Are coprime polynomials also coprime considered in the algebraic closure? 
Question: Let $f_1,f_2\in k[x,y]$ be coprime. If $k\neq\bar{k}$ is not algebraically closed, are $f_1,f_2$ coprime in $\bar{k}[x,y]$? 

In the single-variable case, I know that this result is true, since we can prove the fact that 

$f_1,f_2$ generate $(1)$ in $k[x]$ if any only they generate $(1)$ in $\bar{k}[x]$, 

by showing that a linear system with coefficients in $k$ has a solution if and only it has a solution in any extension of $k$. 
However, this does not seem to help in the multi-variable case, since if we reduce the number of variables by $k[x,y]=k[x][y]$, $k[x]$ is not necessarily a field. So I am wondering whether there is some other trick to answer the question?
Thank you very much for any help!
 A: 
However, this does not seem to help in the multi-variable case

The same trick in fact works as well.
Suppose $f,g$ not coprime over $\overline k$. Then there exist non-zero polynomials $u,v$ with coefficients in $\overline k$, such that $uf=vg$ and $\deg u < \deg g, \deg v<\deg f$, where $\deg$ denotes the total degree.
Now replace the coefficients of $u,v$ with variables. The equation $uf=vg$ then becomes a system of linear equations with coefficients in $k$. The rest is the same argument as you know.
A: Note that $f_1$ and $f_2$ are coprime over a field $K$ iff $f_2$ is not a zero-divisor in the ring $K[x,y]/(f_1)$.  Now observe that $\bar{k}[x,y]/(f_1)\cong k[x,y]/(f_1)\otimes_k \bar{k}$ (proof: they have the same universal property as $k$-algebras).  Since $\bar{k}$ is faithfully flat over $k$, the multiplication by $f_2$ map $k[x,y]/(f_1)\to k[x,y]/(f_1)$ is injective iff it remains injective after tensoring with $\bar{k}$.  That is, $f_2$ is coprime to $f_1$ over $k$ iff it is coprime to $f_1$ over $\bar{k}$.
