Irreducible polynomials in $\mathbb{Q}_p((X))[Y]$ I'm looking for some criteria for the irreducibility of polynomials with coefficients in $\mathbb{Q}_p((X))$. 
In particular, is the polynomial $Y^2+1$ irreducible over $\mathbb{Q}_3((X))$? And how about in $\mathbb{Q}_p((X))$?
 A: Since you ask about $Y^2+1$ over ${\mathbf Q}_p((X))$ and Peter Mueller's answer reduces this to the case of $Y^2+1$ over ${\mathbf Q}_p$, the following broader context might be conceptually useful: if $L/K$ is an extension of fields such that $K$ is algebraically closed in $L$ (that is, any element of $L$ algebraic over $K$ is in $K$), then a polynomial in $K[Y]$ is irreducible in $K[Y]$ if and only if it is irreducible in $L[Y]$.  Trivially, irreducibility in $L[Y]$ implies irreducibility in $K[Y]$. The other direction is more interesting and is left as an exercise, along with the task of showing $F$ is algebraically closed in $F((X))$. Thus for any $c \in {\mathbf Q}_p$, irreducibility of $Y^2-c$ over ${\mathbf Q}_p((X))$ is equivalent to irreducibility of $Y^2-c$ over ${\mathbf Q}_p$.
A: The ring $R=\mathbb Q_p[[X]]$ of formal power series is factorial, and its quotient field is your field $S=\mathbb Q_p((X))$ of coefficients. So the Gauss Lemma holds. In particular, if a polynomial $f(Y)\in\mathbb Q_p[Y]$ factorizes over $S$, then it factorizes over $R$, and upon setting $X=0$ it factorizes over $\mathbb Q_p$. So you are asking if $-1$ is a square in $\mathbb Q_p$, and the answer to that is well known.
