Prove that a reduced Gorenstein ring of Krull dimension 1 is not a complete intersection ring.

Let $I = (x^2 - y^2, x^2 - z^2, \omega x - yz, \omega y - xz, \omega z - xy)$.

1. The homomorphism $\lambda$ is the map sending $x, y, z \mapsto 0$, so $\ker(\lambda) = (x,y,z)$ and so $\ker(\lambda)^2 = (x^2, y^2, z^2, xy, xz, yz)$. Considering the relations in $I$ we have that $\ker(\lambda)^2 = (x^2, \omega x, \omega y, \omega z)$ so now, as an $A$-module, we have that $\omega, x, y, z$ all annihilate $\ker(\lambda)/\ker(\lambda)^2$, so it is actually a $A/(\omega, x, y, z)$ module (that is, a $k$-vector space) of dimension $3$, since $x, y, z$ are a linear basis.

2. As $\mathcal{O}$-module, $A$ is generated by the residue classes of $1$ and any possible monomial. It's clear that $1, x, y, z, x^2$ are all independent. Now, we want to show that any other monomial is a $\mathcal{O}$-multiple of either of these. Consider any other monomial $x^a y^b z^c$, at least one of the exponents non-zero. Now the point is, whenever at least two of the exponents are non-zero, you can use the relations in $I$ to lower the degree, e.g. replacing $yz$ with $\omega x$. Keep iterating until you have a single variable, wlog $x^a$. Now if $a > 2$, considering that $x^2 = y^2 = z^2$ in $A$, you can replace $x^a$ with $x^{a-2} y^2$ and iterate the process, until you are left with either $x^2$ (which is equal to $y^2$ and $z^2$, or one among $x, y, z$. This shows that $1, x, y, z, x^2$ is a $\mathcal{O}$-basis for $A$.

3. The ideal $(\omega, x, y, z)$ clearly contains all the others, and also the quotient $A/(\omega, x, y, z)$ is the same as $\mathcal{O}/(\omega) = k$ which is a field. This implies that the ideal is maximal.

4. By "defining relations" they mean the polynomials in $I$, that is, the relations $x^2 - y^2 = 0$, $x^2 - z^2 = 0$, etc. It's easy to see that \[ R = A/(\omega) = k[x,y,z]/(x^2-y^2, x^2-z^2, xy, xz, yz) \] and by definition the embedding dimension is the dimension of $m/m^2$, where $m = (x,y,z)$ is the maximal ideal. But this is $3$ because it has $x, y, z$ as linear basis.

From https://en.wikipedia.org/wiki/Complete_intersection_ring, we have that if a ring is complete intersection then its embedding dimension is bigger than its actual dimension (we have $\mathsf{emb}\dim R = \dim R + \varepsilon_1(R)$). But here $\dim R = 5$, as the $\mathcal{O}$-basis for $A$ projects to a linear $k$-basis of $R$ (they generate and it's easy to check that they're linearly independent).


5. Cohen-Macauley implies that the depth is the same as the dimension (which is $1$). Using the cited Serre's criterion https://en.wikipedia.org/wiki/Serre%27s_criterion_for_normality, $S_k$ holds because the ring is Cohen-Macauley, and $R_0$ holds because the localization at every minimal prime is a field (the minimal primes are listed, so that's just a straightforward check). Since $R_0$ and $S_1$ hold, $A$ is reduced.

Note that here we need that $2$ is invertible, otherwise if e.g. $2=0$ then $(x+y)^2 = x^2 + y^2 = x^2 - y^2 = 0$, so $x+y$ would be nilpotent.

6. Let $M$ be a submodule of $R = A/(\omega)$ (note we have a presentation in point 4.), and let $p(x,y,z) \in M$. Since $1,x,y,z,x^2$ is a $k$-basis for $R$, then $p(x,y,z) = a + bx + cy + dz + ex^2$, with the coefficients in $k$. Now if $a \neq 0$ then $x^2 p(x,y,z) = ax^2 \in M \cap (x^2)$; if $a=0$ and $b \neq 0$ then $x p(x,y,z) = bx^2 \in M \cap (x^2)$, if $a, b = 0$ and (wlog) $c \neq 0$ (it's the same with $d$) then $y p(x,y,z) = cy^2 = cx^2 \in M \cap (x^2)$, and finally if $a,b,c,d=0$ then $ex^2 \in M \cap (x^2)$, so $(x^2)$ is a socle.

7. According to Wikipedia https://en.wikipedia.org/wiki/Gorenstein_ring, a noetherian ring of dimension $0$ is Gorenstein if and only if it has a simple socle as $R$-module, which is the case here. Wikipedia cites Eisenbud (1995), Proposition 21.5 as source for this statement. Do you need me to check that and write the proof here?

I fixed parts 4 and 5, the answer is now complete.