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Q11.3.11 Artin Algebra 2nd

Let $R$ be a ring, and let $I$ be an ideal of the polynomial ring $R[x]$. Let $n$ be the lowest degree among nonzero elements of $I$. Prove or disprove the following: $I$ contains a monic polynomial of degree $n$ if and only if it is a principal ideal.

I've proved the first direction, now I want to prove:

a principal ideal contains a monic polynomial of least degree $n$

Is it trivial? I kind of missing the starting point, and can not find any resources on this, any help?

Thanks in advance~

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  • $\begingroup$ Are you over an integral domain? $\endgroup$ Oct 24, 2013 at 10:12
  • $\begingroup$ @PrahladVaidyanathan, hey, thanks for comment, we haven't covered that which is in section 11.7, by the definition I just read myself, yes. $\endgroup$
    – Bob
    Oct 24, 2013 at 10:23
  • $\begingroup$ if $I\subset R[x]$ contains a polynomial of degree $n$... It is not always the case that $I$ contains a "Monic" polynomial of degree $n$.. I was a bit confused looking at your question and thought it would be better to write this (at least for me).. $\endgroup$
    – user87543
    Oct 24, 2013 at 10:53
  • $\begingroup$ @PraphullaKoushik,hi, ture, as a principal ideal is generated by a single element of the ring, and n is the lowest degree among nonzero elements of I, so I definitely generated by polynomial of least degree n, but why is ti monic? $\endgroup$
    – Bob
    Oct 24, 2013 at 10:58
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    $\begingroup$ Let f be a monic polynomial of lowest degree n. We claim (f) = I. f ∈ I hence (f) ⊂ I. Now suppose g ∈ I; then, by division with remainder we can write g = fq + r, where if r 6= 0, it has degree lower than f. But then, f,g ∈ I, hence g − fq = r ∈ I, so r = 0, and g ∈ (f). @user428487 $\endgroup$ Apr 21, 2018 at 13:15

2 Answers 2

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If you are not working over a field, then principal ideals of $R[X]$ may not contain any monic polynomials at all, let alone such polynomials of minimal degree. Principal ideals with a monic generator$~P$ obviously do have such an element (namely $P$, which is monic and clearly of minimal degree for a nonzero multiple of$~P$). But these are in fact all cases where it works. A simple example of failure is the ideal generated by $91X^3$ in $\Bbb Z[X]$ (or by $2X$ if you prefer small examples).

My "these are all cases" claim is just the implication: "if $I$ has a monic polynomial$~P$ among its nonzero elements of minimal degree, then $I$ is the principal ideal generated by$~P$", which you say you have already shown (presumably using Euclidean division by$~P$). Note that not all generators of$~I$ are monic (they could have invertible leading coefficient) nor even do they have to be of minimal degree if $R$ has zero divisors (if $n\in R$ is nilpotent then $1-nX^d\in R[X]$ is invertible for any $d>0$, so the ideal it generates is all of $R[X]$, with monic generator $1$ of degree $0<d$).

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If $I$ is the ideal over integers mod 4 generated by $2x^2+2$ then $I$ has no monic polynomials of degree 2.

Note I'm indebted to Daniel Fisher's comment to choose mod 4 rather than mod 3 as I had earlier.

An element of the ideal $I$ is obtained by multiplying $2x^2+2$ either by a polynomial of degree at least one (which makes the degree 3 or more) or else by multiplying $2x^2+2$ by an element of the integers mod 4. Since $2$ has no inverse mod $4$ we cannot choose a constant so as to multiply $2x^2+2$ by it, and obtain a monic polynomial of degree 2. This shows that this $I$ has no monic polynomials of degree 2.

By the way I am taking $n$ as the lowest positive degree. If we allow degree zero, i.e. nonzero constants, as the generating polynomials, we can take the case of the single element $2$ in the polynomials over the integers mod $4$, and generate the ideal $I=\{0,2\}$ as an example, which is an ideal of only constant polynomials, and here $n=0$ and there is no monic polynomial of degree $0$ in $I$ since that monic would have to be $1$.

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    $\begingroup$ $\mathbb{Z}/3\mathbb{Z}$ is a field, choose a different modulus. One that is even. $\endgroup$ Oct 24, 2013 at 10:22

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