$\exists a \in \mathbb{F}_{11}$ such that $\mathbb{F}_{11}[x]/\langle x^5-a\rangle$ is a field 
Prove that there exists an element $a \in \mathbb{F}_{11}$ such that the quotient ring $\mathbb{F}_{11}[x]/\langle x^5-a\rangle$ is a field.

I wrote that it is equivalent to showing that there is an $a$ such that $x^5-a$ is irreducible over $\mathbb{F}_{11}$, but was marked wrong. Why? And how can I even approach it, then?
Edit To clarify, I had no pretention to actually solving it. I only wrote that it is equivalent, and the lecturer wrote that it is not.
 A: Suppose $x^5-a\in\mathbb{F}_{11}[x]$ factors nontrivially as $x^5-a=fg$
for some $f,g\in\mathbb{F}_{11}[x]$. By "nontrivially", we mean that $\deg(f),\deg(g)>0$. Note that $\deg(f)+\deg(g)=5$, so WLOG, the only possibilities are
$$\deg(f)=1,\;\deg(g)=4\qquad \text{or}\qquad\deg(f)=2,\;\deg(g)=3.$$
If $x^5-a$ has a factor of degree $1$, then it has a root in $\mathbb{F}_{11}$. However, there are many values of $a\in\mathbb{F}_{11}$ for which $x^5-a$ has no roots in $\mathbb{F}_{11}$. In particular, $b^5\in\{0,1,10\}$ for any $b\in\mathbb{F}_{11}$, so that  $x^5-a$ has no roots in $\mathbb{F}_{11}$ when $a\notin\{0,1,10\}$. So, now let's only consider $a\notin\{0,1,10\}$.
If $x^5-a$ has an irreducible factor of degree $2$, then it has a root in $\mathbb{F}_{11^2}$. Let's say that $\gamma\in\mathbb{F}_{11^2}$ is a root of $x^5-a$, so that $\gamma^5-a=0$ in the field $\mathbb{F}_{11^2}$. Having required $a\notin\{0,1,10\}$, note that the order of $a$ as an element of the multiplicative group $\mathbb{F}_{11}^\times$ is either $5$ or $10$, so that the order of $\gamma$ as an element of the multiplicative group $\mathbb{F}_{11^2}^\times$ is either $25$ or $50$. However, neither of these numbers divide the order $|\mathbb{F}_{11^2}^\times|=120$, so this is a contradiction.
P.S. Note that, in the ring $\mathbb{F}_{11}[x]$, we have $$\begin{align*}
x^5&=x\cdot x\cdot x\cdot x\cdot x\\\\
x^5-1&=(x-1)(x-3) (x-4) (x-5) (x-9)\\\\
x^5-10&=(x-2)(x-6) (x-7) (x-8) (x-10)\\\\
\end{align*}$$
