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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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.

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.