Intersection of ideals and primary ideals

Question

In T.Y. Lam book

Exercises in Modules and Rings here, page 84

Let $\mathbb{K}$ be a field, $R=\mathbb{K}[X,Y], \space \space $

$ I=(Y^2, XY)$ , $ Q_{1}=(Y) $ and $Q_{2}=(Y^2 , X + tY)$ where $t\in \mathbb{K}.$

He proved that $I\subset Q_{1}\cap Q_{2},$ but I don't understand how to prove

$ Q_{1}\cap Q_{2} \subset I.$

Answer 1

Take an arbitrary element of $Q_1\cap Q_2$. It is possible to write that element as $fY$ for some $f\in\Bbb K[X,Y]$, as it is an element of $Q_1$. Also, as it is an element of $Q_2$, it is a linear combination of the form $gY^2+h(X+tY)$ for $g,h\in\Bbb K[X, Y]$.

Thus we have $$ fY=gY^2+hX+htY\\ hX=(f-gY-ht)Y $$ Since the right-hand side is divisible by $Y$, the same must be true for the left-hand side. In other words, $h=h_0Y$ for some $h_0\in\Bbb K[X,Y]$. This means that our element may be written as $$ gY^2+h_0Y^2+h_0tXY\in I $$

Answer 2

If $f\in Q_1\cap Q_2$ then $f=gY$ and

$f=hY^2+z(X+tY)$ so

$Y|zX$ then $Y| z$ so $z=z’Y$ and

$f=(h+tz’)Y^2+zX= (h+tz’)Y^2+z’YX$