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I am reading Sharp's book "Steps in Commutative Algebra".

But I really had difficulty in the exercise below. So can you give me a hint about the exercise below?

Exercise 16.37
Let $R$ denote the ring of polynomials $\mathbb{Z}_{2\mathbb{Z}}[X]$ in the indeterminate $X$ over the local ring $\mathbb{Z}_{2\mathbb{Z}}$. Use the natural injective ring homomorphism $\mathbb{Z}\rightarrow\mathbb{Z}_{2\mathbb{Z}}$ to identity elements of $\mathbb{Z}$ as elements of $\mathbb{Z}_{2\mathbb{Z}}$. Show that $2$, $X$ form an $R$-sequence which is maximal (in the sense that it cannot be extended to a longer $R$-sequence), and that the single element $1-2X$ also forms a maximal $R$-sequence.

Aloizio Macedo
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Busra
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  • Sharp says "indeterminate". Now, which step is causing you difficulty: (i) is $2,X$ and $R$-sequence (ii) its a maximal $R$-sequence, (iii) $1-2X$ is an $R$-sequence, (iv) it's a maximal $R$-sequence? – Angina Seng Jun 08 '20 at 10:36
  • You alright. I had difficulty in steps (i) and (iii). I find it difficult to show that they are R-sequences. – Busra Jun 08 '20 at 17:48

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I don't much like Sharp's notation. What we have here is $R=S[X]$ where $S$ is the ring of rationals with odd denominators (that is what $\Bbb Z_{2\Bbb Z}$ is in Sharp's baroque notation). Anyway $R$ is an integral domain. Integral domains have no zero divisors, so $a\in R$ forms an $R$-sequence of length $1$ iff $a\ne 0$ and $aR\ne R$, that is $a$ is non-zero and a non-unit. So $2$ forms an $R$-sequence and $1-2X$ forms an $R$-sequence as neither are units in $R$.

Now $R/2R\cong\Bbb F_2[X]$ is a polynomial ring over $\Bbb F_2$ (since $S/2S\cong\Bbb F_2$). Then $X$ is a non-zero non-unit in this ring. This means that $2,X$ forms an $R$-sequence.

To show these sequences are maximal, we need to consider the $R$-modules $R/(2R+XR)$ and $R/(1-2X)R$ and show these are simple modules. As they are rings too, all we need is to prove they are fields. I'll leave this as an "exercise for the reader".

Overall, this is a nice example, showing that even in very "tame" rings there can be maximal $R$-sequences of different lengths.

Angina Seng
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