= Solution
Let $A=A(X)=k[x_1,\ldots,x_m]/I(X)$. <Polynomial> restrictions are <regular functions>, and a restriction is zero precisely when its class in the <coordinate ring> is zero. We prove surjectivity.
Using the usual irreducible-variety convention, view a <regular function> $f$ as an element of the <fraction field> of $A$. On a neighbourhood of every point it has the form $a/b$, where $a,b\in A$ and $b$ does not vanish at that point. Equality on a nonempty open subset of the irreducible variety is equality of rational <functions>. Define
$$
J=\{h\in A:hf\in A\}.
$$
This is an <ideal> and contains a denominator nonzero at each point. Thus its zero locus is empty. The <Hilbert Nullstellensatz> implies $J=A$, so $1\in J$ and $f\in A$. Therefore
$$
\boxed{A(X)=\Gamma(X,\mathcal O_X)}.
$$
The result also holds under a convention allowing reduced reducible affine varieties. One may cover by finitely many principal opens $D(b_i)$, represent $f$ in each localization, and increase powers until the local representatives glue after clearing denominators. Since these opens cover, the <ideal> generated by those powers is the unit <ideal> by the Nullstellensatz; a unit linear combination glues their numerators to one element of $A$. Irreducibility is used only for the shorter fraction-field proof, not for the assertion itself.
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