mjhjmtu

I am wondering if the following is true:

$$mathbbZ[t]/(t,3)congmathbbZ/3.$$

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 $mathbbZ[t]/(t^2,2t)$ further.

If the aforementioned isomorphism is true, that means $mathbbZ[t]/(t,3)cong(mathbbZ[t]/t)/(3)congmathbbZ/3$, where the $(3)$ is an ideal in $mathbbZ[t]/t$. 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. $BbbZ[x]/langle x,3ranglecongBbbZ/langle3rangle$

Well, I'll prove the following steps:

$$BbbZ[x]/langle x,3ranglecong fracBbbZ[x]/langle xranglelangle x,3rangle/langle xrangle congBbbZ/langle 3rangle$$

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




• I think $BbbZ[x]/langle xrangle cong BbbZ$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $BbbZ[x]$ to $BbbZ$ and use 1st Isomorphism Theorem on Rings.




• Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:

$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in BbbZ[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$

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

N.B. To find out the $ker phi$, remember for any $g(x)in BbbZ[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.

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