Bug: "\fbox" is not correctly displayed in my browser if there is a "\tag" inside "\fbox"

Question

Problem

Some formulas I write in \fbox are not displayed strictly for me. The box around the equation has shifted to the left. For example, the equation $$ \fbox{$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right] \tag{4}\\ \end{align*} $} $$ becomes this / looks like this:

(this formula is from my answer here)

It's obvious that this shouldn't be the case... This always occurs when I write \tag inside \fbox. Other examples are e.g.: $$ \begin{align*} \fbox{$ a + b = c \tag{1}$}\\ \fbox{$ i^{2} = -1 \tag{2} $}\\ \fbox{$ i \in \mathbb{C} \tag{3} $}\\ \end{align*} $$

$$ \begin{align*} \fbox{$ \mathscr{B}\left( H \right)\\ q = w + x \cdot i + y \cdot j + z \cdot k \in \mathbb{H}\\ i^{2} = j^{2} = k^{2} = i \cdot j \cdot k = -1 \tag{5} $}\\ \end{align*} $$

Browser

I'm using "Google Chrome" with "Version $112.0.5615.138$ (Offizieller Build) ($64$-Bit)" aka Google Chrome: version $112.0.5615.138$ (official build) ($64$-bit). The only extensions I use is an adblocker and an anti-virus program and the problem isn't with these, because it still shows up erroneously when I disable both.

Answer 1

Taking the tag out of the fbox works for me: $$ \fbox{$ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \end{align*} $}\tag{4} $$

\fbox{$

\begin{align*}

\operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\

\end{align*}

$}\tag{4}

$$ \begin{align*} \fbox{$ \mathscr{B}\left( H \right)\\ q = w + x \cdot i + y \cdot j + z \cdot k \in \mathbb{H}\\ i^{2} = j^{2} = k^{2} = i \cdot j \cdot k = -1$}\\ \end{align*}\tag{5} $$

\begin{align*}

\fbox{$ \mathscr{B}\left( H \right)\\ q = w + x \cdot i + y \cdot j + z \cdot k \in \mathbb{H}\\ i^{2} = j^{2} = k^{2} = i \cdot j \cdot k = -1$}\\

\end{align*}\tag{5}