Polynomial subalgebras 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".]
 A: The question was answered by Mohan in a comment. Here is a copy-and-paste:

Not in general. The simplest example would be to take an abelian variety and remove an ample irreducible divisor. The coordinate ring of the affine open subset has both the properties you mention but any map to a polynomial ring will have image just $K$.

(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.
