Give example of two subrings of R such that their set product and sum is not a subring.

Question

Let $ R$ be a ring. Let $S $ and $T $ be two subrings of $R$ . Define $$S+T=\{ s+t : s \in S, t \in T \}$$

$$S.T=\{s.t: s\in S, t\in T\}$$

I took $\mathbb{Z}, \mathbb{Z}[x]$ as example but was not able to contruct anything.

Edit:

Ring does not need to have identity element. I am following Gallian approach. I have read that in Artin's book, author assumes that ring need to have identity.

Edit 2: Is there a reason I am not able to construct example in $\mathbb{Z}, \mathbb{Z}[x]$ or it is just my mathematicial imaturity.

Thanks in advance.

Answer 1

Consider the ring $\mathbb{Z}[x,y]$ of polynomials in two variables. Then $S:=\mathbb{Z}[x]$ and $T:=\mathbb{Z}[y]$ are subrings such that their sum and product (pairwise defined as you did) are not subrings, as the elements $(x+y)(x+y)=x^2+2xy+y^2\not\in S+T$ and $x^2y+xy^2\not\in ST$ show.

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EDITED IN RESPONSE TO EDITION OF THE OP:

1) In $\mathbb{Z}[x]$ we can give the following example: $S:=\mathbb{Z}[x^2]$, $T:=\mathbb{Z}[x^3]$. Then $(x^2+x^3)^2\not\in S+T$, $x^4x^3+x^2x^6=x^7+x^8\not\in ST$.

2) In $\mathbb{Z}$ this cannot happen: every subring is an ideal, so $S:=n\mathbb{Z}$, $T:=m\mathbb{Z}$ for some $n,m\in\mathbb{Z}$, and then $S+T=\gcd(n,m)\mathbb{Z}$ (by Bézout's identity) and $ST=nm\mathbb{Z}^2=nm\mathbb{Z}$, which are again subrings.

Answer 2

Too long for a comment, about edit 2:

Note Let $R$ be a subring of $\mathbb Z$. Then since $(R,+)$ is a subgroup of $(\mathbb Z,+)$ you get that $R=n \mathbb Z$ for some $n$.

It is easy to see that $$(n\mathbb Z) +(m \mathbb Z)=\mbox{gcd}(m,n) \mathbb Z \\ (m\mathbb Z) (n\mathbb Z) =(mn) \mathbb Z$$

That's why such example cannot exist in $\mathbb Z$. The reason why this proof holds is because the additive group of the ring is cylcic.