Although I agree that there is probably no useful + nontrivial criterion, here's a simple one:

Proposition: Let $R$ be a finitely generated domain over $k$. Then $R$ is a polynomial ring iff some set of $k$-algebra generators of $R$ has size $\operatorname{tr.deg}_k R$.

In fact, if any set of $k$-algebra generators has the "right size" ($= \operatorname{tr.deg}_k R$), then they form a transcendence basis over $k$, and there are no additional algebraic relations in $R$.

The proof is a standard trick in dimension theory: if $a_1, \ldots, a_n$ is a set of $k$-algebra generators, then the assumption implies $n = \dim R$. The map $k[x_1, \ldots, x_n] \twoheadrightarrow R$ sending $x_i \mapsto a_i$ is a surjection between domains of the same dimension, hence is an isomorphism.

To relate this to the defining ideal $I$ of $R$ (w.r.t. some given presentation $R \cong k[y_1, \ldots, y_m]/I$), one can rephrase the above by saying that $R$ is a polynomial ring iff $I$ is prime and $R$ can be generated as a $k$-algebra by $m - \operatorname{ht}(I)$ many elements.

Although I agree that there is probably no useful + nontrivial criterion, here's a simple one:

Proposition: Let $R$ be a finitely generated domain over $k$. Then $R$ is a polynomial ring over $k$ iff some (equivalently, every minimal) set of $k$-algebra generators of $R$ has size $\operatorname{tr.deg}_k R$.

In fact, if any set of $k$-algebra generators has the "right size" ($= \operatorname{tr.deg}_k R$), then they form a transcendence basis over $k$, and there are no additional algebraic relations in $R$.

The proof is a standard trick in dimension theory: if $a_1, \ldots, a_n$ is a set of $k$-algebra generators with $n = \operatorname{tr.deg}_k R$, then by Noether normalization, $n = \dim R$. The map $k[x_1, \ldots, x_n] \twoheadrightarrow R$ sending $x_i \mapsto a_i$ is a surjection between domains of the same dimension, hence is an isomorphism.

To relate this to the defining ideal $I$ of $R$ (w.r.t. some given presentation $R \cong k[y_1, \ldots, y_m]/I$), one can rephrase the above by saying that $R$ is a polynomial ring iff $I$ is prime and $R$ can be generated as a $k$-algebra by $m - \operatorname{ht}(I)$ many elements.

Although I agree that there is probably no useful + nontrivial criterion, here's a simple one:

Proposition: Let $R$ be a finitely generated domain over $k$. Then $R$ is a polynomial ring iff some set of $k$-algebra generators of $R$ has size $\operatorname{tr.deg}_k R$.

In fact, if any set of $k$-algebra generators has the "right size" ($= \operatorname{tr.deg}_k R$), then they form a transcendence basis over $k$, and there are no additional algebraic relations in $R$.

The proof is a standard trick in dimension theory: if $a_1, \ldots, a_n$ is a set of $k$-algebra generators, then the assumption implies $n = \dim R$. The map $k[x_1, \ldots, x_n] \twoheadrightarrow R$ sending $x_i \mapsto a_i$ is a surjection between domains of the same dimension, hence is an isomorphism.

To relate this to the defining ideal $I$ of $R$ (w.r.t. some given presentation $R \cong k[y_1, \ldots, y_m]/I$), one can rephrase the above by saying that $R$ is a polynomial ring if $I$ is prime, and $R$ can be generated as a $k$-algebra by $m - \operatorname{ht}(I)$ many elements.

Although I agree that there is probably no useful + nontrivial criterion, here's a simple one:

Proposition: Let $R$ be a finitely generated domain over $k$. Then $R$ is a polynomial ring iff some set of $k$-algebra generators of $R$ has size $\operatorname{tr.deg}_k R$.

In fact, if any set of $k$-algebra generators has the "right size" ($= \operatorname{tr.deg}_k R$), then they form a transcendence basis over $k$, and there are no additional algebraic relations in $R$.

The proof is a standard trick in dimension theory: if $a_1, \ldots, a_n$ is a set of $k$-algebra generators, then the assumption implies $n = \dim R$. The map $k[x_1, \ldots, x_n] \twoheadrightarrow R$ sending $x_i \mapsto a_i$ is a surjection between domains of the same dimension, hence is an isomorphism.

To relate this to the defining ideal $I$ of $R$ (w.r.t. some given presentation $R \cong k[y_1, \ldots, y_m]/I$), one can rephrase the above by saying that $R$ is a polynomial ring iff $I$ is prime and $R$ can be generated as a $k$-algebra by $m - \operatorname{ht}(I)$ many elements.