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This question is an exact duplicate of:

If $K$ is a field is $K[X^2,X^3]$ a UFD when considered as a subring of $K[X]$?

I know that $K[X]$ is a $PID$ but nothing else ,these ring theory question just dont seem to occur to me..please help...someone told me it is not a $UFD$ and only just a Domain...but i dont know the reason..how should one go about proving these results?..this ring theory course is getting to hard for my understanding..

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marked as duplicate by user26857, hardmath, 3SAT, Patrick Stevens, Alex Provost Mar 22 '16 at 14:11

This question was marked as an exact duplicate of an existing question.

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You might know that UFD's are integrally closed in their field of fractions. I.e. suppose $A$ is a UFD with field of fractions $K$. If an element $a\in K$ satisfies a monic polynomial with entries in $A$, then in fact $a\in A$.

But in your case the element $x$ which is in the field of fractions of $k[x^2,x^3]$. (Check this. It should be easy). But $x$ satisfies the monic polynomial $Y^2-x^2$ which has coefficients in $A$ but $x$ itself is not an element of $A$. So $A$ is not a UFD.

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No, $x^2$ and $x^3$ are non-associate irreducibles in the given ring and $x^8=(x^2)^4=(x^3)^2(x^2)$.

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  • $\begingroup$ What does it mean frank $\endgroup$ – Upstart Mar 21 '16 at 7:11
  • $\begingroup$ What's the definition of UFD? Factorization of each element has to be unique up to associativity. I just write down an element in the ring that has non unique factorization therefore the ring is not a UFD. $\endgroup$ – user175968 Mar 21 '16 at 7:21
  • $\begingroup$ Ohh thanxx ....i got it..i question about associates though.,they are nob associates because they dont divide each other...and $x^3=x.x^2$ but $x$ is not a unit in $K[X^2,X^3]$?? Is it $\endgroup$ – Upstart Mar 21 '16 at 7:24

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