In the set up for the number field sieve, to factor some large $N$, one first takes a monic irreducible $f(x)\in\mathbb{Z}[X]$ and finds some nonzero integer $m$ such that $f(m)\equiv 0\pmod{N}$ in $\mathbb{Z}$. Next, one picks a root $\beta$ of $f(X)$ in an extension field, so that we can take $\mathbb{Z}[\beta]:=\mathbb{Z}[X]/(f(X))$.
Apparently, since $f(m)\equiv 0\pmod{N}$ in $\mathbb{Z}$, then $m\equiv\beta\pmod{N}$ in $\mathbb{Z}[\beta]$. I don't understand, at most I see that $$ f(m)\equiv 0=f(\beta)\pmod{N} $$
in $\mathbb{Z}[\beta]$, but I don't see how you can reduce to seeing $m$ and $\beta$ themselves are congruent.
Edit: I think I have the following idea: There is a map $\mathbb{Z}[X]\to\mathbb{Z}/N\mathbb{Z}$ sending $X\mapsto m+N\mathbb{Z}$. Since $f(X)$ is in the kernel, this descends to a map $$ \mathbb{Z}[\beta]=\mathbb{Z}[X]/(f(X))\to\mathbb{Z}/N\mathbb{Z} $$
Since $\beta=X+(f(X))$, $\beta\mapsto m+N\mathbb{Z}$. Then I think $\beta$ and $m$ both map to $m+N\mathbb{Z}$, so $m-\beta$ is in the kernel. But since $f$ is irreducible, uniqueness of coefficients in the expressions in $\mathbb{Z}[\beta]$ mean that the kernel is precisely expressions with coefficients divisible by $N$, so $m-\beta=Ng(\beta)$ for some $g(\beta)\in\mathbb{Z}[\beta]$. That just means $m\equiv\beta\pmod{N}$ in $\mathbb{Z}[\beta]$.