# Regular ring not UFD [duplicate]

I have to prove that the ring $$R=K[x,y]/(x^2-y^3+y)$$ is not a UFD showing that the prime ideal $$(x,y)R$$ has height 1, but it's not principal.

Do someone know a simple way to prove it? I know there are others way to solve the problem, for example to consider the Picard group to the elliptic curve, but I am interested to solve it in the way I explained. Thanks!

Here is an elementary algebraic solution. I prefer $$R = K[x,y]/(y^2-x^3+x)$$. To show that $$R$$ is not a UFD, it's enough to show that $$y$$ is irreducible in $$R$$ (it's not prime, as $$R/(y)$$ is not a domain; the same reasoning shows it's not a unit). This incidentally also implies that the ideal $$(x,y)$$ isn't principal, as any generator has to divide $$y$$ and $$(x,y) \ne (y)$$.

Suppose $$y$$ factors in $$R$$, so $$y = fg + h(y^2-x^3+x)$$ in $$K[x,y]$$ for some $$f,g,h \in K[x,y]$$. We now use division in $$K[x,y]$$ by the polynomial $$y^2-x^3+x$$ with unique remainder of degree at most 1 in $$y$$.

So WLOG $$f = f_1(x)y+f_2(x)$$ and $$g = g_1(x)y+g_2(x)$$, where $$f_i(x), g_i(x) \in K[x]$$ and we get

$$(f_1(x)y+f_2(x))(g_1(x)y+g_2(x)) \equiv y \pmod{y^2-x^3+x},$$

i.e.

$$f_2 g_1 + f_1 g_2 = 1, \ f_1 g_1 (x^3-x) = - f_2 g_2$$

in $$K[x]$$. We see that $$f_1 = 0$$ iff $$g_2 = 0$$ iff $$f \in K^\times$$. Similarly, $$g_1 = 0$$ iff $$f_2 = 0$$ iff $$g \in K^\times$$. so if our factorisation above is non-trivial, then $$f_1,f_2,g_1,g_2$$ are all non-zero. Moreover, $$\gcd(f_1,f_2) = \gcd(g_1,g_2) = 1$$. Hence $$f_1$$ divides $$g_2$$ and $$g_1$$ divides $$f_2$$ in $$K[x]$$. writing $$g_2 = f_1 v$$ and $$f_2 = g_1 w$$, we get

$$g_1^2 w + f_1^2 v = 1, \ x^3-x = - vw.$$

In particular, $$\deg(v) \not\equiv \deg(w) \pmod 2$$, so $$\deg(g_1^2 w) \not\equiv \deg(f_1^2 v) \pmod 2$$. This implies that either $$g_1^2 w$$ or $$f_1^2 v$$ is of degree 0 and the other vanishes, contradiction.

The ring $K[x,y]$ has Krull dimension $2$. So the ring $R$ has Krull dimension $1$, hence the maximal height of a prime ideal is $1$. And your prime ideal does not have height $0$ because it strictly contains the prime ideal $(x^2-y^3+y)R$.

• I think the more important part is to show that the ideal is not principal.. Aug 10, 2017 at 5:32
• Thank you tiefi. Do you know why is not the Ideal principal? Aug 11, 2017 at 10:18