Let $R$ be a finite ring (with unity) and $S$, $T$ be subrings of $R$. Is $S \cup T$ a subring of R? Let $R$ be a finite ring (with unity) and $S$, $T$ be subrings of $R$. Is $S \cup T$ a subring of R?
(Counterexamples are easy to find to me when $R$ is an infinite ring or a finite rng.)
P.S. I am self-learning undergraduate level mathematics. Sorry if the question is trivial or stupid. Thanks for all answers!
 A: Claim Let $R$ be a ring (with unity) and $S, T$ be subrings of $R$. Then $S \cup T$ is a subring of $R$ if and only if $S \subset T $ or $T \subset S$.
Proof: $\Leftarrow$ is clear, as in this case we have $S \cup T=S$ or $S \cup T=T$
$\Rightarrow$ Assume by contradiction this is not true, then there exists $s \in S \backslash T$ and $t \in T \backslash S$. 
Since $s,t \in S \cup T$ it follows that $s+t \in S \cup T$ and hence $s+t \in S$ or $s+t \in T$.
In the first case we get
$$t=(s+t)-s \in S$$
while in the second we have
$$s=(s+t)-t \in T$$
Contradiction. 
Note It is irrelevant if $R$ is finite or infinite.
P.S. For finite fields, one can use the fact that the multiplicative group is cyclic to prove a stronger result:
If $R$ is a finite ring, it is possible to find subrings $T_1,..,T_n$ so that pairwise they are not included in eachother, and the union is a ring, but we must have $n \geq 3$.
If $K$ is a finite field, and $T_1,..,T_n$ are subfields such that their union is a subfield, then there exists a $j$ so that $T_j$ contains all other $T_i$.
A: Consider the finite field $K=\mathbb{F}_{64}$. It contains the finite fields $\mathbb{F}_8$ and $\mathbb{F}_4$ since $64=2^6$ and $2,3|6$. But since $(2,3)=1$. $\mathbb{F}_8\cap\mathbb{F}_4 = (1)$. Then $\mathbb{F}_8 \cup \mathbb{F}_4$ has size 11, but the smallest ring containing these two is the whole field, since if $\alpha$ is a primitive element for $\mathbb{F}_4$ over $L=\mathbb{F}_2$, then if we let $F=\mathbb{F}_8$, $F[\alpha]$ is a field since $\alpha$ is necessarily algebraic over $L\subset F$, and the field necessarily contains $F$ and $L[\alpha]=\mathbb{F}_4$, which since the size of the field is $2^d$ implies $2,3|d$ and hence $6|d$, so $F[\alpha]=K$ the whole field.
Note that a primitive element definitely exists since the units group of a finite field is cyclic so a generator of the units group of the larger field will be a primitive element for the field over any subfield.
A: Although N.S. has handled the general case, I thought another counterexample might be helpful.
Take any finite ring $A$, and form from it the ring $A[x,y]$ of polynomials in two variables with coefficients in $A$.  Now factor out by the relations $x^n=1$ and $y^m=1$ for some $n,m$.  Now we have $R=A[x,y]/(x^n,y^m)$ -- a finite ring.  Inside of $R$, we find the image of $A[x]$ and call it $S$, and the image of $A[y]$ and call it $T$.
Informally, $S$ "looks like" the set of polynomials in a single variable $x$, with coefficients in $A$, of degree $< n$, and $T$ "looks like" the set of polynomials in a single variable $y$, with coefficients in the same ring, of degree $< m$. 
$S$ and $T$ are subrings of $R$, but $S \cup T$ is not, because it is not closed under addition or multiplication:  $x \in S$ and $y \in T$ but $x+y \notin S \cup T$ and $xy \notin S \cup T$.
