The polynomial $x^{242}=1$ has exactly $242 $ solutions in $K$ True/false

Question

Let $K$ be a field of order $243$ and let $F$ be a subfield of $K $ of order $3$.Pick the correct statements from below

$1.$ There exist $\alpha \in K$ such that $K=F(\alpha)$

$2. $The polynomial $x^{242}=1$ has exactly $242 $ solutions in $K$

$3.$The polynomial $x^{26}=1$ has exactly $26$ roots in $K$

$4.$let $f(x) \in F(x)$ be an irreducible polynomial of degree $5$.Then $f(x)$ has a root in $K$

My attempt

$1.$ True because We know that $K^*$ is cyclic, choose $\alpha\not=0$ such that $K^*=<\alpha>$. Then $K=F(\alpha)$.

$2.$Iam confused

$3.$ False because we have $|K^*|=242=2\cdot 11^2$ hence $x^{26}=1\iff x^2=1\iff x=\pm 1$.

$4.$ True. from isomorphism theorem we have $g:F[x]/(f(x))\rightarrow K \implies g(x) \in K$.

How to solve option $2$

Answer 1

The answer to 2. is TRUE. That $K$ is a field with $243$ elements gives $K\setminus \{0\}$ having $242$ elements. Now $K^*= K\setminus \{0\}$ is a group under multiplication, with $242$ elements. [It's actually cyclic, but to show that the answer is true, all you need is that the equation $|K^*|=242$.] So for each of the $242$ elements $\alpha \in K^*$, it follows that $\alpha^{|K^*|} =$ $\alpha^{242}$ must be the multiplicative identity $1$. So $\alpha^{242}=1$ for each of the $242$ elements $\alpha \in K^*$, or equivalently, $\alpha^{242}-1=0$ for each of the $242$ elements $\alpha \in K^*$. Thus the polynomial $x^{242}-1$ has $242$ roots in $K^*$, namely every $\alpha \in K^*$.

Answer 2

Remember that a polynomial can be written as a product of the difference between x and its roots: $$P_n(x)=a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)...=a_n(x-\frac{r_1}{a_n})\prod^{n}_{r_i=r_2}(x-r_i)$$ Where $n$ is the degree of the polynomial and $a_n$ is the first term. This is true since substituting a root for $x$ would make one of the factors to be zero, making the whole polynomial equal to zero. Now there are $n$ factors here, so there are at most $n$ roots to an $n$th degree polynomial.

Note: by the fundamental theorem of algebra, $x^{242}$ has $242$ (not necessarily distinct) roots.