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I know that I need to find an irreducible element that isn't prime.

I was wondering whether these examples work.

In $\mathbb{Z}[\sqrt{10}]$, $9=(3)(3)=(\sqrt{10} +1)(\sqrt{10} -1)$. $9$ is not prime, but is irreducible.

In $\mathbb{Z}[\sqrt{-17}]$, $18=(6)(3)=(1-\sqrt{-17})(1+\sqrt{-17})$.

Is writing the number as a product of 2 units prove its irreducible? Or do I need to do more? Thanks

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how do your equations show that 9 is irreducible? How do they show it's not prime? Writing a number as a product of two units certainly doesn't show it's irreducible (it shows it's a unit). One direct route is to show that the two factorizations of 9 you presented cannot be further factorized, so you have a direct example of non-unique factorization. –  Alon Amit Jan 10 '11 at 5:10
    
@Alon Amit thanks, i'm not entirely sure i know what you mean. for example for my first statement 9=3*3=(root10 +1)(root10-1): do you mean that i need to show that 3*3 can't be further factorized and (root10 +1)(root10-1) can't be further factorized? thanks :) –  Lucy Marshall Jan 10 '11 at 5:14
    
What Alon is saying is that another route at proving something is not factorial is to show that there is an element that has two essentially distinct factorizations into irreducibles. You've produced, for instance, two factorizations of $9$ in $\mathbb{Z}[\sqrt{10}]$. If these were factorization into irreducibles (if $3$, $\sqrt{10}+1$, and $\sqrt{10}-1$ are each irreducible), then this would show that $\mathbb{Z}[\sqrt{10}]$ is not factorial. The point is that you'd need to show the factorizations cannot be refined; cf. 30 = 3\times 10 = 2\times 15$. Different, but not a problem. –  Arturo Magidin Jan 10 '11 at 6:16

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An element $m$ is irreducible if and only if it is not a unit, not equal to $0$, and whenever $m=ab$, either $a$ is a unit or $b$ is a unit. An element $p$ is prime if and only if $p$ is not a unit, is not zero, and whenever $p|ab$, either $p|a$ or $p|b$.

You are correct that in order to show that these rings are not factorials (unique factorization domain) it suffices to show that there is an irreducible element that is not prime.

However, you did not show that $9$ is irreducible; quite the contrary, you showed that it is reducible: if $3$ is a unit, then $9=3\times 3$ shows that $9$ is a unit (which it is not, since $\mathbb{Z}[\sqrt{10}]$ contains no noninteger rationals); so $3$ is not a unit, which means that you have shown that $9$ can be expressed as a product of two things that are not units. You showed that $9$ is reducible.

(If $u$ and $v$ are units, then $uv$ is a unit: multiply by $v^{-1}u^{-1}$).

So you have not yet produced elements that are irreducible but not prime; neither $9$ nor $18$ are irreducible.

On the other hand, if you can show that $3$ is irreducible in $\mathbb{Z}[\sqrt{10}]$, you're going to be pretty much done; similarly, if you can show either that $3$ or that $2$ are irreducible in $\mathbb{Z}[\sqrt{-17}]$ you'll be very close to done as well along these lines.

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as ever thank you very much! you seem to be the reason i'm not failing algebra at the moment! –  Lucy Marshall Jan 10 '11 at 6:32
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@Lucy: Note that it is not enough to show $3$ is irreducible in $\mathbb{Z}[\sqrt{10}]$, though it will get you very close: you'll know that $3$ divides the product $(\sqrt{10}+1)(\sqrt{10}-1)$, so then you'll want to show that it is not prime by showing it does not divide either factor. Etc. –  Arturo Magidin Jan 10 '11 at 6:35
    
why is it that 3 isn't prime in $\mathbb{Z}[\sqrt{10}]$? similarly for 3 and 2 in $\mathbb{Z}[\sqrt{-17}]$? thanks –  Lucy Marshall Jan 10 '11 at 8:13
    
sorry ignore the above comment. misread what you said. understand it all now :) thanks –  Lucy Marshall Jan 10 '11 at 8:18

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