It is clear that the polynomials \(x^{2}1\) and \(x^{2}+1\) in \(\mathbb{R}[x]\) factor uniquely into monic, linear polynomials in \(\mathbb{C}[x]\). However, the situation is quite different when we look at \(\mathbb{R}[x]\) as a subring of the skew polynomial ring \(\mathbb{C}[x; \sigma]\). Observe that if \(t \in\mathbb{C}\) such that \(t = 1\), and \(\sigma(t)\) is the complex conjugate of t, then
$$x^{2}1 = \bigl(x+\sigma(t)\bigr) (x  t) $$
in \(\mathbb{C}[x; \sigma]\). Thus \(x^{2} 1\) can be factored in an infinite number of ways in \(\mathbb{C}[x; \sigma]\). On the other hand, \(x^{2}+1\) remains irreducible in \(\mathbb{C}[x; \sigma]\). In this paper we contrast how polynomials factor in skew polynomial rings with how they factor in ordinary polynomial rings.
We next introduce the terminology used throughout the remainder of this paper. We let K denote a field and σ an automorphism of K. Define \(\mathsf{K}^{\sigma}= \{ a \in\mathsf{K} \mid \sigma(a) = a\}\) and, when we examine the special case where \(\sigma^{2} = 1\), an important subset of \(\mathsf{K}^{\sigma}\) is the set of norms defined by \(N(\mathsf{K}, \sigma) = \{t \sigma(t) \mid t \in\mathsf{K}\}\). When dealing with the field ℂ, unless stated otherwise, we always assume that σ represents complex conjugation.
For \(f(x) = a_{n} x^{n}+ \cdots+ a_{1}x +a_{0} \in\mathsf{K}[x; \sigma]\) and \(t \in\mathsf{K}\), the (standard) evaluation of \(f(x)\) at \(x = t\) is defined as \(f(t) = a_{n}t^{n} + \cdots+ a_{1}t +a_{0}\). When we study polynomials over fields, one of the first facts we prove is that linear factors correspond to roots. Note that this is not the case in skew polynomial rings: if σ denotes complex conjugation, then
$$f(x) = x^{2}1 = (xi) (xi) $$
in \(\mathbb{C}[x; \sigma]\), yet \(f(i) \ne0\). To better understand the role of linear factors in skew polynomial rings, we define σevaluation as follows.
Definition 2.1
Let \(f(x) = a_{n}x^{n} +a_{n1}x^{n1} + \cdots+ a_{2}x^{2}+a_{1}x +a_{0} \in\mathsf{K}[x; \sigma]\) and \(t \in\mathsf{K}\). Then we define the σevaluation
\(f(t; \sigma) \in\mathsf{K}\) as \(f(t; \sigma) = a_{n}(t\sigma(t) \cdots\sigma^{n1}(t)) + a_{n1}(t\sigma(t) \cdots\sigma^{n2}(t)) + \cdots+a_{2}(t\sigma(t)) + a_{1}t + a_{0}\). If \(f(t; \sigma) = 0\), we say that t is a σroot of \(f(x)\).
Observe that if \(\sigma= \mathrm{id}_{\mathsf{R}}\), and \(\mathsf{K}[x; \sigma ]=\mathsf{K}[x]\) is an ordinary polynomial ring, then a σroot is the same as an ordinary root. Lemma 2.2 and Theorem 2.3 show us that the role of σroots in the skew polynomial case generalizes the role of roots in the ordinary case.
Lemma 2.2
([15])
If
\(f(x) = a_{n}x^{n} + \cdots+ a_{1}x+a_{0} \in\mathsf{K}[x; \sigma]\)
and
\(t \in\mathsf{K}\), then there exists a unique
\(q(x) \in\mathsf {K}[x; \sigma]\)
such that
$$f(x) = q(x) (xt) + f(t; \sigma). $$
When \(f(x) = q(x)(xt)\), instead of simply saying that \(xt\) is a factor of \(f(x)\), we may sometimes use the more precise terminology that \(xt\) is a (right) factor of \(f(x)\). We can now see the relationship between (right) factors and σroots.
