Let $k$ be a field. Let $B = k[X,Y]$ and $A=k + YB$. So we have the ring extension $ A \subseteq B$. Now I can show that the conductor ideal of $A$ in $B$ is $YB$. So if $Q(A), Q(B)$ are the fraction fields of $A,B$ respectively , then $ A \subseteq B \subseteq Q (A) \subseteq Q(B)$. I am trying to show the following facts about $A$ and $B$:
1) $A$ is normal domain, i.e., $A$ is integrally closed in $Q(A)$.
2) The ideal $YB$ in $A$ has height 2.
3) $ \sqrt {YA}=YB$
I haven't been able to prove any of the four facts fully. For (1) I can see that if $g \in Q(A)$ is integral over $A$, then $g$ is integral over the UFD $B$, so $g \in B$, but I am unable to conclude $g \in A$. For (2), the only thing I can see is that $YB = YB \cap A$ is prime ideal in $A$. I have no idea for (3).
Please help for proving the facts. Thanks in advance.
UPDATE ON (1) : Let $g(X,Y) \in B$ be integral over $A$. Now any element $f(X,Y) \in A$ is of the form $a + Y h (X,Y)$ where $a=f(X,0) \in k$. So writing a monic equation for $g$ with coefficients in $A$ and putting $Y=0$ , we see that $g(X,0)$ is constant . Thus $g(X,Y) \in A$ so we are done. I hope this proof is correct.