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Let me say first that we battled my son's addictions to opiates and heroin for over 10 years, so while I have little technical knowledge of addiction and withdrawal, I have a lot of first hand experience. He went through rehab 3 times before it took, so I've nursed him through horrible withdrawal 3 times. I say this so that I won't sound so ignorant in the next paragraph. I know that each drug has its own peculiarities.

On March 3rd this year, I celebrated 30 years since my last cigarette. I quit smoking just a few months before my Ph.D. qualifying exams. So I was in the throes of withdrawal during the crucial study time. (My joke is that I started raising hamsters, so I could kill something once a day.) My main way of coping was to walk. Study for a while and then walk. Then Study. Then walk. I also took up yo-yo, so I had something for my hands to do when I got fidgety. Tommy Smothers had just resurrected the fad and I got quite good.

The good news is that the physical withdrawal is relatively short. So each week you should notice that things are a bit better. Now is the time to develop good habits. I wouldn't push hard, but I'd push medium. Force yourself to do just a bit more than is comfortable, so that you learn that it's possible to concentrate, at least a bit, even with withdrawal distractions.

As someone said in the comments, seeking medical help might be useful. Unless things have changed, the medicos first attempt is to give you Prozac or something similar to calm you down a bit. This didn't work for my son, but everyone responds differently.

Hang in there. As John Belushi said, "When things get tough, the tough get things."

Edit: Another trick is to clear off the top of a 4-drawer filing cabinet and use that for your work surface. When I just couldn't "sit down and study", then I stood up to study.

I think this statement is proven exactly as the primitive element theorem is proven. I'll give you the argument in the case $n=2$, courtesy of J. S. Milne's notes on field theory, from which the general case follows.

Let $E=F[a,b]$ with $b$ separable over $F_u$. Let $f$ and $g$ be the minimum polynomials of $a$ and $b$ respectively over $F$. Let $a=a_1, a_2, \ldots ,a_s$ be the roots of $f$ in some field $K$ containing $E$, and let $b=b_1, b_2, \ldots ,b_t$ be the roots of $g$. For $j \neq 1$, we have $b_j \neq b_1$ so $$a_i +Xb_j=a_1 +Xb_1$$ has exactly one solution: $X= \displaystyle \frac{a_j -a_i}{b_1-b_j}$. This means that $a_i + \frac{u_2}{u_1} b_j \neq a + \frac{u_2}{u_1} b$ in $K_u$ unless $i=1=j$.

Now let $\gamma = a+\frac{u_2}{u_1}b$. Then the polynomials $g(X)$ and $f(\gamma-\frac{u_2}{u_1}X)$ have coefficients in $F_u[\gamma]$, and have $b$ as a root: $$g(b)=0, f(\gamma-\frac{u_2}{u_1}b)=f(a)=0.$$

Moreover, $b$ is their only common root, because $\gamma - \frac{u_2}{u_1}b_j \neq a_i$ unless $i=1=j$. Therefore $$\gcd(g(X),f(\gamma-\frac{u_2}{u_1}X))=X-b.$$

Though we've computed the gcd in some field that splits $fg$, it is a simple fact that the gcd of two polynomials has coefficients in the same field as the coefficients of the polynomials. Hence $b \in F_u[\gamma]$, and this implies that $a=\gamma - \frac{u_2}{u_1}b$ lies in $F_u[\gamma]$. Hence, $F_u[a,b]=F_u[\gamma]=F_u[a+\frac{u_2}{u_1}b]=F_u[au_1+bu_2]$.