Let $S=k[x\_1,\dots,x\_n]$ be a polynomial ring, and $A:=k[x^{u^{(1)}}, \dots x^{u^{(l)}}]$ a monomial subalgebra, generated by monomials $x^{u^{(i)}} = \prod\_{j=1}^n x\_j^{u^{(i)}\_{j}}$ with $u^{(i)} \in \mathbf{N}^n$. What are sufficient criteria for the inclusion $A \to S$ to split? What is the splitting?
Since S is normal, and direct summands of normal rings are normal, normality of A is necessary.



Which monomial subalgebras are direct summands of polynomial rings
**This part answers when $A$ (the monomial algebra) is a direct summand of the polynomial ring as an $A$-module.**


Normality is also sufficient, also this doesn't depend on the field $k$. This is essentially exercise 6.1.10 in "Cohen-Macaulay rings" by Bruns and Herzog.


Let $S\subset T$ be affine semigroups. Call $S$ a full subsemigroup if $S= T\cap \mathbb ZS$. It is not hard to show that if $S$ is a positive affine semigroup then it is normal if and only if it is isomorphic to a full subsemigroup of $\mathbb N^n$, for some $n\geq 0$.


The claim in question follows from the general fact that if $S$ is full in $T$ then $k[S]$ is a direct $k[S]$-summand of $k[T]$. The proof is straightforward, denote by $W$ the $k$-algebra spanned by elements in $T-S$, then clearly $k[T]=k[S]\oplus W$ as vector spaces. To show that $W$ is in fact a $k[S]$ module, it is enough to check that if $\alpha\in S$ and $\beta\in T-S$ then $\alpha+\beta\in T-S$. If we assume for the sake of contradiction that $\alpha'=\alpha+\beta\in S$, then $\beta=\alpha'-\alpha\in T$ which implies $\beta\in S$ since $S$ is full.




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**This part answers when the inclusion $A\to k[x\_1,\dots,x\_n]$ splits as a map of rings.**


I hope I haven't made any silly mistakes, but the final answer seems to be a very restricted class of monomial algebras. Let $I=\lbrace i\_1,\dots i\_m\rbrace \subset \lbrace 1,2,\dots,m\rbrace$ and let $J$ be its complement. I claim that all $A$ in our problem are generated by monomials of the form $x\_{i\_r}f\_r$ where $f\_r$ are monomials in variables indexed by $J$ (for arbitrary choices of $I$ and $f\_r$'s). 


First, to see that the inclusion splits for such monomial algebras notice that we have the map $k[x\_1,\dots,x\_n]\to A$ defined by $x\_{i\_r}\to x\_{i\_r}f\_r$ for all $1\le r\le m$ and $x\_i\to 1$ for all $i\in J$.


Now suppose we have a monomial algebra $A=k[x^{\alpha\_1},\dots,x^{\alpha\_m}]$ and the inclusion $i:A\to k[x\_1,\dots,x\_n]$. In the conditions of our problem we have a map $q:k[x\_1,\dots,x\_n]\to A$ so that $qi=\text{id}\_A$. (Here I am denoting $x^{\alpha\_i}=x\_1^{\alpha\_{i1}}\cdots x\_n^{\alpha\_{in}}$.)


Our conclusion follows because we have $x^{\alpha\_r}=\prod q(x\_i)^{\alpha\_{ri}}$, so that for every $r$ we have for some $i$, $\alpha\_{ri}=1$ and $q(x\_i)=\beta\_r x^{\alpha\_r}$ for some $\beta\_r\in k$ (because $x^{\alpha\_r}$ is a generator so it cannot have a non-trivial factorization). The rest of the variables with $\alpha\_{ri}\geq 1$ are sent to scalars. therefore each $x^{\alpha\_r}$ factors as $x\_{i\_r}f\_r$ where all $i\_r$'s are distinct and all $f\_r$'s are monomials in the rest of the variables.


