplify further:

$$\begin{gathered} 729=729\left(mod6\right) \\ =3 \end{gathered}$$

Now take right hand side of${(a+b)}^{p^n}=a^{p^n}+b^{p^n}$.

$$\begin{gathered} 1^{6^1}+2^{6^1}=1^6+2^6 \\ =1+64 \\ =65 \end{gathered}$$

Simplify further:

$$\begin{gathered} 65=65\left(mod6\right) \\ =5 \end{gathered}$$

This gives right hand side is not equal to left hand side. Therefore in this case Freshman Dream fails.

Now again consider $M\left(Z_2\right)$. Here $M\left(Z_2\right)$is not commutative and$char\left(Z_2\right(p\left)\right)=2$. Here characteristic is a prime.

Let $a=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),b=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$. Noe in this case$n=1,p=2$.

Now take left hand side of ${(a+b)}^{p^n}=a^{pn}+b^{p^n}$.

$$\begin{gathered} {\left(a+b\right)}^{p^n}={\left(\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)+\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\right)}^{2^1} \\ ={\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)}^2 \\ =\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right) \end{gathered}$$

Now take right hand side of${(a+b)}^{p^n}=a^{p^n}+b^{p^n}$.

$$\begin{gathered} a^{p^n}+b^{p^n}={\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)}^{2^1}+{\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)}^{2^1} \\ ={\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)}^2+{\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)}^2 \\ =\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)+\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right) \\ =\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right) \end{gathered}$$

This show that right hand side is not equal to left hand side. Hence Freshman’s Dream fails in both the case. So,${(a+b)}^{p^n}\ne a^{p^n}+b^{p^n}$.