Theorem 2.3
Suppose
\(f(x) \in\mathsf{K}[x; \sigma]\). Then
\(xt\)
is a right factor of
\(f(x)\)
if and only if
t
is a
σroot of
\(f(x)\).
Proof
Using Lemma 2.2, there exists a unique \(q(x) \in\mathsf{K}[x; \sigma]\) such that \(f(x) = q(x)(xt) +f(t; \sigma)\). Therefore, if t is a σroot of \(f(x)\), then \(f(t; \sigma) = 0\), and we can see that \(xt\) is certainly a (right) factor of \(f(x)\). Conversely, if \(xt\) is a (right) factor of \(f(x)\), then \(f(x) = h(x)(xt)\) for some \(h(x) \in\mathsf{K}[x; \sigma]\). But then, by Lemma 2.2, the quotient \(h(x)\) and remainder (0) are unique, so \(f(t;\sigma)=0\). □
For ordinary polynomials of degree 2, being reducible is equivalent to having a root. In our situation, we now show that it is equivalent to having a σroot.
Corollary 2.4
Suppose
\(f(x) \in\mathsf{K}[x; \sigma]\)
has degree 2; then
\(f(x)\)
is reducible in
\(\mathsf{K}[x; \sigma]\)
if and only if
\(f(x)\)
has a
σroot in
K.
Proof
One direction follows immediately from Theorem 2.3, for if \(f(x)\) has a σroot in K, then it has a right factor of degree one in \(\mathsf{K}[x; \sigma]\) and is therefore reducible. In the other direction, if \(f(x)\) is reducible and has degree 2, then it is easy to see that it must have a monic, linear right factor. Theorem 2.3 now asserts that \(f(x)\) has a σroot in K. □
One of the examples that motivated this paper was the fact that \(x^{2}+1\) remains irreducible in \(\mathbb{C}[x; \sigma]\). Thus the situation of quadratic polynomials where \(\sigma^{2} = 1\) is of particular interest. We can now determine precisely that irreducible quadratics in \(\mathsf{K}^{\sigma}[x]\) remain irreducible in \(\mathsf {K}[x; \sigma]\).
Corollary 2.5
Let
\(f(x) = x^{2}+ax+b \in\mathsf{K}^{\sigma}[x]\)
and suppose
\(\sigma^{2} =1\). Then
\(f(x)\)
is reducible in
\(\mathsf{K}[x; \sigma]\)
if and only if either (i) \(f(x)\)
is reducible in
\(\mathsf{K}^{\sigma}[x]\), or (ii) \(a =0 \)
and
\(b \in N(\mathsf{K}, \sigma)\).
Proof
If \(f(x)\) is reducible in \(\mathsf{K}^{\sigma}[x]\), then it is certainly reducible in \(\mathsf{K}[x; \sigma]\). Furthermore, if \(a = 0\) and \(b \in N(\mathsf{K}, \sigma)\), then \(b = t\sigma(t)\) for some \(t \in\mathsf{K}\) and
$$f(x) = x^{2}(b) = x^{2} \bigl(t\sigma(t)\bigr) = \bigl(x+ \sigma(t)\bigr) (xt). $$
This proves one half of the result.
To see the converse, Corollary 2.4 tells us that if \(f(x)\) is reducible in \(\mathsf{K}[x; \sigma]\), then it has a σroot \(t \in \mathsf{K}\). If t is a σroot, then it satisfies the equation
$$t\sigma(t) + at + b = 0. $$
Since \(\sigma^{2} =1\), we know that \(t\sigma(t) \in \mathsf{K}^{\sigma}\). However, \(b\in\mathsf{K}^{\sigma}\), therefore it follows that \(at \in\mathsf{K}^{\sigma}\). If \(a \ne0\), then the fact that \(a \in\mathsf{K}^{\sigma}\) implies that \(t \in\mathsf{K}^{\sigma}\). Since t is an ordinary root of \(f(x)\) in \(\mathsf{K}^{\sigma}\), we see that \(f(x) \) is reducible in \(\mathsf{K}^{\sigma}[x]\). On the other hand, suppose \(a = 0\). Our equation above now tells us that \(b = t\sigma\in N(\mathsf{K}, \sigma)\), thereby concluding the proof. □
The next corollary completely describes the situation for the examples \(x^{2}1, x^{2}+1 \in\mathbb{R}[x] \subseteq\mathbb{C}[x; \sigma]\) that were mentioned at the start of this paper.
