Is $ \mathbb{Z}^+ [ x ] \backslash \{ 0 \} $ a unique factorization monoid with respect to the usual product?

Question

Let $ \mathcal{M} $ be the algebraic structure that consists of $ \mathbb{Z}^+ [ x ] \backslash \{ 0 \} $ equipped with the usual product. Let $ M = \mathbb{Z}^+ [ x ] \backslash \{ 0 \} $.

We have that:

1. $ \mathcal{M} $ is a commutative cancellative monoid;
2. for every nonempty $ S \subseteq M $, $ S $ has a minimal element with respect to the divisibility relation on $ \mathcal{M} $ (i.e., there exists $ s \in S $ such that, for every $ t \in S \backslash \{ s \} $, $ t $ does not divide $ s $ on $ \mathcal{M} $); and
3. every pair of elements of $ M $ has a greatest common divisor on $ \mathcal{M} $.

Therefore, $ \mathcal{M} $ is a unique factorization monoid.

However, $ ( 1 + x + x^2 ) ( 1 + x^3 ) = ( 1 + x^2 + x^4 ) ( 1 + x ) $; and $ 1 + x + x^2 $, $ 1 + x^3 $, $ 1 + x^2 + x^4 $, and $ 1 + x $ are prime on $ \mathcal{M} $.

Where is the mistake?

Answer 1

Let $ a ( x ) = 1 + x $, $ b ( x ) = 1 + x^3 $ and $ c ( x ) = 1 + x + x^2 $.

There does not exist a greatest common divisor of $ a ( x ) b ( x ) $ and $ c ( x ) b ( x ) $ on $ \mathcal{M} $.

The polynomials $ a ( x ) $ and $ b ( x ) $ are common divisors of $ a ( x ) b ( x ) $ and $ c ( x ) b ( x ) $ on $ \mathcal{M} $, but they do not divide each other on $ \mathcal{M} $ (note that $ a ( x ) ( x^2 - x + 1 ) = b ( x ) $, but $ x^2 - x + 1 $ does not belong to $ M $).