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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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    @Leon: Yes, it's j$u$st a minor quibble with words. The difference between "this information uniquel$y$ characterizes the object" and "this is the object." A presentation determines the group *up to isomorphism*; a quotient of a free group by a normal subgroup is a particular, specific, "actual" object.2012-02-28

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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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    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 :).2012-02-28
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$K[x]$ 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 $\cong K$.