Finding irreducible components over $\mathbb Q$ and $\mathbb C$ I want to find the irreducible components over $\mathbb Q$ and $\mathbb C$ for some curves, namely,
(a) $Y^2=X^5$
(b) $Y^2=X^3+1$
Intuitively, it seems to be that if the equation is not homogeneous, then it is immediately irreducible over $\mathbb Q$. If that's the case, then both (a) and (b) are irreducible over $\mathbb Q$. Is this true though?
I don't have much background in algebraic geometry so starting from scratch will be very helpful. Thanks.
 A: a) The polynomial $Y^2-X^5$ is irreducible in $k[X,Y]$ for  any field $k$.
 This is because the polynomial $X^5$ has no square root in $k[X]$: indeed, what would be the degree of such a square root !?
So the corresponding curve $y^2=x^5$ in $\mathbb A^2_k$ is irreducible over any field.  
b) Similarly the curve $y^2=x^3+1$ is irreducible over any field $k$ because the polynomial $X^3+1\in k[X]$ has no square root in $k[X]$.
You  need little algebraic geometry to understand the above. 
The relevant result is   :
 The curve  $C\subset \mathbb A^2_k$ given by the equation $f(x,y)=0$  is irreducible if and only the polynomial $f(X,Y)\in k[X,Y]$ is of the form $f(X,Y)=g(X,Y)^r$ for some integer $r$ and some irreducible polynomial $g(X,Y)$.
 The easiest situation  is when $r=1$ i.e. when $f(X,Y)$ itself is irreducible. This was the case in both examples above.
The slightly awkward formulation involving  some power of an irreducible polynomial is alas necessary: think of the equation $x^r=0$ which defines the  irreducible $y$-axis for any value of $r$.
