How to find irreducible polynomials over $\mathbb{Q}(i)$ with prescribed Galois group? Here is my recent homework question:

For each of the following five fields $F$ and five groups $G$, find an irreducible polynomial in $F[x]$ whose Galois group is isomorphic to $G$. If no example exists, you must justify that.
Fields $F$: $\mathbb{C}$, $\mathbb{R}$, $\mathbb{F}_{11}$, $\mathbb{Q}$, $\mathbb{Q}(i)$
Groups $G$: $C_2$, $C_5$, $C_2\times C_2$, $S_3$, $D_4$

I've found the polynomials for first 4 fields; however, I've got no idea about the $\mathbb{Q}(i)$ one.
Can anyone here help me? Thanks, and regards.
Now, I just found I made mistakes in looking for C2xC2 and C5 one . The polynomial I found are not irreducible ( in fact only with separable irreducible factors ).  Moreover, I don't know how to check the irreducibility of polynomial in F11. So I also can't do this part..
 A: Collecting some of my comments to an answer as a few weeks have passed. I believe that prof. Lubin had some of these examples in mind when posting his comment.
As an example of a quintic cyclic extension of both $\mathbb{Q}$ (resp. $\mathbb{Q}(i)$) I proffer the real subfield of the eleventh cyclotomic field. Let $\zeta=e^{2\pi i/11}$. Then
$K=\mathbb{Q}(\zeta)$ is a cyclic extension of the rationals of degree ten. The Galois group is generated by the automorphism determined by $\sigma:\zeta\mapsto\zeta^2$. We see that $\sigma^5(\zeta)=\zeta^{32}=\zeta^{3\cdot11-1}=\zeta^{-1}=\overline{\zeta}$, so $\sigma^5$ is just the restriction of the complex conjugation to $K$. Therefore $L=K\cap\mathbb{R}$ is
a cyclic quintic extension of the rationals. It can also be constructed as
$$
L=\mathbb{Q}(\zeta+\overline{\zeta})=\mathbb{Q}(2\cos2\pi/11).
$$
Because $L$ is real and Galois over the rationals (aka totally real), all the five automorphisms of $L$ can be extended to $\mathbb{Q}(i)$-automorphisms of $L(i)$, so $L(i)$ is a cyclic quintic extension of $\mathbb{Q}(i)$. The minimal polynomial of $u=\zeta+\zeta^{-1}$ can be found as follows.
Let's write down powers of $u$:
$$
\begin{aligned}
u^5&=\zeta^5+5\zeta^3+10\zeta+10\zeta^{-1}+5\zeta^{-3}+\zeta^{-5}\\
u^4&=\zeta^4+4\zeta^2+6+4\zeta^{-2}+\zeta^{-4}\\
u^3&=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}\\
u^2&=\zeta^2+2+\zeta^{-2}.
\end{aligned}
$$
From this we can easily verify that
$$
u^5+u^4-4u^3-3u^2+3u+1=\sum_{n=-5}^5\zeta^n=\frac{\zeta^{-5}-\zeta^{6}}{1-\zeta}=0,
$$
so the polynomial $x^5+x^4-4x^3-3x^2+3x+1$ works (barring some errors in my calculations).
Devoting this paragraph to the case $F=\mathbb{F}_{11}$. This is a finite field, so all its finite extensions are also finite, and hence have cyclic Galois groups. Therefore only $C_2$ and $C_5$ can be realized as Galois groups in this case. We either know or can quickly check that $-1$ is not a square in $F$, so $F(\sqrt{-1})=\mathbb{F}_{121}$ is a quadratic extension of $F$ and thus $Gal(F(\sqrt{-1})/F)=C_2.$ Finding an irreducible quintic requires more knowledge about finite fields. We observe that $3$ is a fifth root of unity in $F$, because $3\neq1$ but $3^5=243\equiv 1\pmod{11}$. An extenstion field
$\mathbb{F}_{11^n}$ contains a primitive 25th root of unity, if and only if $25\mid 11^n-1$. This follows from the cyclicity of the multiplicative group of finite fields. We observe that $n=5$ is the smallest exponent for which this happens: none of $11^2-1=120$, $11^3-1=1330$, $11^4-1=14640$ are divisible by $25$, but $11^5-1=161050$ is
(it would also be easy to prove in several ways that this happens at latest with $n=5$). 
As $3$ is a fifth root of unity, any solution of
$$
x^5-3=0
$$
must be a 25th root of unity. Therefore any one of those solutions generates the field $\mathbb{F}_{11^5}$, and the polynomial $x^5-3$ must be irreducible over $F$ with Galois group $C_5$.
A: Let me try to give a fairly complete answer, covering those cases that @Jyrki has not; and let’s hope that this will not prove to be too long.
For $\mathbb{C}$, any Galois group is trivial, since the field is algebraically closed. For $\mathbb{R}$, the only nontrivial Galois group is $C_2$, and as we know, the polynomial $X^2+1$ will get the extension. For ${\mathbb{F}}_{11}$, all Galois groups are cyclic, so that $C_2$ and $C_5$ are possible, but the other three are not; and @Jyrki has covered the two possible cases. The remaining two fields are $\mathbb{Q}$ and the Gaussian numbers ${\mathbb{Q}}(i)$, which I will call $k$ from here on.
