$2x+1 \in \mathbb{Z}_4[x]$ has no roots in any ring extension

Question

Show that $f(x)=2x+1 \in \mathbb{Z}_4[x]$ has no roots in any ring $A$ that contains $\mathbb{Z}_4$.

I don't know how to start here... If $\alpha$ is a root of $f$, then $2\alpha+1=0 \Rightarrow \alpha = (2^{-1})(-1)=-2^{-1}$. I know that $2$ has no inverse in $\mathbb{Z}_4$, but I can't see what to do with this information (I don't even know it it's useful in this case...).

It's an exercise of a list of exercises.

The next item is to generalize the result for $\mathbb{Z}_n$, $n$ composite.

I believe that the generalization is that $mx+1$ has no rotos in any ring $A$ that contains $\mathbb{Z}_n$, where $m$ is a factor of $n$.

I'd be glad with a solution or a hint :)

Answer 1

Hint:

In the ring $\Bbb{Z}_4[x]$ (and also over any extension) we have $$ (1+2x)(1-2x)=1-4x^2=1. $$ If $\alpha$ is a zero of the l.h.s. then ...

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For the generalization $1+mx$ with $n=mk$: If $1+m\alpha=0$, then $k+km\alpha=k\cdot0$.

Answer 2

This one, like this one, is also too cute to pass up!

Let $\Bbb Z_n \subset A$ in the sense that $\Bbb Z_n$ is a sub-ring of $A$, and suppose that $m \mid n$; we can then without loss of generality, by flipping a few signs if need be, assume that $1 < m < n$ and that there is an $k$, $1 < k < n$, with $mk = n$ in $\Bbb Z$. Note that I have excluded the case $m = 1$; $x + 1 = 0$ has a solution, $n - 1$, in $\Bbb Z_n \subset A$ and I usually take "factor" to mean "non-unit divisor" in any event. Let us denote the $\Bbb Z_n$-residue class of any $j \in \Bbb Z$ by $j_n$. Suppose then that $a \in A$ solves $m_nx + 1 = 0$, i.e. $m_na + 1 = 0$ or $m_na = -1_n$ in $A$; set $b = -a$ so that $m_nb = 1_n$ holds in $A$, and thus $k_nm_nb = k_n1_n = k_n$. But $k_nm_n = n_n = 0$ in $\Bbb Z_n$, hence in $A$, whence $k_n = n_nb = 0$. But this is impossible in $\Bbb Z_n$ since $1 < k < n$. So $m_nx + 1$ has no zeroes in $A$. QED.

$\Bbb Z_n$ has non-trivial zero divisors; that's what makes this fly.

Note Added Wednesday 10 Septermber 2014 11:14 PM PST:

What This Question Is Really About: Having answered the OP's question, I turned it and my answer over in my mind for several hours afterward. I finally came to the following conclusion, which I'd like to share; hence, this note.

The fundamental issue this question addresses is, in my opinion, stated in the context of unital commutative rings:

A.) Let $R$ be a unital commutative ring; let $\mathscr Z$ be the set of zero divisors in $R$, and let $\mathscr U$ be the set of units. Then

$\mathscr Z \cap \mathscr U = \emptyset. \tag{1}$

We recall that

$I.) \; \; m \in \mathscr Z \Leftrightarrow m \ne 0 \;\text{and} \; \exists t \ne 0, \; mt = 0; \tag{2}$

$II.) \; \; u \in \mathscr U \Leftrightarrow u \mid 1; \text{that is,} \; \exists v \in R \; \text{with} \; uv = 1. \tag{3}$

Then the assertion $\mathscr Z \cap \mathscr U = \emptyset$ may easily be proved as follows: suppose $\exists y \in \mathscr Z \cap \mathscr U$. Then $\exists z \ne 0$ with $yz = 0$, and there exists $w \in R$ with $yw = 1$; but then $ywz = z$ with $ywz = yzw = 0$; thus $z = 0$, a contradiction. So there can be no $y \in \mathscr Z \cap \mathscr U$; $\mathscr Z \cap \mathscr U = \emptyset$. QED.

The significance of (A) for the present purposes lies in its consequence

B.) Let $m \in \mathscr Z$ and $u \in \mathscr U$. Then the polynomial $p(x) = mx + u$ has no zeroes in $R$.

For if $\alpha \in R$ satisifed $m \alpha + u = 0$, or $m \alpha = -u$, then choosing $v$ with $uv = 1$ yields $m \alpha (-v) = (-u)(-v) = 1$, so that $m \mid 1$ or $m \in \mathscr U$; impossible since by assumption $m \in \mathscr Z$. QED.

Now observe that (B) yields as an immediate extension

C.) $R$, etc., as in (B). Suppose $R \subset T$ is a subring of $T$. Then with $m$, $u$ as in (B), $p(x)$ has no zero in $T$.

For a zero divisor in $R$ is a zero divisor in $T$, and a unit in $R$ is a unit in $T$, and (A) and (B) apply to $T$.

Taking $R$ to be $\Bbb Z_n$ and $T$ to be $A$ provides a nice example for the seeker, since $\Bbb Z_n$ is familiar to so many, and all the key concepts are illustrated, if not explicitly, then implicitly through increased familiarity.

OK, I'm done.

End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!