# Proof $\mathbb{A}^n$ is irreducible, without Nullstellensatz

As the title suggests, could anyone either provide me with or direct me to a proof that affine n-space $\mathbb{A}^n$ is irreducible, without using the Nullstellensatz?

This is an exercise in a second course on representation theory, so if there is a reasonably palatable representation-theoretic proof then that's probably what's expected but on the other hand it may just be totally unrelated but presented for the reader's enjoyment. (I'm under no obligation to do it so I wouldn't class it as 'homework'.)

I have spent some time banging my head against the exercise but to no avail, I keep just coming back to the Nullstellensatz proof I already know. As well as a representation-theoretic proof if one exists, any alternative, particularly beautiful proofs would be welcomed for their own sake. Many thanks for the help!

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Every proper closed subset of the affine space has strictly smaller dimension, and the union of two closed sets cannot have greater dimension that the unionands. – Mariano Suárez-Alvarez Feb 27 '12 at 8:39
Irreducible as an algebaic variety ? Isn't it trivial since the underlying ring is a domain ? – Lierre Feb 27 '12 at 8:39
@Lierre: Irreducible in the sense that there is no decomposition into 2 strict algebraic subsets. It may certainly be trivial but I'm afraid my knowledge of algebraic geometry is patchy at best - could you elaborate? – Sandy Feb 27 '12 at 8:53
@Sandy: Dear Sandy, The point is that you have to show somehow that a non-zero polynomial does not vanish at every point of $\mathbb A^n$. One way to do this is via the Nullstellensatz, but it is more elementary than that. But you need to argue something; after all, if we look at $k^n$ with $k$ a finite field then there are non-zero polynomials which vanish at every point of $k^n$. Regards, – Matt E Feb 27 '12 at 15:43
Dear Sandy: To follow up on Matt E's comment: It's mathematically equivalent to prove Matt's claim for $k[x_1,\dots,x_n]$ with $k$ a field, or with $k$ a domain, but it's psychologically easier to do the latter. – Pierre-Yves Gaillard Feb 27 '12 at 16:11

The problem is to show that $\mathbb A^n$ is not the union of two proper algebraic subsets. Any proper algebraic subset is contained in the zero locus $V(f)$ of $f$, for some non-zero polynomial $f$. Also the union of $V(f)$ and $V(g)$ is equal to $V(fg)$. Now $fg$ is non-zero if $f$ and $g$ are so, in conclusion, the problem is to show (writing simply $f$ rather than $fg$) that $V(f)$ is a proper subset of $\mathbb A^n$ provided that $f$ is non-zero.

Concretely, you have to show that if $f$ is a non-zero poynomial in $n$ variables, then $f(x) \neq 0$ for some $x \in \mathbb A^n$. I will leave this as an exercise. (All it requires is that the ground field is infinite; since any algebraically closed field is infinite, this is good enough.)

Added: Reading the comments, there seems to be some uncertainty about what the actual content of this statement is, and why the Nullstellensatz would be invoked at all.

The point is that one has to prove that if $f$ is a non-zero polynomial in $n$ variables, then there is a point of $\mathbb A^n$ at which $f$ doesn't vanish. This certainly follows from the Nullstellensatz (which would imply that since $(f)$ is a non-zero ideal, there is a maximal ideal not containing it, which corresponds to a point at which $f$ doesn't vanish). As I indicate above, the statement is more elementary than the Nullstellensatz, though; e.g. it is true over any infinite field.

But some argument is required. After all, if $k$ is a finite field of order $q$, then $(x_1^q - x_1)\cdots (x_n^q - x_n)$ vanishes at every point of $k^n$, although it is a non-zero polynomial.

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A few friends and I were discussing this and got into a big mess about trying to show precisely this: that a nonzero polynomial (with regards to coefficients) over such a field actually has a point at which it is nonzero! Of course it is not as nice as the case of 1 variable where we may simply factor the polynomial and say there are finitely many roots and infinitely many points to input ($X^2 + Y^2 +1$ gives an example where this approach would clearly fail). I suggested a measure-theoretic view of the zeros would possibly succeed, but would be ugly; do you have a nicer method? – Spyam Feb 27 '12 at 17:06
@Ben: Try induction! – user641 Feb 27 '12 at 17:19
@SteveD (and perhaps @Sandy): Got it! So it's simply a case of supposing that for any n-1-variable polynomial you have infinitely many points nonzero in $\mathbb{A}^{n-1}$ and then treating an n-variable polynomial $f(x_1,\ldots,x_n)$ as $f = g(x_n)$ with coefficients in $k[x_1,\ldots,x_{n-1}]$, then taking an appropriate point so that some $x_n^i$-coefficient of $f$ is nonzero, we get a nonzero polynomial in just $x_n$ and the result falls out nicely. – Spyam Feb 27 '12 at 17:40

The assertion *$\mathbb A^n_k$ is irreducible* means that it is not union of two proper algebraic subsets, i.e. that in the ring $k[x_1,\dotsc,x_n]$, two non-zero ideals cannot have a zero intersection. Let's prove it.

Let $I$ and $J$ two non-zero ideals of the polynomial ring $k[x_1,\dotsc,x_n]$. Let $f\in I$ and $g\in J$, both non-zero. Then $fg$ lies in $I\cap J$. Since the ambient ring is a domain, the product $fg$ is not zero. Thus the ideal $I\cap J$ is not zero.

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It seems like both solutions presented are more or less the same, my 'acceptance' of the other was largely arbitrary - many thanks for the help though, it makes complete sense now. – Sandy Feb 27 '12 at 10:52
@Sandy: Dear Lierre and Sandy, The point is that you have to actually show that since $fg$ is non-zero, there is a point at which it doesn't vanish. This is not difficult, but it is not a complete triviality; see the edits to my answer for more on this. Regards, – Matt E Feb 27 '12 at 15:48
@MattE You are right ! After reading your answer I realized that Sandy is interested in the case where $\mathbb A^n = k^n$, not $\operatorname{Spec} k[x_i]$... – Lierre Feb 28 '12 at 9:23