For $\mathbb{Q}$, all possibilities occur, and as we all know, you an use $X^2+1$ for an extension with Galois group $C_2$; this new field is our $k$. For $C_2\times C_2$, you can take any “biquadratic” field, compositum of two quadratic extensions. Although it’s a very special one, I choose as an example the field ${\mathbb{Q}}(\sqrt{2},i)$, and since it’s going to be important to us later, I choose to name it $E$. The Galois group consists of the identity, complex conjugation (leaving $\sqrt2$ fixed), and the two automorphisms that send $\sqrt2$ to $-\sqrt2$: one leaving $i$ fixed, the other sending $i$ to $-i$. Please notice that $(1+i)/\sqrt2$ is in $E$ and it’s a primitive eighth root of unity. I will call this complex number $\zeta$, and observe that it’s a primitive element for $E$, irreducible polynomial over $\mathbb{Q}$ being $X^4+1$, the eighth cyclotomic polynomial. I will use without proof the fact that $\{1,\zeta,\zeta^2,\zeta^3\}$ form an integral basis for the (algebraic) integers of $E$, that is, an algebraic integer in $E$ is a $\mathbb{Z}$-linear combination of those powers of $\zeta$.
Continuing with $\mathbb{Q}$, the remaining groups $S_3$ and $D_4$ are easy: take the cube root of any squarefree positive integer, say $\alpha$, root of $X^3-m$, and the splitting field has to contain $\omega=(-1+\sqrt{-3})/2$, a primitive cube root of unity. I will not detail the action of the Galois group, nor write down minimal polynomial for $\alpha+\omega$, which will be a primitive element. For the group $D_4$, the dihedral group of order eight, it’s almost the same. Take an integer $m$ that is neither a square nor the negative of a square, and adjoin $\alpha=\root{4}\of{m}$. Again the extension is not normal, but the normal closure must contain $i$. Two automorphisms of the field are $\sigma\colon \{i\mapsto-i, \alpha\mapsto\alpha\}$ and $\tau\colon\{i\mapsto-i, \alpha\mapsto i\alpha\}$. Note that $\tau\circ\sigma$ leves $i$ fixed, but sends $\alpha$ to $i\alpha$, so is an automorphism of period four. This is just what we need for dihedral of order eight. Further details of this case I leave as an exercise.
The most interesting case is our field $k$ of Gaussian numbers. Almost everything goes as for $\mathbb{Q}$, except the problem of finding an extension whose Galois group is $D_4$. I have found one example of very special nature, but there must be many more. The problem, you see, is that if you just adjoin the fourth root of something, you’ll get a cyclic extension of degree four, already normal. Kummer 
Theory tells us that if the characteristic does not divide $m$, and if the $m$-th roots of unity are in the base field, then the adjunction of the $m$-th root of an element, say $u$, gives a normal extension, of degree $m'$ dividing $m$, and with cyclic Galois group. In particular, if $m$ is $4$, and $u$ is not a square, then you get a cyclic extension of degree four.
Thus the standard elementary strategy for finding a $D_4$-extension of $\mathbb{Q}$ will not work for the base field $k$, because $i$ is already there. The nature of the dihedral group is that it has a cyclic normal subgroup $C$ of order four, index two, so that an extension of $k$ with group $D_4$ must start with a quadratic extension of $k$, and then jump up to a cyclic quartic extension of that: $k\subset ?\subset L$, where the group of ? over $k$ is $D_4/C$, and the group of $L$ over ? is $C$. My choice for ? is our field $E$, gotten by adjoining $\sqrt2$ to $k$; recall that $E$ is the field of eighth roots of unity. And the nonsquare element of $E$ whose fourth root I will adjoin to get $L$ is the unit $u=1+\sqrt2$. Let’s call $\beta=\root{4}\of{u}$, so that its minimal polynomial over $E$ is simply $X^4-u$. Now $E(u)$ is normal over $E$, but $k(u)$ is not normal over $k$. Indeed, the normal closure must at least contain a fourth root of $\overline u-1-\sqrt2$. But if $X^4-u$ is the minimal polynomial for $\beta$ over $e$, surely $(X^4-u)(X^4-\overline u)$ must be the minimal polynomial for $\beta$ over $k$. This multiplies out to $X^8-2X^4-1=f(X)$. I’ll show that adjunction of one root of this polynomial to $k$ brings along all other roots, so that we have a normal extension of degree eight, and I’ll show how the Galois group acts.
Consider one root of $f$. Without loss of generality, we may assume that it’s a root of the factor $X^4-(1+\sqrt2)$, even assume that it’s the unique positive real root $\beta$ of this, if you like. Call $L_0=k(\beta)$. With $\beta$ in our field $L_0$, we have its fourth power $u=1+\sqrt2$, and so we have $\sqrt2$, and because $i$ is in the base field, we also have $\zeta$ in $L_0$. That was our eighth root of unity in the first quadrant. Consequently, we find $\zeta/\beta$ in $L_0$. But what is the fourth power of this? It’s $-1/u=\overline u$. So $\zeta/\beta$ is a root of the other factor of $f$. And since $i$ is in $L_0$, We have all roots of both factors of $f$. That is, $L_0$ is our normal extension of $k$ of degree eight. The Galois group? I need to specify two involutions whose composition is of period four. Easy: $\sigma$ sends $\beta$ to $\zeta/\beta$, and $\tau$ sends $\beta$ to $i\zeta/\beta$.
