Is every ideal in $\hat{A}$ extended? Let $A$ be a Noetherian ring, $I\subset A$ an ideal, $\hat{A}$ the $I$-adic completion. Is it true that every ideal of $\hat{A}$ is of the form $\hat{J}$ for some ideal $J\subset A$?
The answer is no. Consider $A=k[x,y]$ over a finite or countable field $k$, $I$ the maximal ideal $(x,y)$. Then $\hat{A}=k[[x,y]]$. Let $f(y)$ be a power series transcendental over $k(y)$ with $f(0)=0$ (such power series exists by cardinality arguments). Now consider the ideal $J&#39;=(x-f(y))k[[x,y]]$. I claim that $J&#39;\cap k[x,y]=0$, which implies that $J&#39;$ can not be a $\hat{J}$. 

Let $P(x,y)\in J&#39;\cap k[x,y]$. Then $P(x,y)=(x-f(y))g(x,y)$ for some $g(x,y)\in k[[x,y]]$. Replacing in this equality $x$ with $f(y)$ (this is legal becase $f(y)\in yk[[y]]$), we get $P(f(y),y)=0$. As $f(y)$ is trenscendental over $k(y)$, this implies that $P(x,y)=0$. 

Update An explicit power series which is transcendental. Consider $e^z\in \mathbb C[[z]]$. If it were algebraic, we would have an integral relation
$$e^{dz}+f_{d-1}(z)e^{(d-1)z}+\dots+f_0(z)=0$$ 
with a positive integer $d$ and $f_i(z)\in \mathbb C(z)$. There exists a positive integer $m$ such that $P_i(z):=z^{m(d-i)}f_i(z)\in\mathbb C[z]$ for all $i\le d-1$. Then
$$(z^me^z)^{d}+P_{d-1}(z)(z^me^z)^{d-1}+\dots+P_0(z)=0$$ 
in $\mathbb C[[z]]$.This implies the same equality in the ring of entire functions. But this is clearly impossible because when the real part of $z$ tends to $+\infty$, $e^{z}$ grows much faster than any polynomial function.

Note that $e^z\in\mathbb Q[[z]]$ is then transcendental over $\mathbb Q((z))$. In fact this holds for any field of characteristic $0$.
That should indeed be true. To prove this take an ideal $\hat{J}$ in $\hat{A}$ and prove that the set of numerators of elements in $\hat{J}$ forms an ideal in $A$.

Edit: Sorry I did not read careful enough. This prove only holds for the localization not the completion.