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I'm currently considering the ideal $(2,X+1)\subset\Bbb{Z}[X]$. I'm trying to figure out if this is a prime ideal, a maximal ideal, or neither. So far I have attempted to look at the quotient ring $\Bbb{Z}[X]/(2,X+1)$. I know that if this quotient ring is a field, then our ideal is maximal, since $\Bbb{Z}[X]$ is commutative. However I'm stuck on how I should be viewing this quotient ring.

Servaes
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ThomasOfSac
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3 Answers3

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A nice thing about having an ideal given by a few generators, is that you can take successive quotients. That is, you can first take the quotient by one generator, and then the quotient by (the image of) the next generator. So $$\Bbb{Z}[X]/(2,X+1)\cong\Big(\Bbb{Z}[X]/(X+1)\Big)/(2),$$ and also $$\Bbb{Z}[X]/(2,X+1)\cong\Big(\Bbb{Z}[X]/(2)\Big)/(X+1).$$ Does either of these give you a better grip on your quotient ring?

Bernard
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Servaes
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  • My thinking so far is that in your second isomorphism X+1 is clearly irreducible. Thus our original quotient ring was indeed a field. This implies that our original ideal is maximal. – ThomasOfSac Mar 30 '20 at 19:20
  • Indeed $X+1$ is irreducible, but the concept of irreducibility only makes sense in integral domains. So you should show or convince yourself that $\Bbb{Z}[X]/(2)$ is an integral domain. (As other answers suggest it is isomorphic to $(\Bbb{Z}/2\Bbb{Z})[X]$.) – Servaes Mar 30 '20 at 19:26
  • On the other hand, in the first isomorphism you could consider what the ring $\Bbb{Z}[X]/(X+1)$ looks like; can you think of a ring homomorphism from $\Bbb{Z}[X]$ with kernel $(X+1)$? Or without isomorphism theorems; what do the cosets of $(X+1)$ look like? – Servaes Mar 30 '20 at 19:27
  • It would be isomorphic to $\mathbf Z$, via the evaluation homomorphism: $P(X)\longmapsto P(-1)$. – Bernard Mar 30 '20 at 19:35
  • So the first isomorphism is equivalent to looking at the quotient ring Z/(2). In this case, Z is clearly an integral domain, but 2 is not irreducible. However, it is prime. Does this imply that our ideal is no maximal? – ThomasOfSac Mar 30 '20 at 19:40
  • @ThomasOfSac What makes you believe $2$ is not irreducible in $\Bbb{Z}$? – Servaes Mar 30 '20 at 19:42
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    I misread the definition. I thought neither of the elements in the product could be units. Obviously 2=2*1, but since only 1 is a unit, it is irreducible. In this case, this would indeed imply that our ideal was maximal. – ThomasOfSac Mar 30 '20 at 19:46
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The third isomorphism theorem says you can divide out by one generator at a time. So start with $\Bbb Z[x]/(2)\cong \Bbb Z_2[x]$. Now look at the ideal $(x-1)$ in this ring, and divide by that to see what you get. Or do it in the other order, if you think that's better.

Arthur
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    This is very helpful. My current thinking is that by using this isomorphism, we can say that (x-1) is irreducible over Z[x]/(2). This implies that our quotient ring is a field and thus our original ideal was maximal. – ThomasOfSac Mar 30 '20 at 19:23
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    @ThomasOfSac If you can't tell immediately what a quotient like $R[x]/(x-a)$ is isomorphic to (for any ring $R$), I suggest you actually look into it. It's a good thing to be aware of, and a nice exercise. – Arthur Mar 30 '20 at 19:29
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Hint: $$\mathbf Z[X]/(2,X+1)\simeq (\mathbf Z/2\mathbf Z)[X ]\big/(X+1).$$

Bernard
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  • Agh okay. If i'm understanding you correctly, then if X+1 is irreducible in Z/2Z then our quotient ring is a field and our original ideal is maximal? I believe X+1 is irreducible here. – ThomasOfSac Mar 30 '20 at 19:14
  • A polynomial with degree $1$ over a field is always irreducible – how would you factor it in a non-trivial way? – Bernard Mar 30 '20 at 19:16
  • Agreed. Then following this logic our original ideal is maximal. – ThomasOfSac Mar 30 '20 at 19:18
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    Also, it is known that maximal ideals in $\mathbf Z[X]$ are generated by a prile $p$ and a polynomial irreducible modulo $p$. – Bernard Mar 30 '20 at 19:51