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First of all, am I right in stating $I=(2x^3-x^2+2x+5)=(2x^3+4x^2+2x)=(x^3+2x^2+x)=\{\alpha(x^3+2x^2+x)\vert \alpha\in\mathbb{Z}_5\}?$ In this case $I$ is proper.

Also, in the solution my professor wrote $I=(x)$ but I can't see how he got that. Analogously, in $\mathbb{Z}_3[x]$ how is $I=(x^2+1)$? Using the (maybe flawed) reasoning above and Fermat I can only get $I=(2x^2+x+2)$.

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    You seem confused by a lot of things. What's $\alpha \in \Bbb R$ doing in a polynomial ring over $\Bbb Z_5$? Also, why do you think the ideal generated by two polynomials $f, g$ is the same as the principal ideal generated by their sum? And how is Fermat related to any of this? – 2017-01-24
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    It is true that the ideals of $\mathbb{Z}_5[x]$ are principal, but you should be thinking of taking the GCD of the generators, not their sum. – 2017-01-24
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    @mathguy: Yeah indeed. Anyway $\alpha\in \mathbb{R}$. Unfortunately my professor wrote in the book that "often we denote $(a,b):=(a)+(b)$ in a generic ring"... I guess I should have seen that this wasn't the case, but how? As for Fermat I thought the coefficients, being elements of $\mathbb{Z_p}$, could be manipulated as per congruences modulo $p$ (e.g. $5$ becomes $2$ in $\mathbb{Z_3}$). – 2017-01-24
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    Arithmetic modulo $p$ is indeed important in this problem, but that is not called "Fermat". "Fermat" is the name of a mathematician, and by extension the name of a very specific theorem about $\Bbb Z_p$, which is not needed or relevant to this problem. – 2017-01-24
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    And the notation $(a)+(b)$ means all sums of a multiple of $a$ and a multiple of $b$ - not just $a+b$ as a single element in the ring. That is a common notation for the sum of two *ideals* in a ring. – 2017-01-24
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    @mathguy: Wait, I used Fermat (with mistakes) on the powers of $x$, is that not allowed? In $$\mathbb{Z_3}[x]$ I would get $I=(-x^2+2x+5,2x)$. Now I understand about $(a,b)$. *earlier I wanted to say $\alpha\in $\mathbb{R}$ was a typi – 2017-01-24
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    @Richard: Yes, that is a mistake. The reduction of powers only works on the *coefficients* of $\mathbb{Z}_5[x]$, not on powers of $x$ an indeterminate. – 2017-01-24
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    @hardmath, mathguy : Thank you both! – 2017-01-24

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First, 5 is the same as 0 in $\Bbb Z_5$. With this in hand, it becomes obvious that both generators are multiples of $x$, and therefore $I$ is contained in the principal ideal $(x)$ (which is already a proper ideal, so $I$ is a proper ideal as well).

It is also trivial to see that in fact $I$ is equal to $(x)$, but why does that matter if the problem was to prove that $I$ is a proper ideal? The difference of the two generators is $x^2$, so $I$ contains $x^2$ and therefore $x^3$ as well, so it also contains $x$ because it contains $x^3+x$. So you get inclusions in both directions.

In $\Bbb Z_3$ 5 = -1, so the difference of the two generators is $x^2+1$. Both generators are divisible by $x^2+1$, so again you get $I=(x^2+1)$ by inclusion in both directions.