# It is possible to realize $\mathbb{Z}$ as $K[x_1,\ldots,x_n]/I$ for some field $K$?

Question: Are the integers $\mathbb{Z}$ an affine $K$-algebra, i.e. does there exist a field $K$, a $n\!\in\!\mathbb{N}$, and an ideal $I\!\unlhd\!K[x_1,\ldots,x_n]\!=\!K[\mathbb{x}]$, such that $\mathbb{Z}\!\cong\!K[\mathbb{x}]/I$, as rings?

If we assume that $I\!=\!0$, then since $\mathbb{Z}$ has two units and $K[\mathbb{x}]$ has $|K|\!-\!1$ units, we must have $K\!=\!\mathbb{Z}_3$. Furthermore, since $\mathbb{Z}$ is a PID, we must have $n\!=\!1$. But $\mathbb{Z}\!\ncong\!\mathbb{Z}_3[x]$, since as an abelian group, $\mathbb{Z}$ is generated by $1$ element, whilst $\mathbb{Z}_3[x]$ is not.

If $I\!\neq\!0$, then $I$ must be prime but not maximal, since $\mathbb{Z}$ is a domain but not a field. Since $\mathbb{Z}$ is not local, $\sqrt{I}$ must not be maximal. This is where I run out of ideas...

Additional question: A group presentation is the free group modulo a normal subgroup, and it is denoted $\langle x_1,\ldots,x_n | w_1,\ldots,w_m\rangle$. An $R$-algebra presentation is the free algebra modulo an ideal, and it is denoted $R\langle x_1,\ldots,x_n|p_1,\ldots,p_m\rangle$. Is the commutative $R$-algebra presentation $R[x_1,\ldots,x_n]/I$, where $I$ is the ideal generated by $p_1,\ldots,p_m$, by any chance denoted by $R[x_1,\ldots,x_n|p_1,\ldots,p_m]$? I have not seen this anywhere in the literature. Is this notation reserved for something else?

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The field $K$ can embed into right hand side but can not embed into left hand side. –  wxu Feb 28 '12 at 15:21
Since $K[X]$ is a principal ideal domain, prime ideals are also maximal. I think this completes your proof. –  Joel Cohen Feb 28 '12 at 16:35
While it's true that a "group presentation" explains how to obtain the group by taking a quotient of a free group modulo a normal subgroup, I would disagree that the latter is what a group presentation "is"; rather, you give a generating set (that generates as a normal subgroup) of the normal subgroup in question. –  Arturo Magidin Feb 28 '12 at 17:53
@Arturo: I don't understand what you're trying to say. If I recall correctly, a group presentation is by definition $F_X/\langle\langle w_i; i\!\in\!I\rangle\rangle$, where $F_X$ is a free group on the set $X$ and $\langle\langle w_i; i\!\in\!I\rangle\rangle$ is the normal subgroup of $F_X$, generated by the words $w_i$. And every group is isomorphic to some group presentation. Anyway, does the notation $R[x_1,\ldots,x_n|p_1,\ldots,p_m]$ ever appear in the literature? –  Leon Feb 28 '12 at 18:12
@Leon: That is: a group presentation describes a group in terms of a quotient of a free group, rather than "being" or "realizing" the group as a quotient of a free group, in that while you have information about who the normal subgroup is, you don't necessarily actually have a way of even recognizing whether a given word is in the subgroup or not. As to your latter question, I have never seen that notation; usually $R[x_1,\ldots,x)n]/(p_1,\ldots,p_n)$ is given. But then, polynomial rings are not commonly thought of as free objects (though they are). –  Arturo Magidin Feb 28 '12 at 18:19

The composition $K \to K[X]/I \to \mathbf Z$ would embed $K$ as a subring of $\mathbf Z$, but this is impossible: $1 \in K$ implies $2 \in K$, which has no inverse in $\mathbf Z$.

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@Leon: An ideal can contain units; the quotient will just not be very interesting. –  Henning Makholm Feb 28 '12 at 15:27
@Dylan: We don't necessarily have $1_\mathbb{Z}\!=\!1_K$. For example, $K\!\times\!2\mathbb{Z}$ has no $1$, but $K\!\times\!0$ does. –  Leon Feb 28 '12 at 18:08
@Leon My impression was that we were assuming everything to be unital, which seems typical in commutative algebra. I'm going to have to reconfigure my head to understand how your example could fit into this situation; this might take a while! –  Dylan Moreland Feb 28 '12 at 18:19
All I'm saying is that if we have unital rings $R\!\subseteq\!S$, then we don't necessarily have $1_R\!=\!1_S$, e.g. in $R\!\times\!0\subseteq R\!\times\!R$, we have $1_{R\times0}=(1_R,0)$ and $1_{R\times R}=(1_R,1_R)$. –  Leon Feb 28 '12 at 18:38
However, if $R\!\subseteq\!S$ are domains, then necessarily $1_R\!=\!1_S$, since $1_R1_S\!=\!1_S1_S$ implies $1_S(1_R-1_S)\!=\!0$, hence $1_R=1_S$. Thus in our case, $1_K=1$, and your proof is solid. Thank you :). –  Leon Feb 28 '12 at 20:23

$K[x]$ inherits the characteristic of $K$, and $K[x]/I$ preserves all of the $1+1+\ldots+1=0$ relations from $K[X]$, so $K$ must have characteristic zero. Therefore $K[X]$ contains $\mathbb Q$, and since $\mathbb Q$ contains an element $a$ such that $a+a=1$, this is also the case in $K[X]/I$. But $\mathbb Z$ has no such element.

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Hint: every nontrivial $K$-algebra contains a subring $\cong K$.

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