0
$\begingroup$

$R$ is a field. Let $\phi: R \to S$ be a homomorphism of rings. Show that $\phi$ is injective. Also provide an example of a non-injective homomorphism if $S$ is a field but $R$ is not a field.

Attempt: So here is what we know already. Since $R$ is a field, it is commutative, unitarian, and has units. $\forall x,y \in R$ $\phi(x+y) = \phi(x) + \phi(y)$ and $\phi(xy) = \phi(x) \cdot \phi(y)$. Now, I know that the kernel of $\phi$ is defined as $ker(\phi) = \{\alpha \in R | \phi(\alpha) = 0' \}$ where $0' \in S$. Thanks for the help.

$\endgroup$
  • $\begingroup$ "and has units" - Every ring has units, such as the multiplicative identity $1$. However, what distinguishes a field is that all nonzero elements are units. $\endgroup$ – Karl Kronenfeld Apr 15 '13 at 5:04
  • 2
    $\begingroup$ The claim is false for $S=0$. $\endgroup$ – Martin Brandenburg Apr 15 '13 at 13:36
1
$\begingroup$

HINTS

  1. What are the ideals of a ring?
  2. What do ideals have to do with ring homomorphisms?
$\endgroup$
1
$\begingroup$

Hint (for the second part): You are given a field $S$ and are to find a ring $R$ and a non-injective homomorphism from $R$ into $S$. A good example of such an $R$ isn't going to be pulled out of thin air, but it's going depend on $S$ somehow. Do you know of a way to construct new rings from old ones?

$\endgroup$

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.