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Which of the following subsets of $R[x]$ are subrings of $R[x]$? Prove or disprove.

All polynomials with constant term $0_R$.

The elements of this set have the form $r_nx^n + r_{n-1}x^{(n-1)} + \dots + r_{1}x$. This is closed, since $R[x]$ does not contain zero divisors.

All polynomials of degree $2$.

This is the subset $F$ with the elements of form $a_2n^2 + a_1n + a_0$. It is clear it is closed under subtraction. It is not closed under multiplication, for multiplying the first two terms of any elements of this set together yields $(a_2n^2)(b_2n^2) = a_2b_2n^4$, a polynomial not in the set $F$.

All polynomials of degree $\leq k$ where $k$ is a fixed positive integer.

Consider the set $G$ of polynomials with $k = 2$. Then there are two polynomials with degree $2$ that yield, under multiplication, an element not in $G$.

All polynomials in which the odd powers of $x$ have zero coefficients. (Consider the even powers as well.)

Call the set of polynomials in which the odd powers of $x$ have zero coefficients $H$. Then $H$ has elements of the form $h_0 + h_2x^2 + h_4x^4 + \dots + h_{2n}x^{2n}$. It is clearly closed under subtraction. Since an even number added to an even number yields an even number, the set $H$ is a subring of $R[x]$.

Call the set of polynomials in which the even powers of $x$ have zero coefficients $J$. Then $J$ has elements of the form $h_1x + h_3x^3 + \dots + h_{2n-1}x^{2n-1}$. It is clearly closed under subtraction. Since an odd number added to an odd number yields an even number, the set $J$ is NOT a subring of $R[x]$.

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  • $\begingroup$ The ring of polynomials whose constant term is $O_R$ is not the zero ring. $\endgroup$ Commented Oct 28, 2013 at 6:13
  • $\begingroup$ $x^2 + x$ has constant term zero. That aside, generally when you say "subring", you want the unit, zero, and multiplication/addition structures of the smaller ring to be inherited from the larger ring.. $\endgroup$
    – Zach L.
    Commented Oct 28, 2013 at 6:15
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    $\begingroup$ Your last one is not right: Note that $(x)(x^3)=x^4$. $\endgroup$ Commented Oct 28, 2013 at 6:17
  • $\begingroup$ @AndréNicolas oh that's right, we are adding, not multiplying. I'm fixing that. $\endgroup$
    – Don Larynx
    Commented Oct 28, 2013 at 6:17
  • $\begingroup$ @PrahladVaidyanathan L. I fixed my proof. Thanks! $\endgroup$
    – Don Larynx
    Commented Oct 28, 2013 at 6:22

1 Answer 1

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Answer updated to match changes to the question.

The set of polynomials with constant term $0_R$ is a subring of $R[x]$, but you’ve not actually proved this. First, $R[x]$ does have zero divisors if $R$ does. Secondly, it’s not clear why zero divisors are relevant in the first place.

The expression $a_2n^2+a_1n+a_0$ is not a polynomial of degree $2$ in $R[x]$; if $n\in R$, it’s simply a constant, and if $n\notin R$, it’s not in $R[x]$ at all. You mean $a_2x^2+a_1x+a_0$. Note that $R$ may have zero divisors, so it’s possible that $(a_2x^2)(b_2x^2)=0_R$, in which case $$(a_2x^2+a_1x+a_0)(b_2x^2+b_1x+b_0)$$ might possibly be a polynomial of degree $2$, depending on whether $a_2b_1+a_1b_2=0_R$ or not. It’s true that the polynomials of degree $2$ aren’t a subring of $R[x]$, but not for the reason that you give. HINT: Contrary to your assertion, this set is not closed under subtraction.

Unlike the set of polynomials of degree $2$, the set of polynomials of degree at most $k$ is closed under subtraction. It’s not closed under multiplication, but your argument doesn’t establish this, for the reasons noted above. Assuming that your ring is unital, however, you could consider the product of $x^k$ with itself.

Your argument to show that $H$ is closed under multiplication is incomplete: the fact that the sum of two even numbers is even is the key ingredient, but in order to make this step completely convincing, you need to show exactly how the exponents in a product of polynomials are related to the exponents in the factors.

Finally, $J$ is not closed under subtraction: what is $x-x$? As you say, $J$ is not closed under multiplication, but it would be better to give an explicit counterexample.

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  • $\begingroup$ I fixed my first question attempt. $\endgroup$
    – Don Larynx
    Commented Oct 28, 2013 at 6:23
  • $\begingroup$ Isn't $R[x]$ the ring of polynomials with real coefficients? Since $R$ has no zero divisors, then... $\endgroup$
    – Don Larynx
    Commented Oct 28, 2013 at 6:33
  • $\begingroup$ @DonLarynx: Not unless you were explicitly told that $R=\Bbb R$. And the use of the symbol $0_R$ strongly suggests that $R$ is intended to be an arbitrary ring, or at least an arbitrary unital ring. $\endgroup$ Commented Oct 28, 2013 at 6:35
  • $\begingroup$ That is specifically what my question said; I had the same thoughts. $\endgroup$
    – Don Larynx
    Commented Oct 28, 2013 at 6:37

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