# Let $\phi: R \rightarrow S$ be an injective homomorphism between two rings $R$ and $S$. Prove or provide a counterexample:

Let $\phi: R \rightarrow S$ be an injective homomorphism between two rings $R$ and $S$. Prove or provide a counterexample:

$R$ is a field implies that $S$ is a field.

$R$ has an identity implies $S$ has an identity.

$R$ is commutative implies $S$ is commutative.

I am using the map $\phi: \mathbb{R} \rightarrow M(\mathbb{R})$ to consider because $\phi$ is an injective homomorphism, but I cannot seem to think of how to prove any of the properties.

• What is the definition of homomorphism? – user60589 Dec 13 '15 at 21:07
• If $R$ and $S$ are rings, then a function $f: R \rightarrow S$ is a homomorphism if $f(a+b)=f(a)+f(b)$ and $f(ab)=f(a)f(b)$ for all $a, b \in \mathbb{R}$. – Bennie Joseph Vassallo Dec 13 '15 at 21:10
• So there is a morphism from the trivial ring to any ring. For the first counterexample: can you construct a ring from a field? – user60589 Dec 13 '15 at 21:11
• For non-trivial counterexample for the last two statements take $R$ to be the additive group generated by $1 \in S$. – user60589 Dec 13 '15 at 21:18
• But I'm not studying group theory. I need to prove using ring theory. – Bennie Joseph Vassallo Dec 13 '15 at 21:23

$R$ is a field implies that $S$ is a field.
Consider $\mathbb{R}$ embedded inside the ring of $2 \times 2$ real matrices or the ring $\mathbb{R}[x]$ of polynomials over $\mathbb{R}$.
$R$ has an identity implies $S$ has an identity.
$R$ is commutative implies $S$ is commutative.