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How can I show that the ring $R=F[x,y,z]/(x^2-yz)$ is an integral domain? (Here $F$ is a field.)

I tried to prove this by contradiction. Suppose there are $\bar f, \bar g \in R-\{0\}$ such that $\bar f\bar g =0 $ in $R$, i.e. $fg$ is divisible by $x^2-yz$ in $F[x,y,z]$. But I don't know how to proceed.

user26857
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user557
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  • Unique factorisation? – Angina Seng Jun 20 '18 at 01:11
  • @LordSharktheUnknown What do you mean by this? $R$ doesn't have unique factorization. $F[x,y,z]$ does, but if you mean that I should apply this, I don't see how. – user557 Jun 20 '18 at 01:14
  • That's just writing out the definition. What have you tried? – anomaly Jun 20 '18 at 01:19
  • @anomaly Which definition do you mean? I wrote out the definition (rather its negation). I can also write out the definition of $fg$ being divisible by $x^2-yz$, but how will this help? – user557 Jun 20 '18 at 01:31
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    Related: https://math.stackexchange.com/questions/1394545/irreducibility-criteria-for-polynomials-with-several-variables – user557 Jun 21 '18 at 22:58
  • Hint to a proof that works for any integral domain $R$: First, construct an $R$-algebra homomorphism $f : R \to F\left[a,b\right]$ that sends $x$, $y$ and $z$ to $ab$, $a^2$ and $b^2$. (This is straightforward using the universal property of $R$.) Then, show that $f$ is injective. (This is tricky. One way is to show that each element of $R$ can be written as $u+vx$ with $u, v \in F\left[y,z\right]$, and then argue that $f$ sends $u+vx$ to $0$ only if both $u$ and $v$ are $0$. This relies on evenness and oddness of exponents in $a^i b^j$.) Combined, these yield ... – darij grinberg Aug 09 '18 at 16:32
  • ... that $F$ embeds as a subring into $F\left[a,b\right]$. Since $F\left[a,b\right]$ is an integral domain, this shows that $F$ is an integral domain. – darij grinberg Aug 09 '18 at 16:33
  • Related: https://math.stackexchange.com/questions/2394949/proving-that-ix-is-irreducible-in-r-i-where-r-kx-y-z-and-i-langle-x – user557 Aug 15 '18 at 14:12

2 Answers2

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Consider $p(x,y,z):=x^2-yz$ as a polynomial in $R[x]$, where $R:=F[y,z]$ is a unique factorization domain. If we can show that $p(x,y,z)$ is irreducible in $R[x]=F[x,y,z]$, then we are done. First, note that $y$ is a prime element of $R$. The constant term of $p(x,y,z)$ as a polynomial in $x$ is $-yz$ which is not in the ideal generated by $y^2$ of $R$. The coefficient of $x$ in $p(x,y,z)$ is $0$, which belongs in the ideal generated by $y$ of $R$. Using Eisenstein's Criterion for integral domains, we conclude that $p(x,y,z)$ is irreducible in $R[x]=F[x,y,z]$. Since $F[x,y,z]$ is a unique factorization domain, $p(x,y,z)=x^2-yz$ is a prime element. That is, $$F[x,y,z]/\langle x^2-yz\rangle$$ is an integral domain.

Batominovski
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    Good idea to use Eisenstein, but to show that a degree 2 polynomial is irreducible can be done by elementary means. Moreover, we can also consider $p$ as a polynomial of degree 1 in $z$. – user26857 Aug 09 '18 at 19:43
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$x^2-yz$ is irreducible, so $x^2-yz$ is prime, so $F[x,y,z]/(x^2-yz)$ is integral domain.

Kenny Lau
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    Why is it irreducible? – user557 Jun 20 '18 at 01:15
  • @user538518 by a degree argument – Kenny Lau Jun 20 '18 at 01:15
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    I tried to express $x^2-yz=(x^2+\beta y+\gamma z + \delta)(x+by+cz+d)$; I got in particular $b\delta+\beta c=-1$ and $\beta b=0, \beta+b=0; \delta+c=0, \delta c = 0$, which are contradictory... – user557 Jun 20 '18 at 01:33
  • I'd just trust my instinct that it is irreducible. – Kenny Lau Jun 20 '18 at 01:36
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    I see. But math is about proving things, not trusting instincts... Thanks anyway. – user557 Jun 20 '18 at 01:44
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    @user538518 By degree couting reason, you can conclude that. Treat $k[x,y,z]=(k[y,z])[x]=R[x]$. If it factors, it can only factor into linear polynomial in $x$ with constant term as element of $R$. You have two possibilities for constant term either all $yz$ or $y$.($z$ and $y$ are symmetric in position. This reduced to 2 cases only.)Then conclude the cross term will not vanish for $y$ only. Similarly for constant term as $xy$ as well. – user45765 Jun 20 '18 at 02:10
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    The ring $R = F[y,z]$ is a UFD and the monic polynomial $x^2 - yz$ in $R[x]$ is Eisenstein at $y$, so it is irreducible in $R[x]$. Since $R[x]$ is a UFD, its irreducible elements are prime, so $x^2 - yz$ is prime and thus $(x^2 - yz)$ is a prime ideal in $R[x]$. The same reasoning shows $(x^n-yz)$ is a prime ideal for all $n \geq 1$, so $R[x]/(x^n-yz)$, which is $F[x,y,z]/(x^n-yz)$, is an integral domain. – KCd Aug 09 '18 at 13:52