I don't know how to answer the question as stated, but I believe that by strengthening the question a bit we can see that there does *not* exist a "reasonable" notion of "polyonomial ring in one-half variable" (unless perhaps we start thinking about enlarging our category or something). That is, let me address the following:

**Modified question:** Let $k$ be a commutative ring. Does there exist a functor $F: CAlg_k \to CAlg_k$ and a natural transformation $\iota: Id \Rightarrow F$ such that $F(F(A))$ is naturally isomorphic to $A[x]$ and such that the composite $A \xrightarrow{\iota_A} F(A) \xrightarrow{\iota_{F(A)}} F(F(A)) = A[x]$ is identified under this isomorphism with the usual map $A \to A[x]$?

It seems to me that this is an extremely minimial, reasonable additional property to ask of a construction deserving the name "polynomial algebra in one half variable". It would be nice to eventually dispense with it, but I hope to shed some light on the original question by adding the assumption of the existence of $\iota$.

**Answer to Modified Question:** No, there is no such functor $F$ and natural transformation $\iota$, at least when $k$ is a field. For if there were, then since the composite $F(k) \to k[x] \to F(k)[x]$ has a retract given by evaluation at zero, it would be the case that $F(k)$ is a retract of $k[x]$. There are no retracts of $k[x]$ besides $F(k) = k$ or $F(k) = k[x]$, neither of which is compatible with $F^2(k) \cong k[x]$ and $F^4(k) \cong k[x,y]$.

**EDIT:** I was suddenly seized with doubt, so here's a proof of the fact I just used:

**Fact:** Let $k$ be a field. Let $k \subseteq R \subseteq k[x]$ be a $k$-algebra which is a retract of the $k$-algebra $k[x]$. Then either $R = k$ or $R = k[x]$.

**Proof:** Let $\pi: k[x] \to R$ be the retraction map, and let $p = \pi(x) \in R \subseteq k[x]$ be the image of $x \in k[x]$ under $\pi$. Since $k[x]$ is a PID, the kernel of $\pi$ is a principal ideal, of the form $(f(x))$ for some $f(x) \in k[x]$. Thus we have $f(p) = 0$. By the following lemma, this implies either that $p \in k$ (in which case $R = k$) or else that $f(x) = 0$ (in which case $R = k[x]$).

**Lemma:** Let $k$ be a commutative ring, and let $p(x) \in k[x]$ be a polynomial. Suppose that the leading coefficient of $p(x)$ is a non-zero-divisor in $k$. Suppose that $p(x)$ is algebraic over $k$, i.e. that $f(p(x)) = 0$ for some $f(y) \in k[y] \setminus \{0\}$. Then $p(x) \in k$ is a degree-zero polynomial.

**Proof:** Suppose for contradiction that $\deg p(x) \geq 1$. Then the leading coefficient of $f(y)$ is $a b^n$ where $a$ is the leading coefficient of $f(y)$ and $b$ is the leading coefficient of $p(x)$ (and $n = \deg f$). This leading coefficient vanishes by hypothesis, so that $b$ is a zero-divisor in $k$, contrary to hypothesis. Therefore $\deg p(x) = 0$ as claimed.

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