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I was recently looking at the irreducibility of polynomial $X^3-Y^2 \in K[X,Y]$ and I stumbled with lots of pages in this site such as this one that from my point of view consider the following statement to be true:

$f \in K[X,Y]$ is irreducible $\iff f \in K[Y][X]$ is irreducible.

I know that $K[X,Y]$ and $K[Y][X]$ are just two ways of looking at the same elements but I have some reserves to apply this argument directly.

Do you a have a formal proof of this fact? How would you discuss then my polynomial above?

My thoughts

For the latest part I would use is Eisenstein criterio assuming $Y^2$ is prime in $K[Y][X]$...

user1868607
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    I'm not sure what there is to prove. Being irreducible is a property of an element in a ring, it does not depend on whether this ring is a polynomial ring over another ring or how it is a polynomial ring over another ring. – Lukas Heger Oct 10 '17 at 23:32
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    By the way, $Y^2$ is not prime in $K[Y][X]$. – Lukas Heger Oct 10 '17 at 23:32
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    It's obvious since $K[X,Y]$ and $K[Y][X]$ are isomorphic. – Bernard Oct 10 '17 at 23:55
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    Instead of Eisenstein I would go "hands-on". If $Y^2-X^3$ factors, it must be either $Y^2-X^3 = (a(X)Y-b(X))(c(X)Y-d(X))$, or $Y^2-X^3 = (a(X)Y^2-b(X))c(X)$. (I guess I'm working in $K[X][Y]$ rather than $K[Y][X]$.) In both cases $a(X)c(X)=1$ so $a,c \in K$; we can assume $a=c=1$ without loss of generality (why?). Then consider $b$ and $d$... etc. – Zach Teitler Oct 11 '17 at 00:33

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