The ring homomorphic image of an ideal is an ideal

Let \varphi : R \rightarrow S be a ring homomorphism.

  1. Prove that if J \subseteq S is an ideal, then \varphi^\ast[J] \subseteq R is an ideal. Apply this to the special case when R is a subring of S and \varphi is the inclusion map to deduce that if J \subseteq S is an ideal and R \subseteq S a subring, then J \cap R \subseteq R is an ideal.
  2. Prove that if \varphi is surjective and I \subseteq R an ideal then \varphi[I] \subseteq S is an ideal. Give an example where this fails if \varphi is not surjective.

  1. Let x,y \in \varphi^\ast[J]. Now 0 \in J and \varphi(0) = 0, so that 0 \in \varphi^\ast[J]. By hypothesis, \varphi(x-y) = \varphi(x) - \varphi(y) \in J, so that \varphi^\ast[J] is closed under subtraction. Now let r \in R. We have \varphi(rx) = \varphi(r)\varphi(x) \in J since J is an ideal of S; thus rx \in \varphi^\ast[J]. Similarly, xr \in \varphi^\ast[J]. Thus \varphi^\ast[J] is an ideal of R.

    The final deduction is immediate since \iota^\ast[J] = J \cap R.

  2. Note that 0 \in I, so that \varphi(0) \in \varphi[I]. Now let x,y \in \varphi[I]. Now x = \varphi(r) and y = \varphi(s) for some r,s \in I. Since x-y = \varphi(r)-\varphi(s) = \varphi(r-s) \in \varphi[I], \varphi[I] is closed under subtraction. Now let s \in S. Since \varphi is surjective, there exists t \in R with \varphi(t) = s. Now sx = \varphi(t) \varphi(r) = \varphi(tr) \in \varphi[I] since I is an ideal; thus sx \in \varphi[I]. Similarly, xs \in \varphi[I]. Thus \varphi[I] is an ideal of S.

    For the counterexample, consider the inclusion \iota : \mathbb{Z} \rightarrow \mathbb{Q}. \mathbb{Z} is an ideal of itself but not of \mathbb{Q} since 2 \cdot \frac{1}{3} = \frac{2}{3} is not an integer.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: