Irreducibility on homogeneous polynomials

Question

I would like to ask you the following irreducibility.

1. Prove that $x^3 - xy^2 + y^3 \in \mathbb{Q}[x, y]$ is irreducible. I know that $t^3 - t + 1 \in \mathbb{Q}[t]$ is irreducible by Gauss's lemma. I wonder if this fact is related to our case.
2. Is $x^3 + y^3 + 7z^3 \in \mathbb{C}[x, y, z]$ irreducible?
3. When is $x^2 + xy + y^2 \in k[x, y]$ irreducible if $k$ is a field? I know that $x^2 + xy + y^2 = (x - y)^2$ if $\mathop{\mathrm{ch}} k = 3$.

Edit: I fixed the statement of 3.

I have learned irreducibility criteria or tools such as famous Eisenstein's criterion or Gauss's lemma. But they are for polynomials of one variables. (But I know the relations such as $k[x, y] = (k[x])[y]$.) I am not used to irreducibility on polynomials of multi variables. The above questions have accumulated. I guess that they are elementary, but I need your help. I would appreciate it if you could help me, thanks.

Answer 1

It is useful to know that homogeneous polynomials decompose into homogemoeous polynomials. In fact, if $f,g$ are not homogeneous and have the (total) degree decomposition $f=f_m+\cdots+f_a,g=g_n+\cdots+g_b$, then $fg$ has the maximal/minimal component $f_mg_n$ and $f_ag_b$ and hence not homogeneous.

In view of this fact, item 1 is clear and irreducibility can be deduced from the inhomogeneous case. For 3, the polynomial decomposes $(x-\omega y)(x-\omega^2 y)$ for $\omega=(-1+\sqrt{-3})/2$ if the field contains $\omega$. Otherwise irreducible. For 2, we can consider it as in $\mathbb{C}[x,y][z]$ and regard $x^3+y^3$ as the constant term, and use Eisenstein with the prime $x+y$. (Or, algebro-geometric nonsingularity proof will be more natural.)

Answer 2

1. irreducibility of $x^{3} - xy^{2} + y^{3}$

View the polynomial in $(\mathbf{Q}[x])[y]$. $$ y^{3} + (-x)y^{2} + x^{3} $$ Since the polynomial has $y$-degree $3$, reducibility would imply the existence of another polynomial $f(x) \in \mathbf{Q}[x]$ such that $$ f(x)^{3} + (-x)f(x)^{2} + x^{3} = 0. $$ For reasons of degree, $f(x)$ would have to be linear, i.e., of the form $ax + b$ for some $a,b \in \mathbf{Q}$. Now, observe that the constant term of the left hand side is precisely $b^{3}$. We thus deduce $b = 0$. We may therefore rewrite the equality as $$ (a^{3} - a + 1)x^{3} = 0. $$ We are now left with the question of whether $a^{3} - a + 1$ is irreducible. As you observe, Gauss's lemma resolves this.

3. irreducibility of $x^{2} + xy + y^{2}$

Once more, view the polynomial in $(k[x])[y]$. $$ y^{2} + xy + x^{2} $$ Reducibility would imply the existence of $f \in k[x]$ such that $$ f(x)^{2} + xf(x) + x^{2} = 0. $$ In fact, $f$ must be of the form $ax$ for some $a \in k$. Substitute $ax$ for $f(x)$ to obtain the necessary condition $$ a^{2} + a + 1 = 0. $$ We find ourselves once more in familiar territory.