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added 36 characters in body
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Dag Oskar Madsen
Dag Oskar Madsen
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Here is a counterexample.

Let $\Bbbk$ be a field and let $A=\Bbbk [x_0,x_1,x_2,\ldots]$ be the polynomial ring in countably many variables with its natural grading. Consider the ideal $I$ generated by $\{x_i x_j \mid i \neq j\}$$\{x_i x_j \mid i,j\in \mathbb N, i \neq j\}$, $\{x_i^2-x_j^2 \mid i \neq j\}$$\{x_i^2-x_j^2 \mid i,j\in \mathbb N, i \neq j\}$ and all monomials of degree three. In short $$I=\langle x_i x_j, x_i^2-x_j^2,A_3\rangle.$$

Then $M=A/I$ is an $A$-module with an induced grading $M=M_0 \oplus M_1 \oplus M_2$, and all submodules of $M$ have a non-zero intersection with the simple socle $\operatorname{Soc} M=M_2$. Therefore $M$ has uniform dimension one.

On the other hand $M/\operatorname{Soc} M$ is isomorphic to $A/\langle A_2 \rangle$ which has an infinite dimensional socle $M_1=A_1$. Therefore $M/\operatorname{Soc} M$ has infinite uniform dimension.

Here is a counterexample.

Let $\Bbbk$ be a field and let $A=\Bbbk [x_0,x_1,x_2,\ldots]$ be the polynomial ring in countably many variables with its natural grading. Consider the ideal $I$ generated by $\{x_i x_j \mid i \neq j\}$, $\{x_i^2-x_j^2 \mid i \neq j\}$ and all monomials of degree three. In short $$I=\langle x_i x_j, x_i^2-x_j^2,A_3\rangle.$$

Then $M=A/I$ is an $A$-module with an induced grading $M=M_0 \oplus M_1 \oplus M_2$, and all submodules of $M$ have a non-zero intersection with the simple socle $\operatorname{Soc} M=M_2$. Therefore $M$ has uniform dimension one.

On the other hand $M/\operatorname{Soc} M$ is isomorphic to $A/\langle A_2 \rangle$ which has an infinite dimensional socle $M_1=A_1$. Therefore $M/\operatorname{Soc} M$ has infinite uniform dimension.

Here is a counterexample.

Let $\Bbbk$ be a field and let $A=\Bbbk [x_0,x_1,x_2,\ldots]$ be the polynomial ring in countably many variables with its natural grading. Consider the ideal $I$ generated by $\{x_i x_j \mid i,j\in \mathbb N, i \neq j\}$, $\{x_i^2-x_j^2 \mid i,j\in \mathbb N, i \neq j\}$ and all monomials of degree three. In short $$I=\langle x_i x_j, x_i^2-x_j^2,A_3\rangle.$$

Then $M=A/I$ is an $A$-module with an induced grading $M=M_0 \oplus M_1 \oplus M_2$, and all submodules of $M$ have a non-zero intersection with the simple socle $\operatorname{Soc} M=M_2$. Therefore $M$ has uniform dimension one.

On the other hand $M/\operatorname{Soc} M$ is isomorphic to $A/\langle A_2 \rangle$ which has an infinite dimensional socle $M_1=A_1$. Therefore $M/\operatorname{Soc} M$ has infinite uniform dimension.

forgot to divide out monomials of degree three
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Dag Oskar Madsen
Dag Oskar Madsen
  • 3.7k
  • 3
  • 28
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Here is a counterexample.

Let $\Bbbk$ be a field and let $A=\Bbbk [x_1,x_2,\ldots]$$A=\Bbbk [x_0,x_1,x_2,\ldots]$ be the polynomial ring in countably many variables with its natural grading. Consider the quadratic ideal $I$ generated by $\{x_i x_j,x_i^2-x_j^2\}_{i \neq j}$$\{x_i x_j \mid i \neq j\}$, $\{x_i^2-x_j^2 \mid i \neq j\}$ and all monomials of degree three. In short $$I=\langle x_i x_j, x_i^2-x_j^2,A_3\rangle.$$

Then $M=A/I$ is an $A$-module with aan induced grading $M=M_0 \oplus M_1 \oplus M_2$, and all submodules of $M$ have a non-zero intersection with the simple socle $\operatorname{Soc} M=M_2$. Therefore $M$ has uniform dimension one.

On the other hand $M/\operatorname{Soc} M$ is isomorphic to $A/\langle A_2 \rangle$ which has an infinite dimensional socle $M_1=A_1$. Therefore $M/\operatorname{Soc} M$ has infinite uniform dimension.

Here is a counterexample.

Let $\Bbbk$ be a field and let $A=\Bbbk [x_1,x_2,\ldots]$ be the polynomial ring in countably many variables with its natural grading. Consider the quadratic ideal $I$ generated by $\{x_i x_j,x_i^2-x_j^2\}_{i \neq j}$.

Then $M=A/I$ is an $A$-module with a grading $M=M_0 \oplus M_1 \oplus M_2$, and all submodules of $M$ have a non-zero intersection with the simple socle $\operatorname{Soc} M=M_2$. Therefore $M$ has uniform dimension one.

On the other hand $M/\operatorname{Soc} M$ is isomorphic to $A/\langle A_2 \rangle$ which has an infinite dimensional socle $M_1=A_1$. Therefore $M/\operatorname{Soc} M$ has infinite uniform dimension.

Here is a counterexample.

Let $\Bbbk$ be a field and let $A=\Bbbk [x_0,x_1,x_2,\ldots]$ be the polynomial ring in countably many variables with its natural grading. Consider the ideal $I$ generated by $\{x_i x_j \mid i \neq j\}$, $\{x_i^2-x_j^2 \mid i \neq j\}$ and all monomials of degree three. In short $$I=\langle x_i x_j, x_i^2-x_j^2,A_3\rangle.$$

Then $M=A/I$ is an $A$-module with an induced grading $M=M_0 \oplus M_1 \oplus M_2$, and all submodules of $M$ have a non-zero intersection with the simple socle $\operatorname{Soc} M=M_2$. Therefore $M$ has uniform dimension one.

On the other hand $M/\operatorname{Soc} M$ is isomorphic to $A/\langle A_2 \rangle$ which has an infinite dimensional socle $M_1=A_1$. Therefore $M/\operatorname{Soc} M$ has infinite uniform dimension.

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Dag Oskar Madsen
Dag Oskar Madsen
  • 3.7k
  • 3
  • 28
  • 51

Here is a counterexample.

Let $\Bbbk$ be a field and let $A=\Bbbk [x_1,x_2,\ldots]$ be the polynomial ring in countably many variables with its natural grading. Consider the quadratic ideal $I$ generated by $\{x_i x_j,x_i^2-x_j^2\}_{i \neq j}$.

Then $M=A/I$ is an $A$-module with a grading $M=M_0 \oplus M_1 \oplus M_2$, and all submodules of $M$ have a non-zero intersection with the simple socle $\operatorname{Soc} M=M_2$. Therefore $M$ has uniform dimension one.

On the other hand $M/\operatorname{Soc} M$ is isomorphic to $A/\langle A_2 \rangle$ which has an infinite dimensional socle $M_1=A_1$. Therefore $M/\operatorname{Soc} M$ has infinite uniform dimension.