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My question pertains to this link (the content of which has been included below in the most recent edit)

The ring of polynomials over a field with no linear term is not a UFD

Let $F$ be a field. Prove that the subset $R \subseteq F[x]$ consisting of all polynomials whose linear coefficient is zero is a subring. Prove also that $R$ is not a unique factorization domain.

To show that $R$ is a subring, we need to see that it is closed under subtraction and multiplication. To that end, let $\alpha = a_0 + > x^2a(x)$ and $\beta = b_0 + x^2 b(x)$ be in R.
Then $\alpha - > \beta$ = $(a_0 - b_0) + x^2(a(x) - b(x)) \in R$ and $\alpha \beta - > a_0b_0 + x^2(b_0a(x)x^2 + a_0b(x)x^2 + x^4a(x)b(x)) \in R$.
So $R$ is a subring.

Next, we claim that $x^2$ and $x^3$ are irreducible in R. To see this, note that no element of $R$ has degree $1$. If $x^2 = p(x)q(x)$ factors in $R$, then computing the degree of both sides we have $\mathsf{deg}(p(x)) > + \mathsf{deg}(q(x)) = 2$. Since neither of these degrees is $1$, one must be $0$, so that without loss of generality, $p(x)$ is constant and thus a unit. So $x^2$ is irreducible. Similarly, $x^3$ is irreducible because if two nonnegative integers sum to $3$ then one of them must be $0$ or $1$. So $x^2$ and $x^3$ are irreducible in $R$.

Now $x^6 = x^2x^2x^2 = x^3x^3$, so that $x^6$ has at least two distinct irreducible factorizations in $R$. Thus $R$ is not a unique factorization domain.

(We are not required to show that this is indeed a subring, only that it is not a unique factorization domain.)

I understand (more or less) everything stated, through "Now $x^6 = x^3x^3 = x^2x^2x^2$".
Does this conclude the proof because we have found an element in R with two distinct factorizations?

How did we use the requirement that the linear terms be zero?

Thanks.

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It's not enough to just find two distinct factorizations (after all, in $\mathbb{Z}$, you have $8 = 2\times2\times 2$ and $8=2\times 4$). What is important is that you find two distinct factorization into irreducibles (that are not related via multiplication by units; so again, $4=2\times 2 = (-2)\times(-2)$ does not disprove that $\mathbb{Z}$ is a UFD).

Here, you need to show that both $x^2$ and $x^3$ are irreducibles in this ring, before you can conclude from $x^6 = x^3x^3 = x^2x^2x^2$ that you have two (essentially) distinct factorizations of $x^6$ into irreducibles. And in order to prove that both $x^2$ and $x^3$ are irreducible in this ring you will have to use that the linear terms of the elements of this ring are zero.

The link you provide proves this before exhibiting the factorization; when it says "since neither of these is degree $1$", that's where they are using there is no linear term.

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  • $\begingroup$ That I am missing such obvious statements is evidence of how lost I am :(... thanks Arturo. I am feeling less lost! $\endgroup$ – The Chaz 2.0 Mar 7 '11 at 3:51
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$F[x^2,x^3]$ is an integral domain with field of fraction $F(x)$. $x \in F(x)$ is integral over $F[x^2,x^3]$ but not in $F[x^2, x^3]$. So it is not a UFD as any UFD is integrally closed domain.

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$\rm (x^2)^3 = (x^3)^2\ $ is a nonunique factorization into irreducibles, because the irreducibles $\rm\:x^2\:,\ x^3$ are nonassociate, i.e. $\rm\ x^3\nmid x^2\ $ (by degrees) and $\rm\: x^2\nmid x^3\: $ since $\rm\ x\not\in R\:.\: $ It is a bit simpler to prove that $\rm\:R\:$ is not a $\rm UFD$ by simply noting that $\rm\:x^2\:$ is irreducible but not prime, since $\rm\ x^2\: | \:(x^3)^2\ $ but $\rm\ x^2\nmid x^3\:.$ Even simpler: $\rm UFD \Rightarrow$ integrally closed, but $\rm\:R\:$ isn't: $\rm\ y^2 - x^2\ $ has the root $\rm\ y = x^3/x^2 = x\not\in R\:.$

That $\rm\:R\:$ is a subring follows easily by the subring test: it contains $1$ and it closed under subtraction and multiplication, since $\rm\:f'(0) = 0 = g'(0)\:$ $\Rightarrow$ $\rm\:(f-g)'(0) = f'(0)-g'(0) = 0\:$ and $\rm\:(fg)'(0) = f'(0)\, g(0) + f(0)\,g'(0) = 0.$

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Because no element of $R$ has degree $1$ which shows that $x^2$ and $x^3$ are irreducible in $R$.

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  • $\begingroup$ (Facepalm)... I just logged back onto M-SE after realizing this! Thanks :) $\endgroup$ – The Chaz 2.0 Mar 7 '11 at 3:48

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