Is $x^2 + xy + y^2$ irreducible in $\mathbb Q[x,y]$? I guess I'm just not sure how to approach factorization outside of $\mathbb Q[x]$. I tried looking at it as a polynomial in $(\mathbb Q[x])[y]$, but the only tricks I know are Eisenstein's Criterion, or the strategy of trying to find zeros. I've only used those techniques in $\mathbb Q[x]$, however, so I have no idea how I would apply them here...
 A: HINT: Suppose it is reducible, so that $$x^2+xy+y^2 = f(x,y)g(x,y)$$
evaluating at $y=1$ you would have
$$x^2+x+1 = f(x,1)g(x,1)$$
A: If $x^2+xy+y^2$ is reducible then there is a factorization into two non-constant polynomials, and these must be of the form
$$
(ax+b+c)(dx+e+f)
$$
with $a,b,c,d,e,f \in \Bbb{Q}$.
By expanding that product you have 
$$ad = 1==be=ae+bd = 1\\
cf = ec+bf = af+cd = 0
$$
the last line says $c$ or $f$ must be zero (since over $\Bbb{Q}$ if the product is zero one of the terms must be zero); wlog take $c=0$.  Then $bf = 0$ and $af = 0$ since $a$ and $b$ cannot both be zero (lest the first polynomial be zero) this shows that $f=0$ and now the second lineis fully satisfied.
Working with the first line we have 
$$d = \frac{1}{a}, e = \frac{1}{b} \\
\frac{a}{b}+\frac{b}{a} = 1
$$
Let $x = \frac{b}{a}$; since $a$ abd $b$ are both in $\Bbb{Q}$, $x \in \Bbb{Q}$.  But 
$$
x + \frac{1}{x} = 1
$$
has no real solutions, much less rational ones, so $x \not \in \Bbb{Q}$.  So the polynomial is irreducible.
A: We can reduce the problem to $\mathbb Q[x]:$
Suppose $f(x,y)=y^2+xy+x^2=(ay+b)(cy+d)$ for $a, b, c, d \in \mathbb Q[x],$ then $ac=1,$ so we can assume that $a=c=1,$ without loss of generality.
Thus $\begin{cases}b+d=x\\bd=x^2\end{cases}.$ So $\begin{cases}b=ux\\d=\frac{1}{u}x\end{cases},$ where $u \in \mathbb Q[x]^{*}=\mathbb Q^{*},$ with $u+\frac{1}{u}=1.$ But $u+\frac{1}{u}\ge2\sqrt{u\times\frac{1}{u}}=2,$ therefore $f$ is irreducible in $\mathbb Q[x, y].$
Hope this helps.  
P.S. This might be seen as a simplified version of the answer by Mark Fischler.
P.P.S. $R^{*}$ denotes the set of units of the ring $R.$
A: Assume it is reducible. Then put $y=1$ then $x^2+x+1$ is reducible (but ıt ıs not reducible also in $R[x]$.)
