Let $K$ be a field and $A$ a finitely generated $K$-algebra such that
$\bullet\ A$ is a domain (i.e. zero is the only zero divisor, and $A$ is not the zero $K$-algebra),
$\bullet$ the units of $A$ are in $K$.
Is there necessarily an injective morphism $A\to K[x_1,\ldots,x_n]$, where $n$ is a nonnegative integer and the $x_i$ are indeterminates?
[In this post "$K$-algebra" means "commutative $K$-algebra with one".]
The question was answered by Mohan in a comment. Here is a copy-and-paste:
(As I said in a comment, I don't know enough algebraic geometry to understand this example, but I trust Mohan.)
EDIT. In a comment user26857 gave a hint for an explicit example. Here is an attempt to follow this hint. (user26857 is not responsible for the possible errors in this edit.)
Let $K$ be a field of characteristic $\ne2$, let $X$ and $Y$ be indeterminates, and set $$ A:=K[X,Y]/(Y^2-X^3+X)=K[x,y], $$ where $x$ and $y$ are the images of $X$ and $Y$, so that we have $$ y^2=x^3-x=x(x-1)(x+1) $$ and $K[x]\simeq K[X]$.
Claim 1
(1a) $A$ is a Dedekind domain (in particular the Krull dimension of $A$ is one),
(1b) the units of $A$ are in $K$,
(1c) $A$ is not a unique factorization domain (more precisely, the elements $x,x-1,x+1$ and $y$ are irreducible).
In view of Theorem 1 in
Eakin P. A note on finite dimensional subrings of polynomial rings. Proceedings of the American Mathematical Society. 1972;31(1):75-80, Google Scholar link,
this will imply that $A$ is not a subalgebra of a polynomial algebra over $K$.
Claim 2 $A$ is the integral closure of $K[x]$ in $K(x,y)$.
This will imply (1a).
It is clear that $A$ is integral over $K[x]$. Let us prove the converse, namely that any element of $K(x,y)$ which is integral over $K[x]$ is in $A$.
Let $u$ be in $K(x,y)$. Then $u=u_1+yu_2$ with $u_i\in K(x)$, and this expression is unique.
In this notation, set $\overline u:=u_1-yu_2$. Then $u\mapsto\overline u$ is an order two automorphism of $K(x,y)$ whose fixed field is $K(x)$.
Let $u=u_1+yu_2\in K(x,y)$ be integral over $K[x]$. We must show $u_i\in K[x]$.
As $u_1=(u+\overline u)/2$ is integral over $K[x]$, which is an integrally closed domain, we have $u_1\in K[x]$.
As $u\overline u=u_1^2-(x^3-x)u_2^2$, the above argument shows $(x^3-x)u_2^2\in K[x]$. Since $x^3-x$ is square free, this implies $u_2\in K[x]$.
This proves Claim 2, and thus (1a).
Define $N:A\to K[x]$ by $N(a):=a\overline a=a_1^2-(x^3-x)a_2^2$. (Recall that $a_i\in K[x]$.) Then $N$ is a morphism of multiplicative monoids.
If $a$ is a unit of $A$ then $N(a)$ is a nonzero constant. This is easily seen to imply (look at the degree of $N(a)$) that $a$ is itself a constant. This proves (1b).
Finally, to prove (1c) it suffices to observe that if $0\ne a\in A$ is not a unit, then the degree of $N(a)$ is at least two, so that, if $a\ne0$ is the product of two non-units, then the degree of $N(a)$ is at least four.