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For a field $F$, a nonzero polynomial $f \in F[x]$ has exactly one monic associate. Show that for other polynomial rings $R[x]$ this is not necessarily true, even if $R$ is an integral domain.

My attempt:

If $R$ is an integral domain, the only non-zero divisor in $R$ is zero. Hence $\mathbb{Z}$ is an integral domain.

edit: I am using $\mathbb{Z}$ as my integral domain example. Is there any other polynomial rings that are integral domains that might be easier?

pf. Suppose $h$ is a nonzero polynomial such that $h \in \mathbb{Z}[x]$. Then $h$ = anxn + ... + a1x + a0 $\in \mathbb{Z}[x]$ such that one of the ai $\neq$ 0. Since the units in $\mathbb{Z}[x]$ are either 1 or -1,

This is where I am stuck. I do not know what to conclude from this last statement. Would highly appreciate any feedback. Thanks!

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  • $\begingroup$ Consider any constant polynomial in $\mathbb{Z}[x]$ that is not a unit. Then there is no monic polynomial associate to it, since the monic polynomial of degree 0 is associate to the units. $\endgroup$
    – user722227
    Nov 23 '19 at 2:25
  • $\begingroup$ Your "attempt" starts out with the assumption that $R$ is an integral domain (which seems a good place to start), but then you say "Hence $\mathbb Z$ is an integral domain." While $\mathbb Z$ certainly is an integral domain, this is unrelated to knowing that $R$ also is. Moreover in $\mathbb Z[x]$, what happens is that non-monic polynomials have a monic associate only if the leading coefficient is $-1$. But if polynomial $p(x)\in \mathbb Z[x]$ has a monic associate, that monic associate is unique. $\endgroup$
    – hardmath
    Nov 23 '19 at 2:46
  • $\begingroup$ @hardmath So for $\mathbb{Z}$, although $\mathbb{Z}$ is an integral domain, a nonzero polynomial h $\in F[x]$ has exactly one monic associate? If that's the case, is there another polynomial ring that's an integral domain that might be easier? $\endgroup$
    – yagayeet
    Nov 23 '19 at 3:13
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    $\begingroup$ @yagayeet: Again you seem to be jumping from $\mathbb Z[x]$ to some unrelated polynomial ring $F[x]$. My point was that a polynomial with leading coefficient not $\pm 1$ like $2x+3 \in \mathbb Z[x]$ has no monic associate in $\mathbb Z[x]$, and thus it is false that every polynomial has exactly one monic associate. $\endgroup$
    – hardmath
    Nov 23 '19 at 3:17
  • $\begingroup$ @hardmath I'm sorry, I meant to write $h \in \mathbb{Z}[x]$. $\endgroup$
    – yagayeet
    Nov 23 '19 at 3:33
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As your opening paragraph states, in a ring of (univariate) polynomials over a field $F$, say $F[x]$, each nonzero polynomial $p(x)$ has exactly one monic associate (a multiple $cp(x)$ where $c$ is a unit in $F[x]$, in this case a nonzero element of $F$).

In a similar polynomial ring $R[x]$ over an arbitrary integral domain $R$ this is no longer true, not because a nonzero polynomial $p(x)$ could have more than one monic associate but because it may have no monic associate at all.

The situation with integral domain $\mathbb Z$ illustrates this. Consider a nonzero first degree polynomial, say $p(x) = mx + b$ with $m,b \in \mathbb Z$. What would it mean for a polynomial to be a monic associate of $p(x)$? We would need to find a unit in $\mathbb Z[x]$ such that multiplying it by $p(x)$ gives us the desired monic associate.

But the only units in $\mathbb Z[x]$ are $\pm 1$. So $p(x)$ has a monic associate only if $m=\pm 1$, so the sought polynomial without a monic associate could be as simple as $p(x) = 2x$.

What is true for integral domain $R$ is that when a polynomial $p(x) \in R[x]$ has a monic associate, it has exactly one monic associate (uniqueness). This follows (with perhaps a little thought) from observing that the units of $R[x]$ are precisely the units of $R$ (assuming the usual inclusion in $R[x]$). It then happens that a polynomial $p(x) \in R[x]$ has a monic associate if and only if the leading coefficient of $p(x)$ is a unit of $R$. Of course this happens for all nonzero polynomials in $R[x]$ only when $R$ is a field (so that all nonzero leading coefficients are units). We've now circled back to the beginning of your problem's setup.

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