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This is question from abstract algebra.

Question : If $p$ is irreducible in UFD $D$, show that $p$ is also irreducible in $D[x]$.

But, I think this question is quite strange. I think every single element in $D$ is irreducible in $D[x]$.

Every element in $D$ cannot be factored into the product of two non-constant polynomials. Because it could be only factored by constants, not polynomials...

Something I misunderstood?

random487510
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1 Answers1

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Recall that an element $a$ of an integral domain $A$ is reducible if it can be written as $a=bc$ with $b,c$ non-units in $A$. Note that e.g $6$ is not irreducible in $\mathbb{Z}[X]$ as $6=3\cdot2$ so your conjecture about all elements being irreducible is false. To answer the question, suppose that $p=fg$ in $D[X]$, then we can reduce this to a question about irreducibility in $D$ by thinking about possible degrees of $f,g$.

CrackedBauxite
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  • Well, irreducible in $D[x]$ means it can't be factored by two non-constant polynomials, isn't it? I understood that irreducible in integral domain and polynomial ring is different...Isn't it? – random487510 Dec 27 '20 at 12:44
  • A polynomial ring over an integral domain is an integral domain. There is no difference between irreducibility in a general ring and in a polynomial ring. Your reasoning about non-constant polynomials does however work over a field, as then all non-zero constants are units. – CrackedBauxite Dec 27 '20 at 12:48