# The image of an ideal under a homomorphism may not be an ideal

This is an elementary question about ideals. Consider a ring homomorphism $$f: \mathbb{Z} \rightarrow \mathbb{Z}[x],$$ and consider the ideal $\left< 2\right>$ in $\mathbb{Z}$. When why is it that $f(\left< 2\right>)$ is not an ideal?

Some websites say that $f(\left< 2\right>)$ is not an ideal because it does not contain nonconstant polynomials. That still doesn't make sense on why it is not an ideal.

Thank you all.

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For it to be an ideal, you must have, for example, $2x \in f((2)),$ which clearly doesn't occur (assuming that $f$ is the natural inclusion). – Andrew Jul 4 '12 at 18:52
Thank you user34377! Since the map $f$ is not given, I wasn't sure whether the map should send $1\mapsto 1$, or $1\mapsto x$. Considering $f$ is the natural inclusion map, why is it that $2x\not\in f((2))$? In the example $(x)\subseteq \mathbb{C}[x,y,z]$, we have $xyz \in (x)$. – math-visitor Jul 4 '12 at 18:55
Since this is a ring homomorphism (of commutative rings), we require that $1\mapsto 1.$ – Andrew Jul 4 '12 at 19:00
@math-visitor, the image of the inclusion map is exactly $\mathbb{Z}$, and $2x$ is not in there. – Adeel Jul 4 '12 at 19:02
@math-visitor, both of the statements in your comment above are correct. But you are not looking at any ring homomorphisms, so the problem from your original question isn't arising. Consider the inclusion homomorphism $f:\mathbb{C}[y]\to\mathbb{C}[x,y,z]$, and the ideal $I=(2)\subset \mathbb{C}[y]$ and the ideal $J=(2)\subset\mathbb{C}[x,y,z]$. Then $$f(I)=\{g(x,y,z)\in\mathbb{C}[x,y,z]\mid g=2h\text{ for some }h\in\mathbb{C}[y]\}$$ which is not equal to $J$, because $2x\in J$ but $2x\notin f(I)$. – Zev Chonoles Jul 4 '12 at 19:11

An ideal $I$ of a ring $R$ is, by definition, "multiplicatively absorptive"; that is, for any $s\in I$ and $r\in R$, we must have $rs\in I$.

Under the usual definition of ring homomorphism, given rings $A$ and $B$, a ring homomorphism $f:A\to B$ must satisfy $f(1_A)=1_B$, so that there is in fact a unique ring homomorphism $f:\mathbb{Z}\to R$ for any ring $R$. In the case of $R=\mathbb{Z}[x]$, this is exactly what you would expect; since $f(1)=1$, we have $f(n)=n$ for all $n\in\mathbb{Z}$.

The set $S=f(\langle 2\rangle)$ in $\mathbb{Z}[x]$ is of course just $$S=\{g\in\mathbb{Z}[x]\mid g=2k\text{ for some }k\in\mathbb{Z}\},$$ i.e. the even integers. For example, $2\in S$. But $x\in \mathbb{Z}[x]$, and $2x\notin S$, so $S$ is not an ideal of $\mathbb{Z}[x]$.

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Thanks for the response Zev! – math-visitor Jul 4 '12 at 19:02
No problem, glad to help! – Zev Chonoles Jul 4 '12 at 19:03
Points for proving that $f$ is the inclusion, which no-one else seemed to do explicitly. – Ben Millwood Jul 4 '12 at 21:16

Strictly speaking, the extension of an ideal $I$ under a homomorphism $f:A \to B$ of rings is defined to be the ideal $f(I)B.$ As this example shows, taking the image of the ideal $I$ might not yield an ideal at all. Fortunately for algebraic geometry, the inverse image of an ideal under a homomorphism is again an ideal.

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Thank you for this insight user34377! – math-visitor Jul 4 '12 at 19:03
You're welcome! – Andrew Jul 4 '12 at 19:04

I just want to reflect this in words, as I think the ideas are quite simple.

Your title refers to the extension of an ideal. In fact, you have extended the ring, but have not extended the ideal. As others point out there is an ideal generated by the image of your original ideal under the inclusion, and this is the proper notion of the extension of your original ideal.

So the main reason that $\left< 2\right>$ is not an ideal in the extended ring $\mathbb Z[x]$ is that it is not big enough. If you think how much you can extend a ring, your ideal can't really hope to keep up unless it is extended too.

It is not always the case that the extension will lead to a proper ideal - think about the inclusion $f: \mathbb Z \to \mathbb Q$

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This perspective is definitely worth thinking about. Thank you Mark! – math-visitor Jul 4 '12 at 19:21

$f : \mathbb{Z} \rightarrow \mathbb{Z}[x]$ is the $f(z) = z$. So $f(\langle 2 \rangle)$ is all the even constant polynomials.

The definition of an ideal is that for all $P \in \mathbb{Z}[x]$ and $Q \in f(\langle 2 \rangle)$, one must have $PQ \in f(\langle 2 \rangle)$. So let $P = x$ and $Q = 2 \in f(\langle 2 \rangle)$. Then $PQ = x\cdot 2 = 2x \notin f(\langle 2 \rangle)$ since $f(\langle 2 \rangle)$ consists of only even constant polynomials. So $f(\langle 2 \rangle)$ is not an ideal of $\mathbb{Z}[x]$.

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You say $P\in \mathbb Z$ and then choose an example where it isn't. – Ben Millwood Jul 4 '12 at 18:55
@BenMillwood Typo by definition, $P$ needs to be in $\mathbb{Z}[x]$. – William Jul 4 '12 at 18:58
I still think your definition of ideal is phrased oddly, so I've proposed an edit that I think improves it. – Ben Millwood Jul 4 '12 at 19:02
Thank you all! =) – math-visitor Jul 4 '12 at 19:09
@BenMillwood I don't see what is so unusual about my definition of an ideal. A ideal of a ring $R$ is just a subring that is closed under multiplication by elements of $R$. Letting $R= \mathbb{Z}[x]$ and assuming $f(\langle 2 \rangle)$ is an ideal, this means exactly that $PQ \in f(\langle 2 \rangle)$. – William Jul 4 '12 at 20:01

$I$ is an ideal of $A$ iff

$\forall i,j \in I, \forall a \in A \implies ai \in I$ and $i+j \in I$.

$f((2))$ is not an ideal of $\mathbb{Z}[x]$, because

$i:=2 \in f((2))$

$a:=x \in \mathbb{Z}[x]$,

but

$ai=x \times 2 \notin f((2))$ because $f((2))=2\mathbb{Z}$.

You have changed of ring. The first ring is $\mathbb{Z}$. The second ring is $\mathbb{Z}[x]$.

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Thank you francis! =) – math-visitor Jul 4 '12 at 19:14