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As the title says is $$x^3+y^2x^2+3yx-y$$ irreducible in $\mathbb{Q}[y][x]$ ? Apparently the answer is that it is indeed irreducible.

My attempt:

My first thoughts are what is $\mathbb{Q}[y][x]$, to which I reckon it's something like $(\mathbb{Q}[y])[x]$, i.e. polynomials in $x$ with coefficients in $\mathbb{Q}[y]$, polynomials in $y$.

Assuming this is correct, then maybe I can use Gauss' Lemma? E.g. I can prove $(\mathbb{Q}[y])[x]$ is irreducible if I can prove $(\mathbb{Z}[y])[x]$ is irreducible. Assuming this is correct, I proceed as follows.

Let $f(x)=x^3+y^2x^2+3yx-y$ and assume $f(x)$ is reducible in $(\mathbb{Z}[y])[x]$, then we must have $$f(x)=(x^2+ax+b)(x+c),$$ where $a,b,c\in\mathbb{Z}[y]$. This implies that $bc=-y$ which means that $$(b,c)\in\{(1,-y),(-1,y),(-y,1),(y,-1)\}$$ since $b,c\in\mathbb{Z}$, i.e. $x=\pm 1$ or $x=\pm y$.

I took a look at $f(1)$ hoping for a contradiction, but found that everything works out if $y=-1$.

But then I see that when $x=1$ and $y-1$ that the original $f(x)=0$, which means $f(x)$ is reducible, doesn't it?

What's the correct way to proceed? Maybe the wrong approach. Maybe there's a way of applying Eisenstein's criterion...

pshmath0
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  • Your interpretation of $\Bbb Q[y][x]$ is correct. It is, for most intents and purposes, the same as $\Bbb Q[x,y]$, but as you say, you think differently about what shape an element takes. – Arthur Nov 09 '16 at 13:50
  • Irreducibility in Q[x][y] is a subcase of irreducibility in R[x][y]

    Brute-force way to prove this irreducibility:

    1. Calculate 6 distinct real points on this curve
    2. check all 15 lines (by 2 points)
    3. check all 6 conics (by 5 points)

    If no one of this curves is not a divisor of this polynome - it is irreducible

     – kotomord Nov 09 '16 at 14:04
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    Eisenstein works fine, $y$ is a prime in $\mathbb{Q}[y]$. – Daniel Fischer Nov 09 '16 at 14:34
  • Thanks @DanielFischer, so would the following be okay? Consider $f(x)=x^3+y^2x^2+3yx-y$ in $(\mathbb{Q}[y])[x]$. $y$ is prime in $\mathbb{Q}[y]$. Clearly $y\mid y^2$, $y\mid (3y)$, $y\mid (-y)$, and $y^2\nmid (-y)$. But what about $y\nmid x^3$ ? Certainly my thinking is wrong, but what if $y=2$, $x=2$ ? Then $y\mid x^3$, or do we have to consider $x$ and $y$ for all values in which case $y\nmid x^3$ in general? – pshmath0 Nov 09 '16 at 14:55
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    The coefficient of $x^3$ is $1$. You need $y \nmid 1$. Note that $x$ and $y$ are indeterminates. – Daniel Fischer Nov 09 '16 at 14:57
  • In Eisenstein's criterion, the coefficients of the polynomial must be in $\mathbb{Z}$, and not in $\mathbb{Q}$. So I would have to use Gauss' lemma to go from $(\mathbb{Q}[y])[x]$ to $(\mathbb{Z}[y])[x]$ and consider $y$ prime in $\mathbb{Z}[y]$, not $y$ prime in $\mathbb{Q}[y]$? – pshmath0 Nov 09 '16 at 15:05

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