Abstract Algebra- Ideals Given an ideal $I\subset R$, define $J=\{\Phi(a):a\in I\}\subset S$. Prove that $J$ is an ideal, provided $\Phi$ maps onto $S$. Give an example to demonstrate that the latter hypothesis is necessary.
Ok, so for this, I feel like a need to make a counterexample (Injection Map) of the form $\Phi(x)=x$ from $R$ to $S$ where $R\subseteq S$. That may help me with the proof, but the thing is I don't know how to do that. Frankly, I don't really know what this question is asking. I know I need to show it holds for multiplication and addition, but I seem to have trouble showing that. Could someone please help with this. I would really appreciate it. Thank You.
 A: We shall assume we're talking/writing/typing about rings, and that $\Phi:R \to S$ is a ring homomorphism. ;-)!!!
This being the case, to show $J$ is an ideal, we need to establish two things:
i.)  for $s_1, s_2 \in J$, $s_1 - s_2 \in J$; and
ii.) for $s \in J$ and $r \in S$, $rs, sr \in J$.
For (i.), since $s_1, s_2 \in J$, there exist $t_1, t_2 \in I$ with $\Phi(t_1) = s_1$ and $\Phi(t_2) = s_2$.  Then $t_1 - t_2 \in I$, whence $s_1 - s_2 = \Phi(t_1) - \Phi(t_2) = \
\Phi(t_1 - t_2) \in J$.
As for (ii.), for $s \in J$ we have some $t \in I$ with $\Phi(t) = s$.  For any $r \in S$ there is a $v \in R$ with $r = \Phi(v)$, since $\Phi$ is surjective.  Then $rs = \Phi(v)\Phi(t) = \Phi(vt) \in J$, since $vt \in I$, $I$ being an ideal in $R$.  A similar argument shows that $sr \in J$.
These demonstrations of (i.) and (ii.) establish that $J$ is a (two-sided) ideal in $S$, assuming $I$ is such in $R$.  If $I$ is merely one-sided, the proof clearly still works; it works no matter what the sided-ness of $I$ may be, left-, right-, or two-!  QED
Nota Bene:  (Edited in 10 November 2013 6:47 PM PST)  Here is an example showing the surjectivity of $\Phi:R \to S$ is essential to the truth of the above assertion:  let $R$ be a ring and let $M_n(R)$ be the ring of $n \times n$ matrices over $R$.  Consider the map $\Phi:R \to M_n(R)$ defined by $\Phi(r) = \text{diag}(r, r, . . . , r) \, $, i.e. for $r \in R$, $\Phi(r)$ is the $n \times n$ matrix each diagonal entry of which is $r$, with zeroes everywhere else.  For any ideal $I \subset R$, $\Phi(I)$ is the set of all matrices $\text{diag}(s, s, . . . ,s)$ where $s \in I$.  Evidently, $\Phi(I)$ is closed under addition and subtraction.  But for the general matrix $A \in M_n(R)$, with $A = [r_{ij}]$, we have $\text{diag}(s, s, . . ., s)A = [sr_{ij}]$ and $A\text{diag}(s, s, . . ., s) = [r_{ij}s]$, neither of which are diagonal in the general case, hence neither of which are in $\Phi(I)$.  Note that $R$ need be neither commutative nor unital for these assertions to bind.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
