Boo. The answer is no. Consider the Markov cluster algebra, whose corresponding matrix is $$\left[\begin{array}{ccc} 0 & -2 & 2 \\ 2 & 0 -2 \\ -2 & 2 & 0 \end{array}\right]$$

For an initial cluster of $\{x_1,x_2,x_3\}$, the mutation relation at 2 is $$ x_2x_2'=x_1^2+x_3^2$$ If the ground field has a square root $i$ of $-1$, then $$ x_2x_2'=(x_1+ix_3)(x_1-ix_3)$$ Thus $x_2$ is not prime.

Boo. The answer is no. Consider the Markov cluster algebra, whose corresponding matrix is $$\left[\begin{array}{ccc} 0 & -2 & 2 \\ 2 & 0 & -2 \\ -2 & 2 & 0 \end{array}\right]$$

For an initial cluster of $\{x_1,x_2,x_3\}$, the mutation relation at 2 is $$ x_2x_2'=x_1^2+x_3^2$$ If the ground field has a square root $i$ of $-1$, then $$ x_2x_2'=(x_1+ix_3)(x_1-ix_3)$$ Thus $x_2$ is not prime.

Boo. The answer is no. Consider the Markov cluster algebra, whose corresponding matrix is $$\left[\begin{array}{ccc} 0 & -2 & 2 \\ 2 & 0 -2 \\ -2 & 2 & 0 \end{array}\right]$$

For an initial cluster of $\{x_1,x_2,x_3\}$, the mutation relation at 2 is $$ x_2x_2'=x_1^2+x_3^2$$ If the ground field has a square root $i$ of $-1$, then $$ x_2x_2'=(x_1+ix_3)(x_1-ix_3)$$ Thus that $x_2$ is not prime.

Boo. The answer is no. Consider the Markov cluster algebra, whose corresponding matrix is $$\left[\begin{array}{ccc} 0 & -2 & 2 \\ 2 & 0 -2 \\ -2 & 2 & 0 \end{array}\right]$$

For an initial cluster of $\{x_1,x_2,x_3\}$, the mutation relation at 2 is $$ x_2x_2'=x_1^2+x_3^2$$ If the ground field has a square root $i$ of $-1$, then $$ x_2x_2'=(x_1+ix_3)(x_1-ix_3)$$ Thus $x_2$ is not prime.