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Let $R$ and $R'$ be rings and let $\phi: R\mapsto R'$ be a ring homomorphism and $N$ an ideal of $R$.

Show that $\phi[N]$ is an ideal of $\phi[R]$, and give an example to show that $\phi[N]$ need not be an ideal of $R'$.

Let $N'$ be an ideal of either $\phi[R]$ or R', and show that $\phi^{-1}[N']$ is an ideal of R.

I just started learning ideals so I am having a lot of trouble with this. I know that the kernel of $\phi$ is an ideal, but I don't know how I can use this.

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  • $\begingroup$ I guess you mean $\phi(N)$ need not to be ideal of $R'$. $\endgroup$ – mesel Mar 5 '14 at 16:49
  • $\begingroup$ that's correct, sorry. $\endgroup$ – terrible at math Mar 5 '14 at 16:54
  • $\begingroup$ $\phi$ is from $R$ to $R'$ , how can you talk about $\phi(R^{'})$ $\endgroup$ – mesel Mar 5 '14 at 16:57
  • $\begingroup$ sorry I made lots of mistakes typing this out, edited. $\endgroup$ – terrible at math Mar 5 '14 at 17:00
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In order to show that $\phi(N)$ is an ideal we must show that it is an additive group and closed under multiplication in $\phi(R)$

Showing $\phi(N)$ is an additive group

Identity

So as $0\in N$ and $\phi$ is a ring homomorphism we have that $\phi(0)=0\in \phi(N)$

Closure

Now take $x,y\in \phi(N)$ then by definition of $\phi(N)$ there must exists $a,b\in N$ such that $\phi(a)=x$ and $\phi(b)=y$. Now as $N$ is an ideal we have that $a+b\in N$ and so $\phi(a+b)\in \phi(N)$, then as $\phi$ is a ring homomorphism we have that $\phi(a+b)=\phi(a)+\phi(b)=x+y\in \phi(N)$.

Inverses

If we have $x\in \phi(N)$ then by definition there must be an $a\in N$ such that $\phi(a)=x$ now as $N$ is an ideal we have that $a^{-1}\in N$ and so $\phi(a^{-1})\in \phi(N)$. Now as $\phi$ is a ring homomorphism we have that $\phi(a^{-1})=\phi(a)^{-1}=x^{-1}\in \phi(N)$

So we have that $\phi(N)$ is an additive subgroup of $\phi(R)$

Now to show that it is an ideal we have to show that if we have $r\in \phi(R)$ then $r\phi(N)\subset \phi(N)$ and $\phi(N)r\subset \phi(N)$

Showing $\phi(N)$ is closed under multiplication of $\phi(R)$

Now if $r\in \phi(R)$ then there must be an $r'\in R$ such that $\phi(r')=r$

Then as $r'N\subset N$ we have that $\phi(r'N)=\phi(r')\phi(N)\subset \phi(R)$ (same argument for $r$ on the right)

Example of image of an ideal is not an ideal

We can define the inclusion map $f:\mathbb{Z}\rightarrow \mathbb{Q}$ and then take an ideal in $\mathbb{Z}$ say $3\mathbb{Z}$ then $f(3\mathbb{Z})$ is not an ideal in $\mathbb{Q}$.

Suppose that it was, then noting that $1\not\in f(3\mathbb{Z})$. But by the definition of ideal we must have, $\frac{1}{3}\in\mathbb{Q}$, $\frac{1}{3}\times 3\in f(3\mathbb{Z})$ which gives a contradiction.

Note: as $\mathbb{Q}$ is a field then it has only the two trivial ideals so it follows directly from this.

Another example is given here: The image of an ideal under a homomorphism may not be an ideal

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  • $\begingroup$ just reading the first couple lines -- is it clear that $\phi(R)$ is a ring? I mean, to show we have an ideal of a ring we need to first establish that we have a ring at all. It's probably true that the image of any ring homomorphism is always a ring, but I just want to be sure. Never hurts in math. $\endgroup$ – terrible at math Mar 5 '14 at 17:07
  • $\begingroup$ @terribleatmath as $R\lhd R$ if we show that $\phi(N)$ is an ideal then this is the same as showing $\phi(R)$ is a ring, pretty much. I mean it is basically what I have done above- you should be able to adapt it pretty easily $\endgroup$ – hmmmm Mar 5 '14 at 17:09
  • $\begingroup$ It's easy to check directly: $\phi(a)\phi(b)=\phi(ab),\phi(a)-\phi(b)=\phi(a-b)$. $\endgroup$ – Kevin Carlson Mar 5 '14 at 17:11
  • $\begingroup$ @hmmmm: if you find an example such that $\phi(N)$ is not ideal of $R^{'}$ then your answer is complete. :) $\endgroup$ – mesel Mar 5 '14 at 17:11
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    $\begingroup$ @terribleatmath it is just running essentially the same argument over and over that things in the image of $\phi$ must have a pre-image in $R$. For these sorts of questions when you say "I don't think I know anything about the inverse image"- you do. You need a bit more confidence to stick with it for longer. You know its the inverse image of a ring homomorphism, so the only thing that you can use is that information- you then just need to play around with it for a while and you will get there :) $\endgroup$ – hmmmm Mar 5 '14 at 23:58

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