Why doesn't adjoining $\sqrt{3}$ to $\mathbb{F}_{11}$ return $\mathbb{F}_{11}$? I am confused about a particular instance where adjoining an element of a field to itself makes it not equal to itself and I am asking for clarification. I can see the result is true, but I can not see why. We are not introducing any new element and we are not setting any new elements equal to zero.
In the finite field $\mathbb{F}_{11}$ we adjoin $\alpha$ where $\alpha^2 - 3 =0$. Because $(\pm 5)^2 -3 =0$ the two square roots of $3$ are already in $\mathbb{F}_{11}$, so we are either adjoining $5$ or $-5$. We do not know which, although both elements are invertible. However we cannot invert $\alpha +5$ because we don't know if $\alpha +5$ or $\alpha-5 = 0$ so $\mathbb{F}_{11}[\alpha]$ is not a field.
By adjoining an ambiguous element of the field to itself I thought maybe we were setting elements equal to zero, but it's not $5 = -5$ because then $10 = 0$ which makes every element $0$.
Source:  This was an example in Algebra by Artin where Artin says $\mathbb{F}_{11}[\alpha] \simeq \mathbb{F}_{11}[x]/(x^2-3)$ is not a field on page 366.
Sorry for edits
Edit 2: If I am understanding what is being said, $\alpha$ must assume a specific value in $\mathbb{F}_{11}[\alpha]$. So it is $not$ true $\mathbb{F}_{11}[\alpha] \simeq \mathbb{F}_{11}[x]/(x^2-3)$ because the kernel of the evaluation homomorphism $\phi: \mathbb{F}_{11}[x] \rightarrow \mathbb{F}_{11}[\alpha]$ is not the ideal $(x^2-3)$, but one of the ideals $(x-5)$ or $(x+5)$.
I will rewrite a comment here hopefully to clarify.
In this image Artin describes $R'$ as &quot;obtained by adjoining an element $\alpha$ to $\mathbb{F}_{11}$&quot;. A page earlier Artin defined &quot;$R[\alpha] = \text{ring obtained by adjoining} \ \alpha \ \text{to} \ R$&quot;. There are also examples of using the evaluation homomorphism to show results such as $\mathbb{R}[x]/(x^2+1)\simeq \mathbb{Q}$ and $R[x,y] \simeq R[x][y]$.
Artin writes
\begin{equation}
R' = \mathbb{F}_{11}[x]/(x^2-3)
\end{equation}
In the same paragraph he says &quot;...procedure applied to $\mathbb{F}_{11}$ does not yield a field&quot;, and &quot;But we haven't told $\alpha$ whether to be equal to $5$ or $-5$. We've only told that its square is $3$.&quot;
With this wording it sounds like the kernel of $\phi$ is not $(x-5)$ or $(x+5)$, but only $(x^2-3)$. Which again confuses me because then $\mathbb{F}_{11}[\alpha]$ is not a field.
There are two different rings being discussed here:

$\mathbb{F}_{11}[\alpha]$, the ring obtained by adjoining some specific $\alpha\in \overline{\mathbb{F}}_{11}$ with $\alpha^2 = 3$;
$\mathbb{F}_{11}[X] / (X^2 - 3)$.

In the first case, $\mathbb{F}_{11}$ already contains $\alpha$; as you point out, $\alpha = \pm 5\in \mathbb{F}_{11}$. In the latter case, we no longer have a field:
\begin{align*}
\mathbb{F}_{11}[X]/(X^2 - 3) = \mathbb{F}_{11}[X]/(X - 5)\oplus \mathbb{F}_{11}[X]/(X + 5).
\end{align*}
If $f\in \mathbb{F}_{11}[X]$ is an irreducible nonconstant polynomial, then the map $\mathbb{F}_{11}[X]/(f) \to \mathbb{F}_{11}[\alpha]$ given by $X \to \alpha$, where $\alpha$ is a zero of $f$ in $\overline{\mathbb{F}}_{11}$, is an isomorphism; for then $\alpha$ is not a zero of any polynomial of degree less than $\deg f$, and comparing dimensions gives the result. That result doesn't hold without the irreducibility assumption, though.
Your sequence
However we cannot invert $\alpha +5$ because we don't know if $\alpha +5$ or $\alpha-5 = 0$ so $\mathbb{F}_{11}[\alpha]$ is not a field.
is confusing.
Either $\alpha = 5$ and in this case $\alpha+5$ is invertible... And $\mathbb F_{11}[\alpha] = \mathbb F_{11}$ is a field. Or $\alpha = -5$ and in that case $\alpha+ 5=0$ is not invertible. But in that case still $\mathbb F_{11}[\alpha] = \mathbb F_{11}$ is a field.