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Can someone explain why $a-b \in F$ means that $a-b = 0$?

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"$a-b$ must be a multiple of the polynomial $p(x)$" means $a-b = q(x) p(x)$ for some $q(x) \in F[x]$. If $q(x) \neq 0$ then deg $p \geq 1$ implies deg $(q p) \geq 1$. But if $a-b \in F$, then deg $pq = 0$, so $q(x)=0$ and hence $a-b=0.$

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    No, it means $\,a-b = p(x)q(x)\,$ for some $\,q(x)\in F[x].\ $ But the degree argument still works, – 2017-02-01
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    @BillDubuque Sorry, you're right. Fixed. – 2017-02-01
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    Why $\deg(pq)=0$ implies that $q(x)=0$? – 2017-02-01
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    If $\text{deg} \thinspace qp = 0,$ then $pq$ is constant. So, both $p$ and $q$ are constant, or either $p(x)=0$ or $q(x)=0$. Since $\text{deg} \thinspace p \geq 1$, then $p$ is non constant and non-zero, hence $q(x)=0.$ – 2017-02-01
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    But if $q(x)=0$, then $pq=0$ and $\deg(0)$ is in general undefined. See here: http://mathworld.wolfram.com/ZeroPolynomial.html. – 2017-02-01
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Note that $a$ and $b$ are elements of the field $F$, so $a-b\in F$ and thus $a-b$ is a constant polynomial in $F[x]$. Now, let's suppose that $a-b\neq 0$, because $p(x)\mid a-b$ there is some polynomial $q(x)\in F[x]$ such that $a-b=p(x)q(x)$ and hence $$\deg(a-b)=0=\deg(pq)=\deg(p)+\deg(q)\ge 1,$$ which is an absurd. Therefore $a-b=0$ (every non-zero polynomial divides the zero polynomial).

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    Thanks your explanation makes sense, I was going for a less rigorous argument along the lines of, "this would mean that $p(x) = (a-b)/q(x)$, which is not a polynomial." Do I have to define a polynomial for this argument to work? – 2017-02-02
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    @Hawk you have to use the fact that $p(x)\mid a-b$ to define $q(x)$, that's all. Besides, the idea to use the degree is a very useful trick that you'll find other times in field theory. – 2017-02-02