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Suppose that $f(x_1,\dots,x_n)$ is a polynomial in variables $x_1,\dots, x_n$. Does there always exist an injective homomorphism $$\phi:\mathbb{Z}[x_1,\dots,x_n]/(f)\to\mathbb{Z}[y_1,\dots,y_m]$$ for some $m$?

If not, are there sufficient conditions one can put on $f$ to ensure that $\phi$ exists?

If $\phi$ exists, is there an algorithm to construct it?

Also, I'm not sure if it matters at all, but it is not incredibly important that the coefficient ring of my polynomial rings be $\mathbb{Z}$. If this question has an easier answer working over a different ring, then I'd be happy to hear it as well.

Edit: As several people pointed out, the answer to the first question is of course no. I have a specific example in mind, maybe I should have just asked it initially. Let $$f(x_1,x_2,x_3,x_4,x_5) = x_1 x_3 x_5 + x_2^2 - x_1^2 x_5 -x_2 x_4.$$

Is there an injective map $$\phi:\mathbb{Z}[x_1,x_2,x_3,x_4,x_5]/(f) \to \mathbb{Z}[y_1,\dots,y_m]$$ for some m?

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There is no such injection when $f$ is reducible such as $x_1^2$, since the LHS is not an integral domain while the RHS is. –  Douglas Zare Mar 24 '11 at 6:26

3 Answers 3

The answer to your first question is of course not. For example, there is no injective morphism from $\mathbb Z[x]/(x^2)$ to any ring of polynomials over any integral domain, because the latter is always a domain, and $\mathbb Z[x]/(x^2)$ obviously has a divisor of zero.

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Let's replace $\mathbb Z$ with $\mathbb C$ to make the question simpler, and assume that $f$ is irreducible, so that the source is a domain and the answer has a chance to be "yes".

The existence of such a surjection then implies that their is a dominant map $\mathbb A^n \to V(f)$, where $V(f)$ is the hypersurface in $\mathbb A^n$ cut out by the equation $f = 0$.

If there is such a dominant map, then $V(f)$ is unirational. Conversely, if $V(f)$ is unirational, then there will be a dominant map from some distinguished open in $\mathbb A^{n-1}$ to $V(f)$, and hence an injection $\mathbb C[x_1,\ldots,x_n]/(f) \to \mathbb C[y_1,\ldots,y_{n-1},1/g]$, for some non-zero polynomial $g(y_1,\ldots,y_{n-1})$.

Now most hypersurfaces are not unirational (e.g. if $f$ has large enough degree and cuts out a smooth hypersurface), and hence such an injection won't exist.

For the particular cubic written down, I don't know the answer. (Rationality of cubic fourfolds is a difficult open question, as a google search will show. The OP is asking for something stronger, and has written down a very particular cubic polynomial, but I didn't think about this particular case at all.)

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HINT $\ $ The target ring is a UFD, but the source ring is not, e.g. $\rm\ x_2\: (x_2 - x_4)\ =\ x_1\: x_5\: (x_1 - x_3)\:.$

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But $\mathbb Z[t^2, t^3]$ is not a UFD, because $t^6=(t^2)^3=(t^3)^2$, yet it is a subring of $\mathbb Z[t]$. So there is something more to it. –  Mariano Suárez-Alvarez Mar 24 '11 at 17:08
Of course, as should be clear, I interpreted the map in the question to be surjective, so your remark does not apply. The OP should again clarify the scope of his question. –  Bill Dubuque Mar 24 '11 at 17:26
I don't understand... What map did you interpret to be surjerctive? –  Mariano Suárez-Alvarez Mar 24 '11 at 17:41
@Bill: But the question explicitly said "injective homomorphism". So you interpret this as the induced map from a surjection $\mathbb{Z}[x_1,\ldots,x_n] \to \mathbb{Z}[y_1,\ldots,y_m]$? –  Arturo Magidin Mar 24 '11 at 18:06
Strange. When I read the question 12 hours ago it was clear that the map was to be injective. Did it flip-flop between the two? –  Mariano Suárez-Alvarez Mar 24 '11 at 19:00

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