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All rings in this question are unitary and commutative and all maps are homomorphisms of commutative rings sending $1$ to $1$.

Let $R$ and $S$ be regular local rings and let $$ \begin{array}{rcl} && R[x,y,z]/(x+y+z-1)\\ &&\qquad\qquad \downarrow\\ S &\xrightarrow{f}& R[x,y,z]/(xy,x+y+z-1) \end{array} $$ be a given diagram. (The vertical map on the right hand side of the diagram is the quotient map.) Can one find a map $S\to R[x,y,z]/(x+y+z-1)$ making the diagram commutative?

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sopot
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Use the ring isomorphisms $R[x,y,z]/(x+y+z-1) \cong R[x,y]$ and $R[x,y,z]/(xy, x+y+z-1) \cong R[x,y]/(xy)$.

Then the statement is true for the following silly reason:

Let $S \rightarrow R[x,y]/(xy)$ be a homomorphism where $S$ is regular local. Suppose $\alpha \in S$ is sent to $f(x,y)$. Then $\alpha$ or $1 + \alpha$ is a unit in $S$, and hence this unit must be sent to a unit in $R[x,y]/(xy)$.

Edit: (as user121097 points) The units in $R[x,y]/(xy)$ are precisely those which arise from units in $R[x,y]$.

In any commutative ring $T$ with 1, the polynomials in $T[x_1, \dots, x_n]$ which are units have an invertible constant coefficient and all other coefficients are nilpotent. Since $R$ is regular, there are no nilpotents, and so $f$ must be the constant polynomial. Therefore the homomorphism factors through $R[x,y]$.

RghtHndSd
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    $S$ is not regular then. – sopot Feb 15 '14 at 20:45
  • Whoops, I missed that condition. I'm fixing it now. – RghtHndSd Feb 15 '14 at 20:59
  • Ack, what I thought would fix it is not a local ring, although I do have a regular ring $S$ for which the map is not possible. I will delete this answer and morph it into a comment. – RghtHndSd Feb 15 '14 at 21:18
  • The example of the non-local but regular $S$ would be very interesting for me. Thank you! – sopot Feb 15 '14 at 21:39
  • @sopot: I was trying to cook up an example with $S = R[x,y]/(y(1+x+y^2) + xy)$ (or something similar), but I must have misplaced a constant term because the map I have written down isn't even well-defined. Not only that, but $S$ isn't even regular. I have the feeling that playing around with something like that will give you a counter-example, but I haven't been able to make it work. – RghtHndSd Feb 15 '14 at 21:49
  • @user121097 Thanks for the corrections! The first was merely a slip of the tongue on my part, but the second is a very good point. I've adjusted the answer. – RghtHndSd Feb 15 '14 at 22:41