Geometrically ireducible curve I know that curve with coefficients in $k$ is geometrically ireducible if it does not factor over algebraic closure of $k$.
I have this curve, for example,
$$2x^2+2x^2y+2y^2+2xy+3xy^2=1.$$
It's coefficients are in $\mathbb{Q}$, so closure is $\mathbb{C}$. How do I check it?
In  Wolfram Alpha it says it is irreducible over all extension (it means in complex numbers) but I am wondering about method to prove it?
EDIT: I'm not sure about curve above if it's easy to solve.  But what about $$x^2+y^2+x^2y^2=1?$$
My textbook just say it is irreducible, but not how to check it?
 A: There are quite a few methods for checking something is irreducible.
One of my favourites is Eistenstein's criterion:

Fact. Let $R$ be a domain, and let $f \in R[x]$; say $f = a_nx^n + \ldots + a_0$. Suppose there exists a prime ideal $\mathfrak p \subseteq R$ such that $a_n \not\in \mathfrak p$, $a_i \in \mathfrak p$ for $i \in \{0,\ldots,n-1\}$, and $a_0 \not\in \mathfrak p^2$. Then $f$ is irreducible in $R[x]$.

Another powerful tool is Gauß's lemma:

Fact. Let $R$ be a UFD with fraction field $K$, and let $f \in R[x]$; say $f = \sum_{i=0}^n a_ix^i$. Then $f$ is irreducible in $R[x]$ if and only if $f$ is irreducible in $K[x]$ and the $a_i$ share no common factor.

Now if you want to check something is irreducible over $\mathbb Q$, you can use Eisenstein's criterion at some prime $(p) \subseteq \mathbb Z$ and then use Gauß's lemma to go from $\mathbb Z[x]$ to $\mathbb Q[x]$. However, you want to prove that your polynomial is irreducible over $\bar{\mathbb Q}$ (or over the much bigger field $\mathbb C$), so we cannot use this method.
However, we can use prime ideals of $\mathbb C[y]$:
Example. Consider the polynomial
$$f = (2+2y)x^2 + (2y+3y^2)x + (2y^2 - 1).$$
This is the same polynomial you wrote down, but I've rewritten it to view it as a polynomial in $x$ over $\mathbb C[y]$. Note that the coefficients do not share a common factor, since $y = -1$ (the only root of the leading coefficient) is not a root of either of the other coefficients. Thus, by Gauß's lemma, $f$ is irreducible in $\mathbb C[y][x] = \mathbb C[x,y]$ if and only if it is irreducible in $\mathbb C(y)[x]$.
But checking whether a quadratic polynomial over a field is irreducible is easy: you just need to check that it doesn't have a root. The quadratic formula tells you that the only thing you need to check is that $b^2 - 4ac$ is not a square in $\mathbb C(y)$ (or equivalently, $\mathbb C[y]$). Do it. $\square$
Example. The other polynomial you ask about is
$$f = (y^2+1)x^2 + (y^2 - 1).$$
But we immediately see that this polynomial is Eisenstein at the prime ideal $(y-1) \subseteq \mathbb C[y]$. $\square$
Alternatively, we could use the same trick as above. To prove it's irreducible over $\mathbb C(y)$, observe that the discriminant is not zero. In this case an even easier argument suffices: irreducibility of $ax^2 + c$ just boils down to checking whether $\frac{c}{a}$ is a square, and $\frac{y^2-1}{y^2+1}$ is not a square. $\square$

Added later:
Remark. Often when you ask whether an element $f \in R$ is irreducible, what you're really interested in is whether the ring $R/(f)$ is a domain, i.e. the ideal $(f)$ is prime. In a UFD (e.g. $k[x_1,\ldots, x_n]$), a principal ideal generated by an irreducible element is prime, but this is not true in general:
Example. Let $R = \mathbb C[x,y]/(y^2 - x^3 - x)$. This is the affine part of an elliptic curve. (Exercise: use one of the methods above to show that $R$ is a domain). Consider the element $x-2$. Note that it is not prime, because
$$R/(x-2) = \mathbb C[x,y]/(y^2 - x^3 - x, x-2) = \mathbb C[y]/(y^2 - 10),$$
which by the Chinese Remainder theorem is isomorphic to
$$\mathbb C[y]/(y - \sqrt{10}) \times \mathbb C[y]/(y + \sqrt{10}) \cong \mathbb C \times \mathbb C.$$
This is not a domain. Geometrically, this corresponds to the observation that the line $x = 2$ intersects the curve in two points, each with multiplicity one.
However, the element $x-2$ is irreducible (this is not obvious, because the obvious degree function is not well-defined on $R$). Indeed, if we could write $x - 2 = a \cdot b$, then
$$\{(2,\sqrt{10}),(2,-\sqrt{10})\} = V(x-2) = V(a) \cup V(b).$$
If both $V(a)$ and $V(b)$ are nonempty, they each have to be a point (of multiplicity one), proving that the prime ideals defining $(2,\pm\sqrt{10})$ are principal. However, for an elliptic curve minus its point at infinity, this is known to be false. More precisely, the map
\begin{align*}
(\operatorname{Spec} R)(\mathbb C) & \to \operatorname{Pic}(\operatorname{Spec} R)\setminus\{0\} \\
P & \mapsto \mathcal O(P) = I_P
\end{align*}
is a bijection, so $I_P$ is never a principal ideal.
(To get the above bijection, note that for the complete elliptic curve $E$ we have an isomorphism $E(\mathbb C) \to \operatorname{Pic}^0(E)$. Then use the isomorphism $\operatorname{Pic}(E\setminus\{\infty\}) \cong \operatorname{Pic}^0(E)$, since any divisor on $E\setminus\{\infty\}$ can be completed uniquely to a degree $0$ divisor on $E$.)
Thus, one of $V(a)$ and $V(b)$ is empty, i.e. one of $a$ and $b$ is a unit, proving that $x-2$ is irreducible. $\square$
Example. If you're a number theorist, you would give the very similar example of the element $3$ in $R = \mathbb Z[\sqrt{-5}]$. The primes of $R$ lying above $(3) \subseteq \mathbb Z$ are $(3,1 + \sqrt{-5})$ and $(3,1-\sqrt{-5})$. Indeed, one computes
$$\mathbb Z[\sqrt{-5}]/(3) = \mathbb F_3[T]/(T^2 + 5) ,$$
and apply the Chinese Remainder theorem to $T^2 + 5 = (T-1)(T+1) \in \mathbb F_3[T]$. This shows that $(3)$ is not prime.
However, it is irreducible: if $3 = a \cdot b$, then taking norms shows that $9 = N(a)N(b)$. The norm is given by $x + y\sqrt{-5} \mapsto x^2 + 5y^2$, so one immediately sees that $3$ cannot be a norm. Similarly, $-3$ cannot be a norm, because it's negative. Thus, from $9 = N(a) N(b)$, we conclude that one of $N(a)$ and $N(b)$ has to be $\pm 1$ (in fact, $1$), hence one of $a$ and $b$ is a unit. This also proves that the prime ideals above $3$ cannot be principal. $\square$
