How many commutative rings of order $p^2$ This question was part of my algebra quiz and I was unable to solve 1 option out of 4. So, I am asking it here.

There exists exactly 2 commutative rings of order $p^2$ are there?

I have a intuition that it's true as only two groups exists and as we are adding a multiplicative operation so that must be true but it's given false.
So, kindly tell what mistake I am making. I studied algebra from Joseph Gallian.
If the ring $R$ has characteristic $p^2$, then it must be $\mathbb Z/(p^2)$. Otherwise $R$ has characteristic $p$, so there’s a ring homomorphism $\mathbb F_p\to R$ (and so $R$ is an $\mathbb F_p$-algebra). Take any element $x$ not in the image of $\mathbb F_p$. Then there is an algebra homomorphism $\mathbb F_p[t]\to R$, $t\mapsto x$, necessarily surjective by comparing vector space dimensions. The kernel is generated by a monic polynomial $f$ of degree 2, and we see that there are three possibilities. Either $f$ is irreducible, or is a square, or it has two distinct factors. Up to isomorphism the quotient ring depends only on these three possibilities, not on the particular choice of $f$. This gives three more rings: a field, the ring $\mathbb F_p[t]/(t^2)$, or $\mathbb F_p\times\mathbb F_p$.
There are (at least) $3$ commutative rings of order $4$.
Aside from $\mathbb F_4$ and $\mathbb Z\big/2\mathbb Z\oplus \mathbb Z\big/2\mathbb Z$ we have the matrices of the form
$$
    \begin{bmatrix}
    x &amp; y  \\
    0 &amp; x  \\
  \end{bmatrix}
$$
Where $x,y \in \mathbb F_2$, the field of order $2$.
Shouldn't be difficult to demonstrate that these are all of them, though that doesn't matter for your problem.