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Let say we have a polynomial $p(x)$ that is irreducible in $F[x]$. Is it also irreducible in $F[x,y]$? ($x$, $y$ are both indeterminates.)

It seems intuitive that this is true, but how do we prove it?

Thanks.

Is this attempted proof ok?

Attempted proof:

Suppose $p(x)$ is not irreducible in $F[x,y]$. Then $p(x)=f(x,y)g(x,y)$ for some non-constant $f,g\in F[x,y]$.

Since $p$ is solely a function of $x$, so are $f, g$. This contradicts that $p$ is irreducible in $F[x]$.

Watson
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yoyostein
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2 Answers2

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In $p(x)=f(x,y)g(x,y)$, think of everything as a polynomial in $y$ with coefficients from $F[x]$. The only way the product of two polynomials in $y$ can be constant is if they are both constant (as long as there are no zero divisors).

Concretely: if the highest power of $y$ in $f$ is $y^n$ and in $g$ is $y^m$ then $fg$ must have $y^{n+m}$.

anon
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    +1, probably the best way to see this. Alternatively, if one had a nontrivial factorization $p(x) = f(x, y)g(x, y)$, then reducing modulo the prime ideal $\langle y \rangle \subset F[x, y]$ would give a nontrivial factorization of $p(x)$ in $F[x]$, a contradiction. – Alex Wertheim Aug 17 '16 at 03:59
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    @AlexWertheim Right, it'd imply one of $f(x,y)$ or $g(x,y)$ was constant with respect to $x$, i.e. purely a function of $y$, and then the same "there is a $y$ somewhere in the expanded product" problem occurs. – anon Aug 17 '16 at 04:02
  • Nice. Is there anything wrong with my proof? Please point it out, thanks! – yoyostein Aug 17 '16 at 05:03
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    @yoyostein : There is nothing wrong with your proof, but one could criticize that you dont give an argument for your claim, that $f,g$ are also functions solely of $x$. – MooS Aug 17 '16 at 06:32
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If $F$ is a factorial ring (a field is a factorial ring), we can invoke the Gauß lemma to see that $F[x,y]$ is factorial, i.e. irreducible and prime coincide and we can test irreducibility by considering the quotient ring. Then we can argue as follows:

$F[x,y]/ \langle p(x) \rangle \cong \Big(F[x]/ \langle p(x) \rangle \Big)[y]$ is a domain, since $F[x]/ \langle p(x) \rangle$ is a domain by the assumption and the polynomial ring over a domain is a domain.

MooS
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