Ideal not reflexive as an R-module I've got the following problem:

Let $R = \mathbb{C}[x,y]$, with ideal $I = (x,y)$.
Use the representation $$R \xrightarrow{(-y,x)} R^2 \xrightarrow{(x \ y)} I \rightarrow 0$$ of $I$ as $R$-module and the left-exactness of $\mathrm{Hom}(-,R)$ to show that $\mathrm{Hom}(I,R) \cong R$ and obtain that $I$, as an $R$-module, is not reflexive.

I guess is have to show that the map $I \rightarrow I^{**}$ is not a bijection, however I have no idea how to attack this problem.
 A: You indicate in the comments that you're struggling to see why the sequence is exact, although you know the definition of exactness. Thus I'll explain that.
Why is the sequence exact?
As you indicate in the comments, a sequence of maps of modules
$$\newcommand\toby\xrightarrow M_1\toby{\phi_1} M_2 \toby{\phi_2} M_3 \quad\quad (*)$$
is exact at $M_2$ if $\ker \phi_2 = \newcommand\im{\operatorname{im}}\im \phi_1$.
A longer sequence of maps like 
$$
R\toby{(-y,x)}R^2\toby{(x,y)} I \to 0
$$
is exact if it's exact at all the internal modules. In this case, we need to show 
that it is exact at $R^2$ and $I$.
Exactness at $I$: The kernel of the map $I\to 0$ is all of $I$, since everything in $I$ maps to $0$. Thus to show that the sequence is exact at $I$ is the same as showing that $R^2\toby{(f,g)\mapsto fx+gy} I$ is surjective. However, $I$ is defined to be the ideal consisting of all sums $fx+gy$ with $(f,g)\in R^2$. I.e., $I$ is defined to be the image of this map already, so the map is definitely surjective.
Exactness at $R^2$: A common tactic for showing exactness in a sequence like (*) 
is to prove two things, 
$\phi_2\circ \phi_1=0$, since this implies $\im\phi_1\subseteq \ker\phi_2$ and that if $m\in \ker \phi_2$, then there exists $n\in M_1$ such that $\phi_1(n)=m$, which implies that $\ker\phi_2\subseteq \im\phi_1$. The two reverse inclusions give equality.
In our case, if $f\in R$, $f\mapsto (-yf,xf)\in R^2$, and then this maps to 
$-yfx + xfy=0$ in $I$. Thus the composite of the maps $R\to R^2$ and $R^2\to I$ is $0$.
Now suppose $(f,g)$ are such that $fx+gy=0$. Then $fx=-gy$, so $y\mid fx$, which implies 
that $y\mid f$. Hence $f=yh$. Then $yhx=-gy$, so $g=-hx$. Thus 
$(f,g) = ((-h)(-y),(-h)x)$ is in the image of the map $R\to R^2$. Thus we have exactness.
Completing the problem:
Since the sequence is exact, and $\newcommand\Hom{\operatorname{Hom}}\Hom$ is left exact, applying $\Hom(-,R)$ gives an exact sequence
$$
0\to I^* \toby{\circ (x,y)} R^{2*} \toby{\circ (-y,x)} R^*.
$$
In other words, $I^*$ is the kernel of the map $R^{2*}\toby{\circ (-y,x)} R^*$.
So we just need to compute this kernel.
Let $\lambda : R^2\to R$ be an element of $R^{2*}$. Since $R^2$ is free, 
$\lambda(x,y) = ax+by$ for some $a,b\in R$. Composing this map with $(-y,x) : R\to R^2$
gives a map $(\lambda\circ (-y,x))(r) = -ayr + bxr$ in $R^*$. This map is $0$ if and only if $(\lambda\circ(-y,x))(1)=-ay+bx=0$. In other words, $\lambda$ is in the kernel 
if and only if $(a,b)$ is a multiple of $(x,y)$. Thus the kernel is isomorphic to $R$.
Finally, $I^*\cong R$, so $I^{**}\cong R^* \cong R$. But $I$ is not principal, whereas $R$ is. Thus $I\not \cong R$.
