Show that $x^2$ is irreducible but not prime in the ring of polynomials with no linear term.

Question

Show that $x^2$ is irreducible but not prime in the ring of polynomials with rational coefficients and no linear term: $$R = \{a_0 + a_2 x^2 + ... a_nx^n \in \mathbb Q[x]\}$$

I am actually very confused, I proved that $x^2$ is in fact a prime. Let $x^2|f(x)g(x)$ then $x^2m(x)=f(x)g(x)$ and so $f(0)g(0)=0$. Now assume that$x^2\not |f(x)$ and $x^2\not|g(x)$ then we know that $f(0)\not=0$ otherwise $f(x)=x^2(a_nx^{n-2}+...+a_2)$ same with $g(0)$ hence $g(0)f(0)\not=0$ thus $x^2$ is prime. Where is my error?

@AnuragA but that has a non zero term in front of $x$ so it does not belong to the ring — 2132123, Aug 18 '20 at 20:48
I found the wording very confusing, especially since $\mathbb Q(x)$ means something else. Please make sure the edit matches what you had in mind! — , Aug 18 '20 at 20:53
I'd like to share the thing that hurts my brain every time i see this example: $x^3\neq 0\pmod {x^2}$ in this ring. — rschwieb, Aug 18 '20 at 21:04
@rschwieb which is key in this problem now that I see it. That is very annoying and should not be a problem on an exam, to easy to make a silly mistake :P — 2132123, Aug 18 '20 at 21:20
This example appears here many times, e.g. see here for further discussion. — Bill Dubuque, Aug 24 '20 at 23:40

score 6 · Accepted Answer · answered Aug 18 '20 at 20:59

Your error is that the factor $a_nx^{n-2}+\dots+a_2$ may have a linear term and thus may not be in $R$. For instance, if $f(x)=g(x)=x^3$ then $x^2$ divides $f(x)g(x)=x^6=x^2\cdot x^4$ but $x^2$ does not divide $f(x)$ or $g(x)$ since they would factor as $x^2\cdot x$ but $x\not\in R$.

Show that $x^2$ is irreducible but not prime in the ring of polynomials with no linear term.

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