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Let $R = \mathbb{Z} + x\mathbb{Q}[x] \subset \mathbb{Q}[x]$. Find the irreducibles of $R$.

Show that the irreducible elements in $R$ are $\pm p$ for prime integers $p$ and the irreducible polynomials $p(x) \in \mathbb{Q}[x]$ whose constant coefficient is $\pm 1$. Prove these irreducibles are prime in $R$.

The first part is pretty clear. ($\pm p$ for prime integers $p$), if $\alpha \in R$ s.t. $\deg(\alpha)=0$ then $\alpha$ is a constant polynomial thus in $\mathbb{Z}$. If $\alpha$ is a prime then it is irreducible.

Having trouble showing that for $p(x) \in R$ s.t. $\deg(p(x)) > 0$ with constant term $\pm 1$ are irreducible.

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  • $\begingroup$ Do you mean $\subset \mathbb Q[x]$? $\endgroup$ – Thomas Andrews May 2 '15 at 21:54
  • $\begingroup$ I don't think the statement is true. For example, $\pm p$ is irreducible but does not have constant term $\pm 1$. Or, $(1+x)(1+x)$ is a polynomial whose constant coefficient is $1$, so it is not irreducible (assuming that you can see that $1+x$ is not a unit). $\endgroup$ – rogerl May 2 '15 at 22:25
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    $\begingroup$ @rogerl $(1+x)^2$ is not irreducible in $\mathbb Q[x]$. $\endgroup$ – user26857 May 2 '15 at 22:26
  • $\begingroup$ Related: math.stackexchange.com/questions/1264077/… $\endgroup$ – user26857 May 3 '15 at 8:37
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It's obvious that an irreducible polynomial $p\in\mathbb Q[x]$ with $p(0)=\pm 1$ remains so in $R$.

The question is to prove that these are the only irreducible elements of $R$ of degree $\ge 1$.

Let $p\in R$ irreducible with $\deg p\ge1$.

If $k=|p(0)|>1$, then $p(x)=kq(x)$ with $q\in R$, so $p$ is reducible, a contradiction.
If $p(0)=0$, then $p(x)=xq(x)$ with $q\in\mathbb Q[x]$. Write $q(0)=\frac mn$, notice that $p(x)=(\frac 1nx)(nq(x))$ and $\frac 1nx,nq(x)\in R$, a contradiction.

Now suppose $p(0)=\pm1$, but $p$ is reducible in $\mathbb Q[x]$. Thus $p(x)=u(x)v(x)$ with $u,v\in\mathbb Q[x]$ with $\deg u,\, \deg v\ge1$. In particular, $u(0)v(0)=\pm1$, so we can assume $u(0)=v(0)=\pm1$ (why?). This shows that $p$ is reducible in $R$, a contradiction.

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  • $\begingroup$ your answer is slightly confusing. can you explain $k = |p(x)| > 0$ then $p(x) = kq(x)$ $\endgroup$ – oliverjones May 2 '15 at 22:45
  • $\begingroup$ @oliverjones Well, $k=|p(0)|>1$ (please read the answer carefully !). Then you can force $k$ as a common factor and write $p(x)=k\frac 1kp(x)$. Set $q(x)=\frac 1kp(x)$ and note that $q(0)=\frac 1kp(0)=1$. $\endgroup$ – user26857 May 2 '15 at 22:49
  • $\begingroup$ are you using k is the absolute value of $p(0)$ or are you taking some sort of norm? you also say if $p(0)=0$, then ( then what?). What if $p(x)=x+5$ ? is that not irreducible? $\endgroup$ – oliverjones May 2 '15 at 23:02
  • $\begingroup$ $p(x)=x+5 \in R$, $|p(0)|=|0+5|=5$ this is irreducible in $R$ and does not have constant $\pm 1$ so I am still not following $\endgroup$ – oliverjones May 2 '15 at 23:28
  • $\begingroup$ $x+5$ is reducible in $R$: write $x+5=5(1+\frac 15x)$ and notice that $5,1+\frac 15x\in R$ are non-invertible. (In fact, $R^\times=\{\pm1\}$. It's important to know the invertibles of a ring you want to study its arithmetics!) $\endgroup$ – user26857 May 2 '15 at 23:30

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