Is the following proof that no element of $(x^2, x^3)$ in $k[x^2, x^3]$ is prime valid?
I'm also curious whether I'm making a big assumption (like uniqueness of factorization) that is not safe to sweep under the rug.
I read this question yesterday and the ring that the author mentioned is quite interesting.
$S$ is defined as $k[x^2, x^3]$ where $k$ is an arbitrary field.
I have never seen a ring like this before.
It is not hard to show that every monomial $x^n$ where $n$ is greater than or equal to $2$ is expressible as a sum of a finite number of $2$s and $3$s.
Because of this $S = k[x^2, x^3]$ consists of elements of $k[x]$ of the form $\sum a_ix^i $ where all but finitely many $a_i$ are zero and $a_1$ is constrained to be zero. Any monomial besides $x^1$ can be represented by combining $x^2$ and $x^3$ and we get $1 = x^0$ for free.
That question asks us to consider whether the ideal $(x^2, x^3)$ in $S$ and whether it contains any prime elements.
We can prove that the ideal $(x^2, x^3)$ is prime because the only elements not in the ideal are the nonzero elements of $k$, which are closed under multiplication.
I tried for a bit yesterday to prove that $(x^2, x^3)$ contains or does not contain a prime element, but then I hit a wall and gave up.
Earlier today, I looked at the element $x^2 + x^3$.
This element is not prime:
- $x^2 + x^3 \nmid (x^2 + x^3)x \;\;\text{because $x$ is not in $S$}$
- $x^2 + x^3 \nmid (x^2 + x^3)x$
- BUT $x^2 + x^3 \mid (x^2 + x^3)(x^2 + x^3)(x^2)$
This makes me wonder if the same strategy for witnessing the non-primality of an element always works for elements of $(x^2, x^3)$.
Theorem 101: No element of $(x^2, x^3)$ is prime.
An element of $(x^2, x^3)$ has the form $ax^2 + bx^3$ where $a$ and $b$ are nonzero polynomials in $k[x]$.
- $ax^2 + bx^3 \nmid (ax^2 + bx^3)x \;\; \text{because $x$ is not in $S$}$
- $ax^2 + bx^3 \nmid (ax^2 + bx^3)x$
- BUT $ax^2 + bx^3 \mid (ax^2 + bx^3)(ax^2 + bx^3)(x^2)$ because $(ax^2 + bx^3)(x^2)$ is in $S$.
Suppose the element is $0$, then we're done because zero is not prime.
Suppose the element has the form $ax^2$. Note that elements of the form $bx^3$ also have the form $(bx)x^2$.
- $ax^2 \nmid (ax^2)x$
- $ax^2 \nmid (ax^2)x$
- BUT $ax^2 \mid (ax^2)(ax^2)(x^2)$
This exhausts the possibilities.
End of proof of Theorem 101