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I am wondering if the following is true:

$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$

I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.

If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:

1) Is this argument legitimate?

2) If it is true, is there any theorem justifying the aforementioned isomorphism?

Yes your intuition is correct and I try to explain necessary steps:

Claim. $Bbb{Z}[x]/<x,3>congBbb{Z}/<3>$

Well, I prove the following steps:

$$Bbb{Z}[x]/<x,3>cong frac{Bbb{Z}[x]/<x>}{<x,3>/<x>} congBbb{Z}/<3>$$

• The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.




• I think $Bbb{Z}[x]/<x> cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.




• Now for the equality $<x,3>/<x>cong <3>$ Try to define the homomorphism as follows:

$phi :<x,3> to <3>$ by mapping any element $xf(x)+3g(x)in <x,3>$ to $3g(0)$ in $<3>$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$

See that this is an epimorphism. Also $ker phi=<x>$. Then by 1st Isomorphism theorem $<x,3>/ker phi cong <3>$. Which justifies all your steps...!!

N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin <x>$.

It's legitimate and a consequence of the isomorphism theorems.