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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).2012-07-04
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    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 $xyz \in (x)$.2012-07-04
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    Since this is a ring homomorphism (of commutative rings), we *require* that $1\mapsto 1.$2012-07-04
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    @math-visitor, the image of the inclusion map is exactly $\mathbb{Z}$, and $2x$ is not in there.2012-07-04
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    I'm still a tad bit confused because in a different example, consider $(x)\subseteq \mathbb{C}[x,y,z]$; then $xyz\in (x)$, right? Or considering $(2)\subseteq \mathbb{C}[x,y,z]$, don't we also have $2x\in (2)$?2012-07-04
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    @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 $2x\notin f(I)$.2012-07-04
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    Oh... I see. =D Thank you Zev! Your examples are certainly helpful!2012-07-04
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    @math-visitor May we add the tag 'commutative-algebra' to your question?2012-07-04
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    Yes @MattN. please do!2012-07-04

5 Answers 5

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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!2012-07-04
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    No problem, glad to help!2012-07-04
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    Points for proving that $f$ is the inclusion, which no-one else seemed to do explicitly.2012-07-04
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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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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!2012-07-04
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$f : \mathbb{Z} \rightarrow \mathbb{Z}[x]$ 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.2012-07-04
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    @BenMillwood Typo by definition, $P$ needs to be in $\mathbb{Z}[x]$.2012-07-04
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    I still think your definition of ideal is phrased oddly, so I've proposed an edit that I think improves it.2012-07-04
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    @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 $PQ \in f(\langle 2 \rangle)$.2012-07-04
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    @William: it's odd because it's not clear what exactly you are defining to be an ideal - you just give an implication. I'd be happy if you said "If $f(\langle 2 \rangle)$ were an ideal, then for all..." but that's not what you're saying, you're purporting to give a general definition of the concept.2012-07-04
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$I$ 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 $\mathbb{Z}[x]$.