0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is the definition of homomorphism? $\endgroup$ – user60589 Dec 13 '15 at 21:07
  • 1
    $\begingroup$ 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 $a, b \in \mathbb{R}$. $\endgroup$ – Bennie Joseph Vassallo Dec 13 '15 at 21:10
  • $\begingroup$ So there is a morphism from the trivial ring to any ring. For the first counterexample: can you construct a ring from a field? $\endgroup$ – user60589 Dec 13 '15 at 21:11
  • $\begingroup$ For non-trivial counterexample for the last two statements take $R$ to be the additive group generated by $1 \in S$. $\endgroup$ – user60589 Dec 13 '15 at 21:18
  • $\begingroup$ But I'm not studying group theory. I need to prove using ring theory. $\endgroup$ – Bennie Joseph Vassallo Dec 13 '15 at 21:23
0
$\begingroup$

$R$ 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.

The trivial ring embeds inside any ring.

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

Again, consider the trivial ring, or the example with real matrices above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.