# Show that $K[X^2,X^3]$ is not factorial. [duplicate]

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How to show that when $$K$$ is a field, $$K[X^2,X^3]$$ is not factorial? What in fact is the ring $$K[X^2,X^3]$$, is this all the polynomials with exponents $$2,3,6,8,9,...$$?

We need to show that there is some $$f\in K[X^2,X^3]$$ such that:

$$f\neq uf_1f_2f_3....f_n$$

Where $$u$$ is a unit.

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 29 at 20:56

• $K[X^2, X^3]$ is the subring of $K[X]$ consisting of all polynomials which can by "built" using $X^2$ and $X^3$. In particular, this also involves $X^5 = X^2 \cdot X^3$, which is missing in your list. – red_trumpet Jan 29 at 20:45
• @red_trumpet Taking the thought one step further, you will get every $x^n$ where $n>1$. – Aaron Jan 29 at 20:48
• $K[X^2,X^3]$ is not integrally closed ($X\in \text{Frac}(K[X^2,X^3])$ is integral over $K[X^2,X^3]$ but doesn't belong to it). In particular it can't be factorial. – yamete kudasai Jan 29 at 20:51
Hint: Consider $$(X^2)^3=X^6=(X^3)^2$$.
It seems that you're trying to find an element that cannot be decomposed into irreducibles. You won't find any because $$K[X]$$ is factorial. You'll need to disprove uniqueness of factorization. Hence the hint.