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The "ring $B$ is perfect" assertion implicitly stated, other small details added.
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Leonid Positselski
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Let $k$ be a field. Consider the commutative algebra of polynomials $A=k[x_1,x_2,x_3,\dotsc]$ in a countable set of variables $x_1$, $x_2$, $x_3,\,\dots$ with coefficients in $k$. Let $I$ be the ideal in $A$ generated by the following elements:

$x_n^2$, for every $n\ge1$,;

$x_ix_j$, for every $i\ge1$ and $j\ge2i$.

Then the quotient algebra $B=A/I$ is a commutative local $k$-algebra whose maximal ideal $J$ (generated by the elements $x_1$, $x_2$, $x_3,\,\dots$) is $T$-nilpotent, but not nilpotent.

Indeed, every monomial in the variables $x_1$, $x_2$, $x_3,\,\dots$ divisible by $x_n$ and having length exceeding $n$ for some $n\ge1$ vanishes in $B$; so. So the ideal $J\subset B$ is $T$-nilpotent and the ring $B$ is perfect. But

But the monomial $x_nx_{n+1}\dotsm x_{2n-1}$ of length $n$ is nonzero in $B$ for every $n\ge1$, so. So the idealJacobson radical (= nilradical) $J$ of the ring $B$ is not nilpotent.

(I haven't invented this example for the occasion, but rather have recently seen it somewhere. But I forgot the source.)

Let $k$ be a field. Consider the commutative algebra of polynomials $A=k[x_1,x_2,x_3,\dotsc]$ in a countable set of variables $x_1$, $x_2$, $x_3,\,\dots$ with coefficients in $k$. Let $I$ be the ideal in $A$ generated by the following elements:

$x_n^2$, for every $n\ge1$,

$x_ix_j$, for every $i\ge1$ and $j\ge2i$.

Then the quotient algebra $B=A/I$ is a commutative local $k$-algebra whose maximal ideal $J$ (generated by the elements $x_1$, $x_2$, $x_3,\,\dots$) is $T$-nilpotent, but not nilpotent.

Indeed, every monomial in the variables $x_1$, $x_2$, $x_3,\,\dots$ divisible by $x_n$ and having length exceeding $n$ for some $n\ge1$ vanishes in $B$; so the ideal $J\subset B$ is $T$-nilpotent. But the monomial $x_nx_{n+1}\dotsm x_{2n-1}$ of length $n$ is nonzero in $B$ for every $n\ge1$, so the ideal $J$ is not nilpotent.

(I haven't invented this example for the occasion, but rather have seen it somewhere. But I forgot the source.)

Let $k$ be a field. Consider the commutative algebra of polynomials $A=k[x_1,x_2,x_3,\dotsc]$ in a countable set of variables $x_1$, $x_2$, $x_3,\,\dots$ with coefficients in $k$. Let $I$ be the ideal in $A$ generated by the following elements:

$x_n^2$, for every $n\ge1$;

$x_ix_j$, for every $i\ge1$ and $j\ge2i$.

Then the quotient algebra $B=A/I$ is a commutative local $k$-algebra whose maximal ideal $J$ (generated by the elements $x_1$, $x_2$, $x_3,\,\dots$) is $T$-nilpotent, but not nilpotent.

Indeed, every monomial in the variables $x_1$, $x_2$, $x_3,\,\dots$ divisible by $x_n$ and having length exceeding $n$ for some $n\ge1$ vanishes in $B$. So the ideal $J\subset B$ is $T$-nilpotent and the ring $B$ is perfect.

But the monomial $x_nx_{n+1}\dotsm x_{2n-1}$ of length $n$ is nonzero in $B$ for every $n\ge1$. So the Jacobson radical (= nilradical) $J$ of the ring $B$ is not nilpotent.

(I haven't invented this example for the occasion, but rather have recently seen it somewhere. But I forgot the source.)

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Leonid Positselski
  • 15.6k
  • 1
  • 57
  • 95

Let $k$ be a field. Consider the commutative algebra of polynomials $A=k[x_1,x_2,x_3,\dotsc]$ in a countable set of variables $x_1$, $x_2$, $x_3,\,\dots$ with coefficients in $k$. Let $I$ be the ideal in $A$ generated by the following elements:

$x_n^2$, for every $n\ge1$,

$x_ix_j$, for every $i\ge1$ and $j\ge2i$.

Then the quotient algebra $B=A/I$ is a commutative local $k$-algebra whose maximal ideal $J$ (generated by the elements $x_1$, $x_2$, $x_3,\,\dots$) is $T$-nilpotent, but not nilpotent.

Indeed, every monomial in the variables $x_1$, $x_2$, $x_3,\,\dots$ divisible by $x_n$ and having length exceeding $n$ for some $n\ge1$ vanishes in $B$; so the ideal $J\subset B$ is $T$-nilpotent. But the monomial $x_nx_{n+1}\dotsm x_{2n-1}$ of length $n$ is nonzero in $B$ for every $n\ge1$, so the ideal $J$ is not nilpotent.

(I haven't invented this example for the occasion, but rather have seen it somewhere. But I forgot the source.)