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I am trying to show that $R=\Bbb Z[x,y,z,w]/(xw-zy)$ is not a UFD.

Let $I=(xw-zy)$.

Let $X=x+I$, $Y=y+I$, $Z=z+I$, and $W=w+I$. My guess is that $X$ is irreducible and therefore $(X)$ is a prime ideal. Now, $XW=ZY$, so that $ZY\in(X)$. I want to show that $Z,Y\notin(X)$ and conclude that $R$ is not a UFD.

However, I cannot confirm my guess because I don't even know how $R$ looks like and how to find an argument to show $X$ is irreducible. I think it suffices to prove that $X$ is irreducible. If it is so, the argument can be applied to $Y$, $Z$, and $W$. Therefore, $Z,Y\notin(X)$.

Any help?

user26857
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YYF
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2 Answers2

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Since $xw = zy$ in $R$, it suffices to show that $x, w, z, y$ are all irreducible. Note that $(xy - zw)$ is a graded prime ideal in $\mathbb{Z}[x,y,z,w]$ (apply Eisenstein at the prime $w$ in the ring $\mathbb{Z}[w,y,z][x]$), so $R$ is a positively graded domain, with degree $0$ subring $\mathbb{Z}$, and if we write $\deg(a)$ for the degree of the highest nonzero component in $a$, then $\deg(ab) = \deg a + \deg b$ for any $a, b \in R$.

Now if $x = rs$ for some $r, s \in R$, then $1 = \deg x = \deg r + \deg s$, but this implies one of $\deg r, \deg s = 0$, say $\deg r = 0$. Then $r \in \mathbb{Z}$, but the only elements of $\mathbb{Z}$ dividing $x$ in $R$ are $\pm 1$ (verify this!). Thus $r$ is a unit, so $x$ is irreducible, and by symmetry, $y, z, w$ are all irreducible. Thus $xw = zy$ are two distinct factorizations into irreducibles, so $R$ is not a UFD.

By the way, $(x)$ is not a prime ideal in $R$:

$$R/(x) \cong \mathbb{Z}[x,y,z,w]/(x, xw - zy) = \mathbb{Z}[x,y,z,w]/(x, zy) \cong \mathbb{Z}[y,z,w]/(yz)$$

is not a domain. This gives another way to see that $R$ is not a UFD, as $x$ is an irreducible element that is not prime.

zcn
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    There is no need to invoke fancy theorems like Eisenstein's: the polynomial is of degree two, so if it factors, it factors as a product of two linear polynomials, and it clearly doesn't :-) – Mariano Suárez-Álvarez Jul 27 '14 at 08:52
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    @MarianoSuárez-Alvarez: Checking that it doesn't factor as a product of two linear polynomials is still too much casework for me - ostensibly one has $(a_0 + a_1x + a_2y + a_3z + a_4w)(b_0 + b_1x + \ldots) = \ldots$ Besides, I find that Eisenstein is incredibly useful to know and use :) – zcn Jul 27 '14 at 08:53
  • Well, one can think a bit before doing that. Since the polynomial has no constant or linear terms, there are no constant terms in the factors; each variable appears in exactly one of them, $x$ and $w$ appear in different factors and $y$ and $z$ appear in different factors with inverse coefficients, so the factorization is as a product of two binomials with disjoint variables: but any such product has four terms. :-) – Mariano Suárez-Álvarez Jul 27 '14 at 09:47
  • Thank you for your answer, but I have not learned graded ring yet. Do you have some other way to show $x$ is irreducible? – YYF Jul 27 '14 at 14:59
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    @MarianoSuárez-Alvarez: Yes, my earlier comment was entirely in jest - your analysis was precisely the casework I claimed to be too lazy to do :) – zcn Jul 27 '14 at 18:09
  • @Y.Fan: One actually doesn't have to know about graded rings in full generality - we're just talking about the polynomial ring here. Every element of $R$ is of the form $f + I$ for some $f = \sum a_{i,j,k,l} x^iy^jz^kw^l \in \mathbb{Z}[x,y,z,w]$, and degree is defined as usual for a polynomial. One can then do direct (but rather tedious) computations to show that $x$ cannot factor nontrivially in $R$ (using that $I$ is prime, and every nonzero term in $I$ has all terms of degree $\ge 2$). Using graded rings is simply a way of organizing the computation nicely – zcn Jul 27 '14 at 18:10
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Consider the element $a = xw$. Then $$ xw = xw - zy + zy \overset{mod \ I}{\equiv} 0 + zy = zy $$ so in $R$ we have $xw = a = zy$. Hence, $R$ is not a UFD.

LinAlgMan
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