I would like to understand some trivialities about log-structures. Given a log-scheme $(X,M\_X)$ the log-structure $M\_X$ is defined via push-out. Are there stupid examples in which this push-out is actually a direct sum $\mathbb{G}\_m\oplus P$ for some monoid $P$? 


Furthermore assume we have a ring $R$, an element $t\in R$ defining a log-structure on $Spec(R)$ via $\mathbb{N}\rightarrow t^n$. Consider $X=Spec(R[x,y]/(xy-t))$ with the standard log-structure on the two brances $x$ and $y$. Let $p:Y\rightarrow X$ be the blow-up of $X$ at $(x,y,t)$ and put a log-structure on $Y$ over the preimage $\tilde{x},\tilde{y}$ of the two branches $x,y$ but not on the exceptional divisor. Namely locally on $Y$ we take $\mathbb{N}\rightarrow \mathcal{O}\_Y$ where we send $1$ to $\tilde{x}$ (resp. to $\tilde{y}$). Call this log-structure $M\_Y$. Is it true that $p\_{\*}M\_Y$ splits as a direct sum? If so how do the summands look like?



trivialities on log-structures
For the first question, given a sharp monoid $P$ (i.e., the group of units of $P$ is the
zero group), there is a log structure on any scheme $X$ given by $M\_X={\mathcal O}\_X^\*\oplus P$ with the structure map given by $(f,p)\mapsto f$ if $p=0$ and $(f,p)\mapsto 0$ if $p\not =0$. This comes from the prelog structure $P\rightarrow {\mathcal O}\_X$ given by 
$p\mapsto 1$ if $p=0$, $p\mapsto 0$ if $p\not=0$. I wouldn't call this a stupid log structure: it in fact can be quite useful. For example, the standard log point is defined as the log structure on the scheme $Spec(k)$ with $P$ the natural numbers as above.


For the second question, I'm not quite sure what direct sum decomposition you are expecting, but I can't think of a natural direct sum decomposition here. If you are thinking that 
$p\_\*M\_Y$ will decompose as a direct sum of the sheaf ${\mathcal O}\_X^\*$ and a sheaf supported
on the branches $x$ and $y$, this is not the case. To take a simpler example, taking
the divisorial log structure associated to $0 \in \mathbb{A}^1$, the log structure does
not decompose as $O\_{A^1}\oplus {\mathbb N}\_0$.
Indeed, there is no
section of $M\_{\mathbb{A}^1}$ with support at $0$. In the example you describe, there will
be an exact sequence
$$
1 \rightarrow {\mathcal O}\_Y^\*\rightarrow M\_Y \rightarrow {\mathbb{N}}\_x \oplus {\mathbb{N}}\_y\rightarrow 0
$$
where the subscripts $x$ and $y$ denote which branch the constant sheaf with coefficients
${\mathbb N}$ is sitting on. I believe when pushing down, one obtains an exact sequence
$$
1\rightarrow {\mathcal O}\_X^\*\rightarrow p\_\*M\_Y\rightarrow {\mathcal F} \rightarrow 0
$$
where ${\mathcal F}$ is the sheaf supported on $D$, the vanishing locus of $t$, which
is obtained by taking the constant sheaf ${\mathbb N}$ on $D-\{0\}$ and extending by zero
across $0$. This is because there are no functions on the inverse image of a open neighbourhood of $0$ which do not vanish off the two branches, but do vanish one or both of the branches.


