It has been 5 days now so you probably have figured it out by now, but still, here is my take. First, we find the units . Since is a subring of any unit of is also a unit of . But we know (from a theorem) that the units of are the non-zero elements . Thus we see that .
Thus, to show that is irreducible in , we need to show that if for , then or is in . So, let for some . We can view any polynomial of as a polynomial in and thus we'll work over instead. There, we have that
for some . Multiplying these out gives us
Since , we see that it must be zero for the LHS and RHS to agree. Since is an integral domain, or . We choose , that is
Since is integral, we can cancel out the . We are left with
and thus . Thus we see that and for some . Therefore, is a unit, and we are done.