@article {IOPORT.06021300,
    author       = {Ishaq, Muhammad},
    title        = {Upper bounds for the Stanley depth.},
    year         = {2012},
    journal      = {Communications in Algebra},
    volume       = {40},
    number       = {1},
    issn         = {0092-7872},
    pages        = {87-97},
    publisher    = {Taylor \& Francis, Philadelphia, PA},
    doi          = {10.1080/00927872.2010.523642},
    abstract     = {Let $S=k[x_1,\dots,x_n]$ with $k$ a field. In [Invent. Math. 68, 175--193 (1982; Zbl 0516.10009)] {\it R. P. Stanley} conjectured that for any finitely generated $\mathbb{Z}^n$-graded $S$ module $M$ one always has that $\mathrm{sdepth}(M)\geq \mathrm{depth}(M)$ i.e. that the Stanley depth is always greater than or equal to the depth. This is known as the Stanley conjecture and remains open although it has been proven to hold for relevant cases, see [J. Algebra 318, No. 2, 1027--1031 (2007; Zbl 1132.13009)] and [Bull. Math. Soc. Sci. Math. Roum., Nouv. S\'er. 51(99), No. 3, 205--211 (2008; Zbl 1174.13033)] for instance. A fruitful line of research towards the solution of the Stanley conjecture is by finding bounds for the Stanley depth. The present paper advances in this line. The main results of the paper are: Theorem 2.1: $\mathrm{sdepth}(J/I)\leq \mathrm{sdepth}(\sqrt{J}/\sqrt{I})$ for $I$ and $J$ monomial ideals. Theorem 2.8: Let $Q$ and $Q'$ be two primary ideals with $\sqrt{Q}=(x_1,\dots,x_t)$ and $\sqrt{Q'}=(x_{t+1,\dots,x_n})$ where $t\geq 2$ and $n\geq 4$. Then $$ \mathrm{sdepth}(Q\cap Q')\leq \frac{n+2}{2} $$ Theorem 2.19: Let $Q$ and $Q'$ be two primary monomial ideals with $\sqrt{Q}=(x_1,\dots,x_t)$ and $\sqrt{Q'}=(x_{r+1,\dots,x_p})$ where $1<r\leq t<p\leq n$ and $n\geq 4$. Then $$ \mathrm{sdepth}(Q\cap Q')\leq \min\{ \frac{2n+t-p-r+2}{2},n-\lfloor \frac{t}{2}\rfloor,n-\lfloor\frac{p-t}{2}\rfloor\}. $$ The inequality becomes equality if $t=r$, $n$ is odd and $Q, Q'$ are irreducible.},
    reviewer     = {Eduardo Saenz-de-Cabezon (Logro\~no)},
    identifier   = {06021300},
}