Show that the rings $R_1=F_5[x]/(x^2)$ and $R_2=F_5\times F_5$ are not isomorphic.

Question

Show that the rings $R_1=F_5[x]/(x^2)$ and $R_2=F_5\times F_5$ are not isomorphic.($F_5$ is the field with $5$ elements.)

My Work:

Since $(0,1)$ does not have an inverse, $F_5\times F_5$ is not a field. Since $x^2$ is reducible in $F_5[x]$, $F_5[x]/(x^2)$ is not a field. Also both $R_1,R_2$ have same number of elements. So, I was fail to prove that they are not isomorphic. Is there any property that do not satisfied by them together?

Answer 1

Hint: one ring has non-zero nilpotent elements, the other one hasn't.

Answer 2

(1). Let $A$ and $B$ be two rings (commutative with $1$) and let $\phi: A \rightarrow B$ be a homomorphism. If $\phi$ is an isomorphism then every non-zero nilpotent element of $A$ will map to a non-zero nilpotent element of $B.$

In this case, $A:= R_1$ contains a non-zero nilpotent element, $x + (x^2),$ but $B:= R_2$ has no non-zero nilpotent element.

(2). $R_1$ has only one prime ideal, the ideal generated by $x + (x^2)$ where as $R_2$ has two prime ideals, $ 0 \times F_5$ and $F_5 \times 0.$