Corollary 2.6
Let
\(f(x) = x^{2}+ax+b \in\mathbb{R}[x]\).

(i)
\(f(x)\)
is reducible in
\(\mathbb{C}[x; \sigma]\)
if and only if
\(f(x)\)
is reducible in
\(\mathbb{R}[x]\).

(ii)
If
\(f(x)\)
is reducible, then the factorization of
\(f(x)\)
in
\(\mathbb{C}[x; \sigma]\)
into monic, linear factors is unique when
\(a \ne0\), whereas
\(f(x)\)
factors an infinite number of ways into monic, linear factors when
\(a =0\).
Proof
In light of Corollary 2.5, in order to prove part (i), it suffices to show that if \(a = 0\) and \(b \in N(\mathbb{C}, \sigma)\), then \(f(x)\) is reducible in \(\mathbb{R}[x]\). However, in this situation, \(N(\mathbb{C}, \sigma)\) is equal to the set of positive real numbers. Therefore, if \(b \in N(\mathbb{C}, \sigma)\), then \(b = c^{2}\) for some \(c \in\mathbb{R}\). Thus
$$f(x) = x^{2}  (b) = (x+c) (xc), $$
as required.
For the proof of part (ii), let \(\operatorname{cis}(\theta) = \cos\theta+ i \sin\theta\) for every \(\theta\in\mathbb{R}\). Then, when we are in the case where \(a = 0\), we can rewrite the equation above as
$$f(x) = x^{2}(b) = (x+c) (xc) = \bigl(x+c\bigl(\operatorname{cis}( \theta)\bigr)\bigr) \bigl(x  c\bigl(\operatorname{cis}(\theta)\bigr)\bigr). $$
By letting θ range between 0 and 2π, we can see that \(f(x)\) does factor an infinite number of ways in \(\mathbb{C}[x; \sigma]\).
Finally, suppose \(a \ne0\); the argument in the proof of Corollary 2.5 indicates that any factorization of \(f(x)\) actually takes place in \(\mathbb{R}[x]\). Since there is only one way to factor \(f(x)\) in \(\mathbb{R}[x]\), it follows that the factorization of \(f(x)\) in \(\mathbb{C}[x; \sigma]\) is also unique. □
When we look back at Corollaries 2.5 and 2.6, it becomes natural to look for an example where reducibility in \(\mathsf{K}[x; \sigma]\) is not equivalent to reducibility in \(\mathsf{K}^{\sigma}[x]\).
Example 2.7
Let \(\mathsf{K}= \mathbb{Q}(i)\) and once again let σ denote complex conjugation. In this case, \(x^{2}2\) and \(x^{2}5\) are irreducible in \(\mathsf{K}^{\sigma}[x] = \mathbb{Q}[x]\). However, since \(2, 5 \in N(\mathbb{Q}, \sigma)\), these polynomials become reducible in \(\mathbb{Q}(i)[x; \sigma]\) as we have
$$x^{2}2 = \bigl(x+(1i)\bigr) \bigl(x(1+i)\bigr) \quad \mbox{and} \quad x^{2} 5 =\bigl(x+(2i)\bigr) \bigl(x(2+i)\bigr). $$
On the other hand, since \(2, 3 \notin N(\mathbb{Q}(i), \sigma)\), the polynomials \(x^{2}+2\) and \(x^{2}3\) are irreducible in \(\mathbb{Q}[x]\) and remain irreducible in \(\mathbb{Q}(i)[x; \sigma]\).