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[Edit] My answer is unnecessarily complicated regarding that of Manny.

The answer is yes. First suppose $A$ is noetherian. You can find $t_1,\dots, t_n\in A$ algebraically independent over $k$. Then you get a dominant morphism $\operatorname{Spec}A\to \mathbb A^n_k$ of integral noetherian schemes which is generically finite. This then implies that $\dim A\le \dim \mathbb A^n_k$ (see this answer to related question).

If $A$ is not noetherian, write $A$ as an inductive limit of noetherian subdomains $A_i$ with $\mathrm{Frac}(A_i)=\mathrm{Frac}(A)$. Let $\mathfrak p_0\subset \cdots \subset \mathfrak p_r$ be a chain of prime ideals in $A$. Then for some $A_i$, the $\mathfrak p_j\cap A_i$ is a chain of prime ideals (take $A_i$ big enough to contain an element of $\mathfrak p_j\setminus \mathfrak p_{j-1}$ for all $j\le r$). Then by the above $r\le \dim A_i\le n$. Hence $\dim A\le n$.

[Edit] My answer is unnecessarily complicated regarding that of Manny.

The answer is yes. First suppose $A$ is noetherian. You can find $t_1,\dots, t_n\in A$ algebraically independent over $k$. Then you get a dominant morphism $\operatorname{Spec}A\to \mathbb A^n_k$ of integral noetherian schemes which is generically finite. This then implies that $\dim A\le \dim \mathbb A^n_k$ (see this answer to related question).

If $A$ is not noetherian, write $A$ as an inductive limit of noetherian subdomains $A_i$ with $\mathrm{Frac}(A_i)=\mathrm{Frac}(A)$. Let $\mathfrak p_0\subset \cdots \subset \mathfrak p_r$ be a chain of prime ideals in $A$. Then for some $A_i$, the $\mathfrak p_j\cap A_i$ is a chain of prime ideals (take $A_i$ big enough to contain an element of $\mathfrak p_j\setminus \mathfrak p_{j-1}$ for all $j\le r$). Then by the above $r\le \dim A_i\le n$. Hence $\dim A\le n$.

[Edit] My answer is unnecessarily complicated regarding that of Manny.

The answer is yes. First suppose $A$ is neotherian. You can find $t_1,\dots, t_n\in A$ algebraically independent over $k$. Then you get a dominant morphism $\mathrm{Spec}A\to \mathbb A^n_k$ of integral noetherian schemes which is generically finite. This then implies that $\dim A\le \dim \mathbb A^n_k$ (see this answer to related question).

If $A$ is not neotherian, write $A$ as an inductive limit of noetherian subdomains $A_i$ with $\mathrm{Frac}(A_i)=\mathrm{Frac}(A)$. Let $\mathfrak p_0\subset ... \subset \mathfrak p_r$ be a chain of prime ideals in $A$. Then for some $A_i$, the $\mathfrak p_j\cap A_i$ is a chain of prime ideals (take $A_i$ big enough to contain an element of $\mathfrak p_j\setminus \mathfrak p_{j-1}$ for all $j\le r$. Then by the above $r\le \dim A_i\le n$. Hence $\dim A\le n$.

[Edit] My answer is unnecessarily complicated regarding that of Manny.

The answer is yes. First suppose $A$ is noetherian. You can find $t_1,\dots, t_n\in A$ algebraically independent over $k$. Then you get a dominant morphism $\operatorname{Spec}A\to \mathbb A^n_k$ of integral noetherian schemes which is generically finite. This then implies that $\dim A\le \dim \mathbb A^n_k$ (see this answer to related question).

If $A$ is not noetherian, write $A$ as an inductive limit of noetherian subdomains $A_i$ with $\mathrm{Frac}(A_i)=\mathrm{Frac}(A)$. Let $\mathfrak p_0\subset \cdots \subset \mathfrak p_r$ be a chain of prime ideals in $A$. Then for some $A_i$, the $\mathfrak p_j\cap A_i$ is a chain of prime ideals (take $A_i$ big enough to contain an element of $\mathfrak p_j\setminus \mathfrak p_{j-1}$ for all $j\le r$). Then by the above $r\le \dim A_i\le n$. Hence $\dim A\le n$.

The answer is yes. First suppose $A$ is neotherian. You can find $t_1,\dots, t_n\in A$ algebraically independent over $k$. Then you get a dominant morphism $\mathrm{Spec}A\to \mathbb A^n_k$ of integral noetherian schemes which is generically finite. This then implies that $\dim A\le \dim \mathbb A^n_k$ (see this answer to related question).

If $A$ is not neotherian, write $A$ as an inductive limit of noetherian subdomains $A_i$ with $\mathrm{Frac}(A_i)=\mathrm{Frac}(A)$. Let $\mathfrak p_0\subset ... \subset \mathfrak p_r$ be a chain of prime ideals in $A$. Then for some $A_i$, the $\mathfrak p_j\cap A_i$ is a chain of prime ideals (take $A_i$ big enough to contain an element of $\mathfrak p_j\setminus \mathfrak p_{j-1}$ for all $j\le r$. Then by the above $r\le \dim A_i\le n$. Hence $\dim A\le n$.

[Edit] My answer is unnecessarily complicated regarding that of Manny.

The answer is yes. First suppose $A$ is neotherian. You can find $t_1,\dots, t_n\in A$ algebraically independent over $k$. Then you get a dominant morphism $\mathrm{Spec}A\to \mathbb A^n_k$ of integral noetherian schemes which is generically finite. This then implies that $\dim A\le \dim \mathbb A^n_k$ (see this answer to related question).

If $A$ is not neotherian, write $A$ as an inductive limit of noetherian subdomains $A_i$ with $\mathrm{Frac}(A_i)=\mathrm{Frac}(A)$. Let $\mathfrak p_0\subset ... \subset \mathfrak p_r$ be a chain of prime ideals in $A$. Then for some $A_i$, the $\mathfrak p_j\cap A_i$ is a chain of prime ideals (take $A_i$ big enough to contain an element of $\mathfrak p_j\setminus \mathfrak p_{j-1}$ for all $j\le r$. Then by the above $r\le \dim A_i\le n$. Hence $\dim A\le n$.