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Identify the ring $\Bbb Z[x]/(2x^2-4,4x-5)$.

I've read a proof for this problem and it goes like this.

Let $I=(2x^2-4,4x-5)$, then $x-3=x^2(6)-(3x+1)(2x-1) \in I$. Define $\varphi : \Bbb Z[x] \to\Bbb Z$ by $\varphi(f) =f(3)$. Then $\varphi$ is surjective and it's kernel $(x-3)$ is contained in $I$. Therefore $$\mathbb{Z}[x]/I \cong \mathbb{Z}/\varphi(I) = \mathbb{Z}/(\varphi(2x^2 - 4), \varphi(4x - 5)) = \mathbb{Z}/(14, 7) = \mathbb{Z}/7\mathbb{Z}.$$

What confuses me about this proof is that where on earth did this $x-3$ term come in the beginning? Why would anyone consider this element in the first place?

Second confusing part is where they in a sense distribute $\varphi$ on $I$. They seem to state that $$\varphi(I)=\varphi((2x^2-4, 4x-5))=(\varphi(2x^2-4), \varphi(4x-4))$$

but I cannot find a proof for this result anywhere. Is it always true?

Jacob Sanders
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    Step by step: $2(2x^2 - 4) - x(4x-5) = 5x-8 $ [kill the quadratic term], thus $5x-8 \in I$; $(5x-8 )- (4x-5) = x-3$, thus $x - 3 \in I$; $4x-5 - 4(x-3) = 7$[again, kill the linear term], then $7, x-3\in I $. Thus $\Bbb Z[x]/(7, x-3) \cong \Bbb F_7[x]/(x-3) \cong \Bbb F_7 $. – xbh Mar 28 '22 at 19:22
  • They seem to be implicitly using a slight generalization of a standard ring isomorphism theorem (which is not always explicitly presented in textbooks) I will explain shortly. – Bill Dubuque Mar 28 '22 at 20:00
  • @xbh Thanks for the comment. I see that all these computations checks out, but I don't know where these coefficients for the generator are coming from. For example in $2(2x^2 - 4) - x(4x-5) = 5x-8$ where does the $2$ and $x$ come from and why these? – Jacob Sanders Mar 28 '22 at 20:34
  • I see. I was wondering if there is the usage of the Euclidean algorithm behind this and somewhat confused as if there was a way to skip it. I'll be waiting for your answer! – Jacob Sanders Mar 28 '22 at 20:41
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    In the argument by @xbh the calculations only show that $(7, x-3) \subseteq I$. It remains to show the opposite inclusion $I \subseteq (7, x-3)$ by checking that both $2x^2-4$ and $5x-8$ can be written as combinations of 7 and $x-3$. – Daniel Schepler Mar 28 '22 at 22:25
  • @Daniel Correct, but $,I = (7,x-3),$ has already been proved in my answer a way that does both directions simultaneously (with less work) – Bill Dubuque Mar 28 '22 at 22:35
  • @JacobSanders Like I said, I always want simpler generators, so I try to kill the leading terms of highest degrees using the original generators. – xbh Mar 29 '22 at 01:45

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It seems they skipped explicit mention of an intermediate step that applies the below extension of a common ring isomorphism theorem (proved e.g. in this answer)

If $\varphi: R \to S$ is a surjective ring hom then $\frac{R/ \ker\varphi}{(I+\ker\varphi)/ \ker\varphi} \cong \frac{S}{\varphi(I)}$ for any ideal $I$ of $R.$

Applied to OP this yields $\ \dfrac{\Bbb Z[x]/(x\!-\!3)}{(I+(x\!-\!3))/(x\!-\!3)} \cong \dfrac{\Bbb Z}{\varphi(I)}\cong\, \Bbb Z/7\Bbb Z$

where on earth did this $x-3$ term come in the beginning?

Again they seem to have skipped some steps, pulling $\,x\!-\!3\,$ out of a hat like magic. But we can do it algorithmically - as here we can use the a twist on the Euclidean algorithm to show $\,I = (x\!-\!3,7).\,$ Note $\,(2,4x\!-\!5) = (2,5)=1$ so $(f,\,4x\!-\!5) = (2f,\,4x\!-\!5).\,$ We use this in our first step below.

$$\begin{align} I = (2x^2\!-4,\,4x-5) &= (\color{#0a0}{4x}x-8,\, \color{#0a0}{4x}-\color{#c00}5)\\ &=(\color{#c00}5x-8,\ \ \ 4x-5)\ \ \ \,{\rm by}\ \ \color{#0a0}4x\equiv\color{#c00} 5\\ &= (x-3,\ \ \ \ \ 4x-5)\ \ \ \,{\rm by}\ \ \color{c00}4x\equiv 5\\\ &= (x-3,\ \ \ \ \ 7)\qquad\ \ \ \ \ {\rm by}\ \ \ x\equiv 3\ \end{align}\qquad\qquad$$

where we used an ideal analog of gcd mod reduction (Euclidean algorithm reduction step), i.e.

$$ (a,b,c,\ldots) = (a,\,b\bmod a,\, c\bmod a,\ldots)\quad\ $$

e.g. see here and here.

Bill Dubuque
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  • I tend to think of this as more of a version of Buchberger's algorithm for finding Groebner bases of ideals in polynomial rings -- the idea being to try taking combinations with smallest powers which cancel the leading terms, mod out by the others, and see if you get something new; then repeat this until no pair gives something new when you try it. (Though for the case of $\mathbb{Z}[x_1, \ldots, x_n]$ as opposed to the usual case of polynomial rings over a field, what you actually end up needing to do is taking a smallest (monic) monomial times an element with lower leading term, then... – Daniel Schepler Mar 28 '22 at 22:03
  • mod out by the other elements starting with the one with higher leading term you were trying to match.) – Daniel Schepler Mar 28 '22 at 22:04
  • @Daniel The reason it works is explained in the linked answer. If you peruse the many Linked questions there you will see that this method applies to many questions asked here. Grobner bases are overkill for simple cases like these, esp. for hand computation (not to mention far too advanced for many of these elementary questions). – Bill Dubuque Mar 28 '22 at 22:28
  • Sorry, I missed that you did discuss briefly why in this case it's OK to write $(f, 4x-5) = (2f, 4x-5)$ where $f = 2x^2-4$. (Namely, to put it slightly differently, that $2(2x-2) \equiv 1 \pmod{4x-5}$ so 2 is a unit modulo the rest of the ideal generators.) – Daniel Schepler Mar 28 '22 at 22:51